Rd Sharma XII Vol 1 2019 Solutions for Class 12 Commerce Math Chapter 9 Continuity are provided here with simple step-by-step explanations. These solutions for Continuity are extremely popular among Class 12 Commerce students for Math Continuity Solutions come handy for quickly completing your homework and preparing for exams. All questions and answers from the Rd Sharma XII Vol 1 2019 Book of Class 12 Commerce Math Chapter 9 are provided here for you for free. You will also love the ad-free experience on Meritnation’s Rd Sharma XII Vol 1 2019 Solutions. All Rd Sharma XII Vol 1 2019 Solutions for class Class 12 Commerce Math are prepared by experts and are 100% accurate.

#### Question 1:

Test the continuity of the function on f(x) at the origin:

Given:

We observe
(LHL at = 0) =
=​

(RHL at x = 0)​ =
=​$\underset{h\to 0}{\mathrm{lim}}\frac{h}{\left|h\right|}=\underset{h\to 0}{\mathrm{lim}}\frac{h}{h}=\underset{h\to 0}{\mathrm{lim}}1=1$

Hence, $f\left(x\right)$ is discontinuous at the origin.

#### Question 2:

A function f(x) is defined as,

Show that f(x) is continuous that x = 3.

Given:

We observe
(LHL at = 3) =
=​

And, (RHL at = 3)​ =
=​

Also, $f\left(3\right)=5$

Hence, $f\left(x\right)$ is continuous at $x=3$.

#### Question 3:

A function f(x) is defined as

Show that f(x) is continuous at x = 3

Given:

We observe
(LHL at x = 3) = $\underset{\mathrm{x}\to {3}^{-}}{\mathrm{lim}}f\left(x\right)=\underset{h\to 0}{\mathrm{lim}}f\left(3-h\right)$
=   ​$\underset{h\to 0}{\mathrm{lim}}\frac{{\left(3-h\right)}^{2}-9}{\left(3-h\right)-3}=\underset{h\to 0}{\mathrm{lim}}\frac{{3}^{2}+{h}^{2}-6h-9}{3-h-3}=\underset{h\to 0}{\mathrm{lim}}\frac{{h}^{2}-6h}{-h}==\underset{h\to 0}{\mathrm{lim}}\frac{h\left(h-6\right)}{-h}=\underset{h\to 0}{\mathrm{lim}}\left(6-h\right)=6$

(RHL at x = 3) = $\underset{\mathrm{x}\to {3}^{+}}{\mathrm{lim}}f\left(x\right)=\underset{h\to 0}{\mathrm{lim}}f\left(3+h\right)$
=   ​

Given:
$f\left(3\right)=6$

$\therefore \underset{\mathrm{x}\to {3}^{-}}{\mathrm{lim}}f\left(x\right)=\underset{\mathrm{x}\to {3}^{+}}{\mathrm{lim}}f\left(x\right)=f\left(3\right)$

Hence, $f\left(x\right)$ is continuous at $x=3$.

#### Question 4:

If
Find whether f(x) is continuous at x = 1.

Given:

We observe
(LHL at x = 1) = $\underset{\mathrm{x}\to {1}^{-}}{\mathrm{lim}}f\left(x\right)=\underset{h\to 0}{\mathrm{lim}}f\left(1-h\right)$
=   ​$\underset{h\to 0}{\mathrm{lim}}\frac{{\left(1-h\right)}^{2}-1}{\left(1-h\right)-1}=\underset{h\to 0}{\mathrm{lim}}\frac{1+{h}^{2}-2h-1}{1-h-1}=\underset{h\to 0}{\mathrm{lim}}\frac{{h}^{2}-2h}{-h}=\underset{h\to 0}{\mathrm{lim}}\frac{h\left(h-2\right)}{-h}=\underset{h\to 0}{\mathrm{lim}}\left(2-h\right)=2$

(RHL at x = 1) = $\underset{\mathrm{x}\to {1}^{+}}{\mathrm{lim}}f\left(x\right)=\underset{h\to 0}{\mathrm{lim}}f\left(1+h\right)$
=   ​$\underset{h\to 0}{\mathrm{lim}}\frac{{\left(1+h\right)}^{2}-1}{\left(1+h\right)-1}=\underset{h\to 0}{\mathrm{lim}}\frac{1+{h}^{2}+2h-1}{1+h-1}=\underset{h\to 0}{\mathrm{lim}}\frac{{h}^{2}+2h}{h}=\underset{h\to 0}{\mathrm{lim}}\frac{h\left(h+2\right)}{h}=\underset{h\to 0}{\mathrm{lim}}\left(2+h\right)=2$

Given:
$f\left(1\right)=2$

$\therefore \underset{\mathrm{x}\to {1}^{-}}{\mathrm{lim}}f\left(x\right)=\underset{\mathrm{x}\to {1}^{+}}{\mathrm{lim}}f\left(x\right)=f\left(1\right)$

Hence, $f\left(x\right)$ is continuous at $x=1$.

#### Question 5:

If
Find whether f(x) is continuous at x = 0.

Given:

We observe
(LHL at x = 0) = $\underset{\mathrm{x}\to {0}^{-}}{\mathrm{lim}}f\left(x\right)=\underset{h\to 0}{\mathrm{lim}}f\left(0-h\right)=\underset{h\to 0}{\mathrm{lim}}f\left(-h\right)$
=   ​$\underset{h\to 0}{\mathrm{lim}}\frac{\mathrm{sin}\left(-3h\right)}{-h}=\underset{h\to 0}{\mathrm{lim}}\frac{-\mathrm{sin}\left(3h\right)}{-h}=\underset{h\to 0}{\mathrm{lim}}\frac{3\mathrm{sin}\left(3h\right)}{3h}=3\underset{h\to 0}{\mathrm{lim}}\frac{\mathrm{sin}\left(3h\right)}{3h}=3·1=3$

(RHL at x = 0) = $\underset{\mathrm{x}\to {0}^{+}}{\mathrm{lim}}f\left(x\right)=\underset{h\to 0}{\mathrm{lim}}f\left(h\right)$
=   ​$\underset{h\to 0}{\mathrm{lim}}\frac{\mathrm{sin}3h}{h}=\underset{h\to 0}{\mathrm{lim}}\frac{3\mathrm{sin}3h}{3h}=3\underset{h\to 0}{\mathrm{lim}}\frac{\mathrm{sin}\left(3h\right)}{3h}=3·1=3$

Given:
$f\left(0\right)=1$

It is known that for a function $f\left(x\right)$ to be continuous at xa,
$\underset{\mathrm{x}\to {a}^{-}}{\mathrm{lim}}f\left(x\right)=\underset{\mathrm{x}\to {a}^{+}}{\mathrm{lim}}f\left(x\right)=f\left(a\right)$

But here,
$\underset{\mathrm{x}\to {0}^{-}}{\mathrm{lim}}f\left(x\right)=\underset{\mathrm{x}\to {0}^{+}}{\mathrm{lim}}f\left(x\right)\ne f\left(0\right)$

Hence, $f\left(x\right)$ is discontinuous at $x=0$.

#### Question 6:

If find whether f is continuous at x = 0.

Given:

We observe
(LHL at x = 0) = $\underset{\mathrm{x}\to {0}^{-}}{\mathrm{lim}}f\left(x\right)=\underset{h\to 0}{\mathrm{lim}}f\left(0-h\right)=\underset{h\to 0}{\mathrm{lim}}f\left(-h\right)$
=   ​$\underset{h\to 0}{\mathrm{lim}}{e}^{\frac{-1}{h}}=\underset{h\to 0}{\mathrm{lim}}\left(\frac{1}{{e}^{\frac{1}{h}}}\right)=\frac{1}{\underset{h\to 0}{\mathrm{lim}}{e}^{\frac{1}{h}}}=0$

(RHL at x = 0) = $\underset{\mathrm{x}\to {0}^{+}}{\mathrm{lim}}f\left(x\right)=\underset{h\to 0}{\mathrm{lim}}f\left(h\right)$
=   ​$\underset{h\to 0}{\mathrm{lim}}{e}^{\frac{1}{h}}=\infty$

Given:
$f\left(0\right)=1$

It is known that for a function $f\left(x\right)$ to be continuous at x = a,
$\underset{\mathrm{x}\to {a}^{-}}{\mathrm{lim}}f\left(x\right)=\underset{\mathrm{x}\to {a}^{+}}{\mathrm{lim}}f\left(x\right)=f\left(a\right)$

But here,
$\underset{\mathrm{x}\to {0}^{-}}{\mathrm{lim}}f\left(x\right)\ne \underset{\mathrm{x}\to {0}^{+}}{\mathrm{lim}}f\left(x\right)$

Hence, $f\left(x\right)$ is discontinuous at $x=0$.

#### Question 7:

Let
Show that f(x) is discontinuous at x = 0.

Given:

Consider:
$\underset{x\to 0}{\mathrm{lim}}f\left(x\right)=\underset{x\to 0}{\mathrm{lim}}\left(\frac{1-\mathrm{cos}x}{{x}^{2}}\right)\phantom{\rule{0ex}{0ex}}⇒\underset{x\to 0}{\mathrm{lim}}f\left(x\right)=\underset{x\to 0}{\mathrm{lim}}\left(\frac{2{\mathrm{sin}}^{2}\frac{x}{2}}{{x}^{2}}\right)\phantom{\rule{0ex}{0ex}}⇒\underset{x\to 0}{\mathrm{lim}}f\left(x\right)=\underset{x\to 0}{\mathrm{lim}}\left(\frac{2{\mathrm{sin}}^{2}\frac{x}{2}}{4\left(\frac{{x}^{2}}{4}\right)}\right)\phantom{\rule{0ex}{0ex}}⇒\underset{x\to 0}{\mathrm{lim}}f\left(x\right)=\underset{x\to 0}{\mathrm{lim}}\left(\frac{2{\left(\mathrm{sin}\frac{x}{2}\right)}^{2}}{4{\left(\frac{x}{2}\right)}^{2}}\right)\phantom{\rule{0ex}{0ex}}⇒\underset{x\to 0}{\mathrm{lim}}f\left(x\right)=\frac{2}{4}\underset{x\to 0}{\mathrm{lim}}{\left(\frac{\mathrm{si}\mathrm{n}\frac{x}{2}}{\frac{x}{2}}\right)}^{2}\phantom{\rule{0ex}{0ex}}⇒\underset{x\to 0}{\mathrm{lim}}f\left(x\right)=\frac{1}{2}·{1}^{2}=\frac{1}{2}$

Given:
$f\left(0\right)=1$

∴ ​$\underset{x\to 0}{\mathrm{lim}}f\left(x\right)\ne f\left(0\right)$

Thus, f(x) is discontinuous at x = 0
.

#### Question 8:

Show that
is discontinuous at x = 0.

The given function can be rewritten as:

$⇒$

We observe
(LHL at x = 0) = $\underset{x\to {0}^{-}}{\mathrm{lim}}f\left(x\right)=\underset{h\to 0}{\mathrm{lim}}f\left(0-h\right)=\underset{h\to 0}{\mathrm{lim}}f\left(-h\right)$$=\underset{h\to 0}{\mathrm{lim}}\left(-h\right)=0$

(RHL at x = 0) = $\underset{x\to {0}^{+}}{\mathrm{lim}}f\left(x\right)=\underset{h\to 0}{\mathrm{lim}}f\left(0-h\right)=\underset{h\to 0}{\mathrm{lim}}f\left(h\right)$$\underset{h\to 0}{\mathrm{lim}}0=0$

And, $f\left(0\right)=2$
∴ ​

Thus, f(x) is discontinuous at x = 0
.

#### Question 9:

Show that
is discontinuous at x = a.

The given function can be rewritten as:

$⇒$

$⇒$

We observe
(LHL at x = a) = $\underset{x\to {a}^{-}}{\mathrm{lim}}f\left(x\right)=\underset{h\to 0}{\mathrm{lim}}f\left(a-h\right)$$=\underset{h\to 0}{\mathrm{lim}}\left(-1\right)=-1$

(RHL at x = a) = $\underset{x\to {a}^{+}}{\mathrm{lim}}f\left(x\right)=\underset{h\to 0}{\mathrm{lim}}f\left(a+h\right)$$\underset{h\to 0}{\mathrm{lim}}\left(1\right)=1$

∴ ​$\underset{x\to {a}^{-}}{\mathrm{lim}}f\left(x\right)\ne \underset{x\to {a}^{+}}{\mathrm{lim}}f\left(x\right)$

Thus, f(x) is discontinuous at x = a.

#### Question 10:

Discuss the continuity of the following functions at the indicated point(s):
(i)

(ii)

(iii)

(iv)

(v)

(vi)

(vii)

(viii)

(i) Given:

We observe

$⇒\underset{x\to 0}{\mathrm{lim}}f\left(x\right)=f\left(0\right)$
Hence, f(x) is continuous at x = 0.

(ii) Given:

We observe

$⇒\underset{x\to 0}{\mathrm{lim}}f\left(x\right)=f\left(0\right)$
Hence, f(x) is continuous at x = 0.

(iii) Given:

Putting xa = y, we get

$⇒\underset{x\to a}{\mathrm{lim}}f\left(x\right)=f\left(a\right)=0$
Hence, f(x) is continuous at x = a.

(iv) Given:

We observe
$\underset{x\to 0}{\mathrm{lim}}f\left(x\right)=\underset{x\to 0}{\mathrm{lim}}\frac{{e}^{x}-1}{\mathrm{log}\left(1+2x\right)}\phantom{\rule{0ex}{0ex}}⇒\underset{x\to 0}{\mathrm{lim}}f\left(x\right)=\underset{x\to 0}{\mathrm{lim}}\frac{{e}^{x}-1}{\frac{2x\mathrm{log}\left(1+2x\right)}{2x}}\phantom{\rule{0ex}{0ex}}⇒\underset{x\to 0}{\mathrm{lim}}f\left(x\right)=\frac{1}{2}\underset{x\to 0}{\mathrm{lim}}\frac{\left(\frac{{e}^{x}-1}{x}\right)}{\left(\frac{\mathrm{log}\left(1+2x\right)}{2x}\right)}\phantom{\rule{0ex}{0ex}}⇒\underset{x\to 0}{\mathrm{lim}}f\left(x\right)=\frac{1}{2}×\frac{\left(\underset{x\to 0}{\mathrm{lim}}\frac{{e}^{x}-1}{x}\right)}{\left(\underset{x\to 0}{\mathrm{lim}}\frac{\mathrm{log}\left(1+2x\right)}{2x}\right)}=\frac{1}{2}×\frac{1}{1}=\frac{1}{2}$
And, $f\left(0\right)=7$
$⇒\underset{x\to 0}{\mathrm{lim}}f\left(x\right)\ne f\left(0\right)$

Hence, f(x) is discontinuous at x = 0.

(v) Given:

Here, $f\left(1\right)=n-1$

$\underset{x\to 1}{\mathrm{lim}}f\left(x\right)=\underset{x\to 1}{\mathrm{lim}}\frac{1-{x}^{n}}{1-x}\phantom{\rule{0ex}{0ex}}⇒\underset{x\to 1}{\mathrm{lim}}f\left(x\right)=\underset{x\to 1}{\mathrm{lim}}\left[{\left(1-x\right)}^{n-1}+{}^{n}C_{1}{\left(1-x\right)}^{n-2}x+{}^{n}C_{2}{\left(1-x\right)}^{n-3}{x}^{2}+...+{}^{n}C_{n-1}{\left(1-x\right)}^{0}{x}^{n-1}\right]$
$⇒\underset{x\to 1}{\mathrm{lim}}f\left(x\right)=0+0...+{\left(1\right)}^{n-1}=1\ne f\left(1\right)$

Thus, .

(vi) Given:

We observe
(LHL at = 1) = $\underset{x\to {1}^{-}}{\mathrm{lim}}f\left(x\right)=\underset{h\to 0}{\mathrm{lim}}f\left(1-h\right)=\underset{h\to 0}{\mathrm{lim}}-\left(1-h\right)-1=\underset{h\to 0}{\mathrm{lim}}-2+h=-2$
And, $f\left(1\right)=2$

$⇒\underset{x\to {1}^{-}}{\mathrm{lim}}f\left(x\right)\ne f\left(1\right)$

Hence, f(x) is discontinuous at x = 1.

(vii) Given:

We observe

(LHL at x = 0) = $\underset{x\to {0}^{-}}{\mathrm{lim}}f\left(x\right)=\underset{h\to 0}{\mathrm{lim}}f\left(-h\right)=\underset{h\to 0}{\mathrm{lim}}\left[-h-2\right]=-2$
(RHL at x = 0) = $\underset{x\to {0}^{+}}{\mathrm{lim}}f\left(x\right)=\underset{h\to 0}{\mathrm{lim}}f\left(h\right)=\underset{h\to 0}{\mathrm{lim}}\left(2+\mathrm{h}\right)=2$

$⇒\underset{x\to {0}^{-}}{\mathrm{lim}}f\left(x\right)\ne \underset{x\to {0}^{+}}{\mathrm{lim}}f\left(x\right)$

Hence, f(x) is discontinuous at x = 0.

(viii) Given:

We observe

(LHL at x = a) = $\underset{x\to {a}^{-}}{\mathrm{lim}}f\left(x\right)=\left(-a+a\right)\mathrm{sin}\left(\frac{1}{-a+a}\right)=0$
(RHL at x = a) = $\underset{x\to {a}^{+}}{\mathrm{lim}}f\left(x\right)=\left(a-a\right)\mathrm{sin}\left(\frac{1}{a-a}\right)=0$

$⇒\underset{x\to {a}^{-}}{\mathrm{lim}}f\left(x\right)=\underset{x\to {a}^{+}}{\mathrm{lim}}f\left(x\right)=f\left(a\right)$

Hence, f(x) is continuous at x = a.

#### Question 11:

Show that is discontinuous at x = 1.

Given:

We observe
(LHL at x = 1) = $\underset{x\to {1}^{-}}{\mathrm{lim}}f\left(x\right)=\underset{h\to 0}{\mathrm{lim}}f\left(1-h\right)$$=\underset{h\to 0}{\mathrm{lim}}\left(1+{\left(1-h\right)}^{2}\right)=\underset{h\to 0}{\mathrm{lim}}\left(2+{h}^{2}-2h\right)=2$

(RHL at x = 1) = $\underset{x\to {1}^{+}}{\mathrm{lim}}f\left(x\right)=\underset{h\to 0}{\mathrm{lim}}f\left(1+h\right)$$=\underset{h\to 0}{\mathrm{lim}}\left(2-\left(1+h\right)\right)=\underset{h\to 0}{\mathrm{lim}}\left(1-h\right)=1$

∴ ​$\underset{x\to {1}^{-}}{\mathrm{lim}}f\left(x\right)\ne \underset{x\to {1}^{+}}{\mathrm{lim}}f\left(x\right)$

Thus, f(x) is discontinuous at x = 1.

#### Question 12:

Show that

Given:

We observe
(LHL at x = 0) = $\underset{x\to {0}^{-}}{\mathrm{lim}}f\left(x\right)=\underset{h\to 0}{\mathrm{lim}}f\left(0-h\right)=\underset{h\to 0}{\mathrm{lim}}f\left(-h\right)$
$=\underset{h\to 0}{\mathrm{lim}}\left(\frac{\mathrm{sin}3\left(-h\right)}{\mathrm{tan}2\left(-h\right)}\right)=\underset{h\to 0}{\mathrm{lim}}\left(\frac{\mathrm{sin}3h}{\mathrm{tan}2h}\right)=\underset{h\to 0}{\mathrm{lim}}\left(\frac{\frac{3\mathrm{sin}3h}{3h}}{\frac{2\mathrm{tan}2h}{2h}}\right)\phantom{\rule{0ex}{0ex}}=\frac{\underset{h\to 0}{\mathrm{lim}}\left(\frac{3\mathrm{sin}3h}{3h}\right)}{\underset{h\to 0}{\mathrm{lim}}\left(\frac{2\mathrm{tan}2h}{2h}\right)}=\frac{3\underset{h\to 0}{\mathrm{lim}}\left(\frac{\mathrm{sin}3h}{3h}\right)}{2\underset{h\to 0}{\mathrm{lim}}\left(\frac{\mathrm{tan}2h}{2h}\right)}=\frac{3×1}{2×1}=\frac{3}{2}$

(RHL at x = 1) = $\underset{x\to {0}^{+}}{\mathrm{lim}}f\left(x\right)=\underset{h\to 0}{\mathrm{lim}}f\left(0+h\right)=\underset{h\to 0}{\mathrm{lim}}f\left(h\right)$
$=\underset{h\to 0}{\mathrm{lim}}\left(\frac{\mathrm{log}\left(1+3h\right)}{{e}^{2h}-1}\right)=\underset{h\to 0}{\mathrm{lim}}\left(\frac{3h\frac{\mathrm{log}\left(1+3h\right)}{3h}}{\frac{2h\left({e}^{2h}-1\right)}{2h}}\right)\phantom{\rule{0ex}{0ex}}=\frac{3}{2}\underset{h\to 0}{\mathrm{lim}}\left(\frac{\frac{\mathrm{log}\left(1+3h\right)}{3h}}{\frac{\left({e}^{2h}-1\right)}{2h}}\right)=\frac{3}{2}\frac{\underset{h\to 0}{\mathrm{lim}}\left(\frac{\mathrm{log}\left(1+3h\right)}{3h}\right)}{\underset{h\to 0}{\mathrm{lim}}\left(\frac{\left({e}^{2h}-1\right)}{2h}\right)}=\frac{3×1}{2×1}=\frac{3}{2}$

And, $f\left(0\right)=\frac{3}{2}$

∴ ​$\underset{x\to {0}^{-}}{\mathrm{lim}}f\left(x\right)=\underset{x\to {0}^{+}}{\mathrm{lim}}f\left(x\right)=f\left(0\right)$

Thus, f(x) is continuous at x = 0.

#### Question 13:

Find the value of 'a' for which the function f defined by
$f\left(x\right)=\left\{\begin{array}{ll}a\mathrm{sin}\frac{\mathrm{\pi }}{2}\left(x+1\right),& x\le 0\\ \frac{\mathrm{tan}x-\mathrm{sin}x}{{x}^{3}},& x>0\end{array}\right\$is continuous at x = 0.

Given:
We have

(LHL at x = 0) =

(RHL at x = 0) =

#### Question 14:

Examine the continuity of the function

Also sketch the graph of this function.

The given function can be rewritten as:

We observe

(LHL at x = 0) = $\underset{x\mathit{\to }{0}^{-}}{\mathrm{lim}}f\left(x\right)=\underset{h\mathit{\to }0}{\mathrm{lim}}f\left(0-h\right)=\underset{h\mathit{\to }0}{\mathrm{lim}}f\left(-h\right)$ = ​$\underset{h\mathit{\to }0}{\mathrm{lim}}3\left(-h\right)-2=-2$

(RHL at x = 0) =  $\underset{x\mathit{\to }{0}^{+}}{\mathrm{lim}}f\left(x\right)=\underset{h\mathit{\to }0}{\mathrm{lim}}f\left(0+h\right)=\underset{h\mathit{\to }0}{\mathrm{lim}}f\left(h\right)$ = ​$\underset{h\mathit{\to }0}{\mathrm{lim}}\left(h+1\right)=1$

$\therefore \underset{x\mathit{\to }{0}^{-}}{\mathrm{lim}}f\left(x\right)\ne \underset{x\mathit{\to }{0}^{+}}{\mathrm{lim}}f\left(x\right)$

Hence, $f\left(x\right)$ is discontinuous at $x=0$.

#### Question 15:

Discuss the continuity of the function f(x) at the point x = 0, where

Given:

(LHL at x = 0) = $\underset{x\mathit{\to }{0}^{-}}{\mathrm{lim}}f\left(x\right)=\underset{h\mathit{\to }0}{\mathrm{lim}}f\left(0-h\right)=\underset{h\mathit{\to }0}{\mathrm{lim}}f\left(-h\right)$ = ​$\underset{h\mathit{\to }0}{\mathrm{lim}}-\left(-h\right)=0$

(RHL at x = 0) =  $\underset{x\mathit{\to }{0}^{+}}{\mathrm{lim}}f\left(x\right)=\underset{h\mathit{\to }0}{\mathrm{lim}}f\left(0+h\right)=\underset{h\mathit{\to }0}{\mathrm{lim}}f\left(h\right)$ = ​$\underset{h\mathit{\to }0}{\mathrm{lim}}\left(h\right)=0$

And, ​$f\left(0\right)=1$

$\therefore \underset{x\mathit{\to }{0}^{-}}{\mathrm{lim}}f\left(x\right)=\underset{x\mathit{\to }{0}^{+}}{\mathrm{lim}}f\left(x\right)\ne f\left(0\right)$

Hence, $f\left(x\right)$ is discontinuous at $x=0$.

#### Question 16:

Discuss the continuity of the function f(x) at the point x = 1/2, where
$f\left(x\right)=\left\{\begin{array}{c}x,\\ 1/2,\\ 1-x,\end{array}\begin{array}{c}0\le x<1/2\\ x=1/2\\ 1/2

Given:

We observe

(LHL at x = $\frac{1}{2}$) = $\underset{x\mathit{\to }{\frac{1}{2}}^{-}}{\mathrm{lim}}f\left(x\right)=\underset{h\mathit{\to }0}{\mathrm{lim}}f\left(\frac{1}{2}-h\right)$ = ​$\underset{h\mathit{\to }0}{\mathrm{lim}}\left(\frac{1}{2}-h\right)=\frac{1}{2}$

(RHL at x = $\frac{1}{2}$) = $\underset{x\mathit{\to }{\frac{1}{2}}^{\mathit{+}}}{\mathrm{lim}}f\left(x\right)=\underset{h\mathit{\to }0}{\mathrm{lim}}f\left(\frac{1}{2}+h\right)$$\underset{h\mathit{\to }0}{\mathrm{lim}}\left(1-\left(\frac{1}{2}+h\right)\right)=\frac{1}{2}$

Also, ​$f\left(\frac{1}{2}\right)=\frac{1}{2}$

Hence, $f\left(x\right)$ is continuous at $x=\frac{1}{2}$.

#### Question 17:

Discuss the continuity of

Hence, f(x) is discontinuous at x = 0.

#### Question 18:

For what value of k is the following function continuous at x = 1?

Given:
If $f\left(x\right)$ is continuous at x = 1, then

$k=2$

#### Question 19:

Determine the value of the constant k so that the function

Given:

If $f\left(x\right)$ is continuous at x = 1, then,

$k=-1$

#### Question 20:

For what value of k is the function

Given:

If $f\left(x\right)$ is continuous at x = 0, then

$k=\frac{5}{3}$

#### Question 21:

Determine the value of the constant k so that the function

Given:

If $f\left(x\right)$ is continuous at x = 2, then
...(1)

Now,

And, $f\left(2\right)=3$

From (1), we have

$4k=3\phantom{\rule{0ex}{0ex}}⇒k=\frac{3}{4}$

#### Question 22:

Determine the value of the constant k so that the function

Given:

If $f\left(x\right)$ is continuous at x = 0, then

$k=\frac{2}{5}$

#### Question 23:

Find the values of a so that the function

Given:

We observe
(LHL at x = 2) = $\underset{\mathrm{x}\to {2}^{-}}{\mathrm{lim}}f\left(x\right)=\underset{h\to 0}{\mathrm{lim}}f\left(2-h\right)$ =  ​$\underset{h\to 0}{\mathrm{lim}}a\left(2-h\right)+5=2a+5$

(RHL at x = 2) = $\underset{\mathrm{x}\to {2}^{+}}{\mathrm{lim}}f\left(x\right)=\underset{h\to 0}{\mathrm{lim}}f\left(2+h\right)$ =  ​$\underset{h\to 0}{\mathrm{lim}}\left(2+h-1\right)$ = $1$

And, $f\left(2\right)=a\left(2\right)+5=2a+5$

Since $f\left(x\right)$ is continuous at x = 2, we have
$\underset{\mathrm{x}\to {2}^{-}}{\mathrm{lim}}f\left(x\right)=\underset{\mathrm{x}\to {2}^{+}}{\mathrm{lim}}f\left(x\right)=f\left(2\right)$
$2a+5=1$
$2a=-4$
$a=-2$

#### Question 24:

Prove that the function

remains discontinuous at x = 0, regardless the choice of k.

The given function can be rewritten as:

We observe
(LHL at x = 0) = $\underset{x\to {0}^{-}}{\mathrm{lim}}f\left(x\right)=\underset{h\to 0}{\mathrm{lim}}f\left(0-h\right)=\underset{h\to 0}{\mathrm{lim}}f\left(-h\right)$
=$\underset{h\to 0}{\mathrm{lim}}\frac{1}{-2h-1}=-1$

(RHL at x = 0) = $\underset{x\to {0}^{+}}{\mathrm{lim}}f\left(x\right)=\underset{h\to 0}{\mathrm{lim}}f\left(0+h\right)=\underset{h\to 0}{\mathrm{lim}}f\left(h\right)$
=$\underset{h\to 0}{\mathrm{lim}}\frac{1}{2h+1}=1$

So, ​$\underset{x\to {0}^{-}}{\mathrm{lim}}f\left(x\right)\ne \underset{x\to {0}^{+}}{\mathrm{lim}}f\left(x\right)$ such that are independent of k.

Thus, f(x) is discontinuous at x = 0, regardless of the choice of k.

#### Question 25:

Find the value of k if f(x) is continuous at x = π/2, where

Given:

If f(x) is continuous at x = $\frac{\mathrm{\pi }}{2}$, then

$\underset{x\to \frac{\mathrm{\pi }}{2}}{\mathrm{lim}}\frac{k\mathrm{cos}x}{\mathrm{\pi }-2x}=3$     ...(1)

Putting $\frac{\mathrm{\pi }}{2}-x=h$, we get

From (1), we have

$⇒k×1=6$
$⇒k=6$

Hence, for $k=6$ , f(x) is continuous at x = $\frac{\mathrm{\pi }}{2}$.

#### Question 26:

Determine the values of a, b, c for which the function
is continuous at x = 0.

The given function can be rewritten as:

We observe
(LHL at x = 0) = $\underset{x\to {0}^{-}}{\mathrm{lim}}f\left(x\right)=\underset{h\to 0}{\mathrm{lim}}f\left(0-h\right)=\underset{h\to 0}{\mathrm{lim}}f\left(-h\right)$

(RHL at x = 0) = $\underset{x\to {0}^{+}}{\mathrm{lim}}f\left(x\right)=\underset{h\to 0}{\mathrm{lim}}f\left(0+h\right)=\underset{h\to 0}{\mathrm{lim}}f\left(h\right)$
=$\underset{h\to 0}{\mathrm{lim}}\left(\frac{\sqrt{1+bh}-1}{bh}\right)=\underset{h\to 0}{\mathrm{lim}}\left(\frac{bh}{bh\left(\sqrt{1+bh}+1\right)}\right)=\underset{h\to 0}{\mathrm{lim}}\left(\frac{1}{\left(\sqrt{1+bh}+1\right)}\right)=\frac{1}{2}$

And, $f\left(0\right)=c$

If $f\left(x\right)$ is continuous at x = 0, then

​$\underset{x\to {0}^{-}}{\mathrm{lim}}f\left(x\right)=\underset{x\to {0}^{+}}{\mathrm{lim}}f\left(x\right)=f\left(0\right)$

$⇒a=\frac{-3}{2}$

Now, $\frac{\sqrt{1+bx}-1}{bx}$ exists only if $bx\ne 0⇒b\ne 0$.

$b\in R-\left\{0\right\}$

#### Question 27:

If

Given:
If $f\left(x\right)$ is continuous at x = 0, then
...(1)

Consider:

$⇒\underset{x\to 0}{\mathrm{lim}}f\left(x\right)=\frac{2{k}^{2}}{4}×1=\frac{{k}^{2}}{2}$

From equation (1), we have
$\frac{{k}^{2}}{2}=f\left(0\right)$
$⇒\frac{{k}^{2}}{2}=\frac{1}{2}\phantom{\rule{0ex}{0ex}}⇒k=±1$

#### Question 28:

If is continuous at x = 4, find a, b.

Given:

We observe
(LHL at x = 4) = $\underset{x\to {4}^{-}}{\mathrm{lim}}f\left(x\right)=\underset{h\to 0}{\mathrm{lim}}f\left(4-h\right)$
$=\underset{h\to 0}{\mathrm{lim}}\left(\frac{4-h-4}{\left|4-h-4\right|}+a\right)=\underset{h\to 0}{\mathrm{lim}}\left(\frac{-h}{\left|-h\right|}+a\right)=a-1$

(RHL at x = 4) = $\underset{x\to {4}^{+}}{\mathrm{lim}}f\left(x\right)=\underset{h\to 0}{\mathrm{lim}}f\left(4+h\right)$
$=\underset{h\to 0}{\mathrm{lim}}\left(\frac{4+h-4}{\left|4+h-4\right|}+b\right)=\underset{h\to 0}{\mathrm{lim}}\left(\frac{h}{\left|h\right|}+b\right)=b+1$

And $f\left(4\right)=a+b$

If f(x) is continuous at x = 4, then

​$\underset{x\to {4}^{-}}{\mathrm{lim}}f\left(x\right)=\underset{x\to {4}^{+}}{\mathrm{lim}}f\left(x\right)=f\left(4\right)$
$⇒a-1=b+1=a+b$

#### Question 29:

For what value of k is the function

continuous at x = 0?

Given:

If f(x) is continuous at x = 0, then

$\underset{x\to 0}{\mathrm{lim}}f\left(x\right)=f\left(0\right)$
$⇒\underset{x\to 0}{\mathrm{lim}}\frac{\mathrm{sin}2x}{x}=k$
$⇒\underset{x\to 0}{\mathrm{lim}}\frac{2\mathrm{sin}2x}{2x}=k$
$⇒2\underset{x\to 0}{\mathrm{lim}}\frac{\mathrm{sin}2x}{2x}=k$
$⇒2×1=k$
$⇒k=2$

#### Question 30:

Let $f\left(x\right)=\frac{\mathrm{log}\left(1+\frac{x}{a}\right)-\mathrm{log}\left(1-\frac{x}{b}\right)}{x}$, x ≠ 0. Find the value of f at x = 0 so that f becomes continuous at x = 0.

Given:

If f(x) is continuous at x = 0, then

$\underset{x\to 0}{\mathrm{lim}}f\left(x\right)=f\left(0\right)$
$⇒\underset{x\to 0}{\mathrm{lim}}\left(\frac{\mathrm{log}\left(1+\frac{x}{a}\right)-\mathrm{log}\left(1-\frac{x}{b}\right)}{x}\right)=f\left(0\right)$
$⇒\underset{x\to 0}{\mathrm{lim}}\left(\frac{\mathrm{log}\left(1+\frac{x}{a}\right)}{\frac{ax}{a}}-\frac{\mathrm{log}\left(1-\frac{x}{b}\right)}{\frac{bx}{b}}\right)=f\left(0\right)$
$⇒\frac{1}{a}\underset{x\to 0}{\mathrm{lim}}\left(\frac{\mathrm{log}\left(1+\frac{x}{a}\right)}{\frac{x}{a}}\right)-\left(-\frac{1}{b}\right)\underset{x\to 0}{\mathrm{lim}}\left(\frac{\mathrm{log}\left(1-\frac{x}{b}\right)}{\frac{-x}{b}}\right)=f\left(0\right)$

#### Question 31:

If is continuous at x = 2, find k.

Given:

If f(x) is continuous at x = 2, then

$\underset{x\to 2}{\mathrm{lim}}f\left(x\right)=f\left(2\right)\phantom{\rule{0ex}{0ex}}⇒\underset{x\to 2}{\mathrm{lim}}\frac{{2}^{x+2}-16}{{4}^{x}-16}=f\left(2\right)\phantom{\rule{0ex}{0ex}}⇒\underset{x\to 2}{\mathrm{lim}}\frac{4\left({2}^{x}-4\right)}{\left({2}^{x}-4\right)\left({2}^{x}+4\right)}=k\phantom{\rule{0ex}{0ex}}⇒\underset{x\to 2}{\mathrm{lim}}\frac{4}{\left({2}^{x}+4\right)}=k\phantom{\rule{0ex}{0ex}}⇒\frac{4}{\left({2}^{2}+4\right)}=k\phantom{\rule{0ex}{0ex}}⇒\frac{4}{8}=k\phantom{\rule{0ex}{0ex}}⇒k=\frac{1}{2}$

#### Question 32:

If is continuous at x = 0, find k.

Given:

If f(x) is continuous at x = 0, then

$\underset{x\to 0}{\mathrm{lim}}f\left(x\right)=f\left(0\right)\phantom{\rule{0ex}{0ex}}⇒\underset{x\to 0}{\mathrm{lim}}\frac{{\mathrm{cos}}^{2}x-{\mathrm{sin}}^{2}x-1}{\sqrt{{x}^{2}+1}-1}=k\phantom{\rule{0ex}{0ex}}⇒\underset{x\to 0}{\mathrm{lim}}\frac{1-{\mathrm{sin}}^{2}x-{\mathrm{sin}}^{2}x-1}{\sqrt{{x}^{2}+1}-1}=k\phantom{\rule{0ex}{0ex}}⇒\underset{x\to 0}{\mathrm{lim}}\frac{-2{\mathrm{sin}}^{2}x}{\sqrt{{x}^{2}+1}-1}=k\phantom{\rule{0ex}{0ex}}⇒\underset{x\to 0}{\mathrm{lim}}\frac{-2\left({\mathrm{sin}}^{2}x\right)\left(\sqrt{{x}^{2}+1}+1\right)}{\left(\sqrt{{x}^{2}+1}-1\right)\left(\sqrt{{x}^{2}+1}+1\right)}=k\phantom{\rule{0ex}{0ex}}⇒\underset{x\to 0}{\mathrm{lim}}\frac{-2\left({\mathrm{sin}}^{2}x\right)\left(\sqrt{{x}^{2}+1}+1\right)}{{x}^{2}}=k\phantom{\rule{0ex}{0ex}}⇒-2\underset{x\to 0}{\mathrm{lim}}\frac{\left({\mathrm{sin}}^{2}x\right)\left(\sqrt{{x}^{2}+1}+1\right)}{{x}^{2}}=k\phantom{\rule{0ex}{0ex}}⇒-2\underset{x\to 0}{\mathrm{lim}}{\left(\frac{\mathrm{sin}x}{x}\right)}^{2}\underset{x\to 0}{\mathrm{lim}}\left(\sqrt{{x}^{2}+1}+1\right)=k\phantom{\rule{0ex}{0ex}}⇒-2×1×\left(1+1\right)=k\phantom{\rule{0ex}{0ex}}⇒k=-4$

#### Question 33:

Extend the definition of the following by continuity
at the point x = π.

Given:

If f(x) is continuous at x = $\mathrm{\pi }$, then

$\underset{x\to \mathrm{\pi }}{\mathrm{lim}}f\left(x\right)=f\left(\mathrm{\pi }\right)\phantom{\rule{0ex}{0ex}}⇒\underset{x\to \mathrm{\pi }}{\mathrm{lim}}\frac{1-\mathrm{cos}7\left(\mathrm{x}-\mathrm{\pi }\right)}{5{\left(\mathrm{x}-\mathrm{\pi }\right)}^{2}}=f\left(\mathrm{\pi }\right)\phantom{\rule{0ex}{0ex}}⇒\frac{2}{5}\underset{x\to \mathrm{\pi }}{\mathrm{lim}}\frac{{\mathrm{sin}}^{2}\left(\frac{7\left(\mathrm{x}-\mathrm{\pi }\right)}{2}\right)}{{\left(\mathrm{x}-\mathrm{\pi }\right)}^{2}}=f\left(\mathrm{\pi }\right)\phantom{\rule{0ex}{0ex}}⇒\frac{2}{5}×\frac{49}{4}\underset{x\to \mathrm{\pi }}{\mathrm{lim}}\frac{{\mathrm{sin}}^{2}\left(\frac{7\left(\mathrm{x}-\mathrm{\pi }\right)}{2}\right)}{\frac{49}{4}{\left(\mathrm{x}-\mathrm{\pi }\right)}^{2}}=f\left(\mathrm{\pi }\right)\phantom{\rule{0ex}{0ex}}⇒\frac{2}{5}×\frac{49}{4}\underset{x\to \mathrm{\pi }}{\mathrm{lim}}\frac{{\mathrm{sin}}^{2}\left(\frac{7\left(\mathrm{x}-\mathrm{\pi }\right)}{2}\right)}{{\left(\frac{7}{2}\left(\mathrm{x}-\mathrm{\pi }\right)\right)}^{2}}=f\left(\mathrm{\pi }\right)\phantom{\rule{0ex}{0ex}}⇒\frac{2}{5}×\frac{49}{4}\underset{x\to \mathrm{\pi }}{\mathrm{lim}}{\left[\frac{\mathrm{sin}\left(\frac{7\left(\mathrm{x}-\mathrm{\pi }\right)}{2}\right)}{\left(\frac{7}{2}\left(\mathrm{x}-\mathrm{\pi }\right)\right)}\right]}^{2}=f\left(\mathrm{\pi }\right)\phantom{\rule{0ex}{0ex}}⇒\frac{2}{5}×\frac{49}{4}×1=f\left(\mathrm{\pi }\right)\phantom{\rule{0ex}{0ex}}⇒\frac{1}{5}×\frac{49}{2}×1=f\left(\mathrm{\pi }\right)\phantom{\rule{0ex}{0ex}}⇒\frac{49}{10}=f\left(\mathrm{\pi }\right)$

Hence, the given function will be continuous at $x=\mathrm{\pi }$, if $f\left(\mathrm{\pi }\right)=\frac{49}{10}$.

#### Question 34:

If x ≠ 0 is continuous at x = 0, then find f (0).

Given:

If f(x) is continuous at x = 0, then

$\underset{x\to 0}{\mathrm{lim}}f\left(x\right)=f\left(0\right)\phantom{\rule{0ex}{0ex}}⇒\underset{x\to 0}{\mathrm{lim}}\frac{2x+3\mathrm{sin}x}{3x+2\mathrm{sin}x}=f\left(0\right)\phantom{\rule{0ex}{0ex}}⇒\underset{x\to 0}{\mathrm{lim}}\frac{x\left(2+3\frac{\mathrm{sin}x}{x}\right)}{x\left(3+2\frac{\mathrm{sin}x}{x}\right)}=f\left(0\right)\phantom{\rule{0ex}{0ex}}⇒\underset{x\to 0}{\mathrm{lim}}\frac{\left(2+3\frac{\mathrm{sin}x}{x}\right)}{\left(3+2\frac{\mathrm{sin}x}{x}\right)}=f\left(0\right)\phantom{\rule{0ex}{0ex}}⇒\frac{\underset{x\to 0}{\mathrm{lim}}\left(2+3\frac{\mathrm{sin}x}{x}\right)}{\underset{x\to 0}{\mathrm{lim}}\left(3+2\frac{\mathrm{sin}x}{x}\right)}=f\left(0\right)\phantom{\rule{0ex}{0ex}}⇒\frac{2+3\underset{x\to 0}{\mathrm{lim}}\left(\frac{\mathrm{sin}x}{x}\right)}{3+2\underset{x\to 0}{\mathrm{lim}}\left(\frac{\mathrm{sin}x}{x}\right)}=f\left(0\right)\phantom{\rule{0ex}{0ex}}⇒\frac{2+3×1}{3+2×1}=f\left(0\right)\phantom{\rule{0ex}{0ex}}⇒\frac{5}{5}=f\left(0\right)\phantom{\rule{0ex}{0ex}}⇒f\left(0\right)=1$

#### Question 35:

Find the value of k for which
is continuous at x = 0;

Given:

If f(x) is continuous at x = 0, then

#### Question 36:

In each of the following, find the value of the constant k so that the given function is continuous at the indicated point;
(i) at x = 0

(ii) at x = 1

(iii) at x = 0

(iv) at x = π

(v) at x = 5

(vi) at x = 5

(vii) at x = 1

(viii)

(ix)

(i) Given:

If f(x) is continuous at x = 0, then
$\underset{x\to 0}{\mathrm{lim}}f\left(x\right)=f\left(0\right)\phantom{\rule{0ex}{0ex}}⇒\underset{x\to 0}{\mathrm{lim}}\frac{1-\mathrm{cos}2kx}{{x}^{2}}=8\phantom{\rule{0ex}{0ex}}⇒\underset{x\to 0}{\mathrm{lim}}\frac{2{k}^{2}{\mathrm{sin}}^{2}kx}{{k}^{2}{x}^{2}}=8\phantom{\rule{0ex}{0ex}}⇒2{k}^{2}\underset{x\to 0}{\mathrm{lim}}{\left(\frac{\mathrm{sin}kx}{kx}\right)}^{2}=8\phantom{\rule{0ex}{0ex}}⇒2{k}^{2}×1=8\phantom{\rule{0ex}{0ex}}⇒{k}^{2}=4\phantom{\rule{0ex}{0ex}}⇒k=±2$

(ii) Given:

If f(x) is continuous at x = 1, then

(iii) Given:

We have
(LHL at x = 0) = $\underset{x\to {0}^{-}}{\mathrm{lim}}f\left(x\right)=\underset{h\to 0}{\mathrm{lim}}f\left(0-h\right)=\underset{h\to 0}{\mathrm{lim}}f\left(-h\right)=\underset{h\to 0}{\mathrm{lim}}k\left({h}^{2}+2h\right)=0$
(RHL at x = 0) = $\underset{x\to {0}^{+}}{\mathrm{lim}}f\left(x\right)=\underset{h\to 0}{\mathrm{lim}}f\left(0+h\right)=\underset{h\to 0}{\mathrm{lim}}f\left(h\right)=\underset{h\to 0}{\mathrm{lim}}\mathrm{cos}h=1$

$\therefore \underset{x\to {0}^{-}}{\mathrm{lim}}f\left(x\right)\ne \underset{x\to {0}^{+}}{\mathrm{lim}}f\left(x\right)$

(iv) Given:

We have
(LHL at x = $\mathrm{\pi }$) = $\underset{x\to {\mathrm{\pi }}^{-}}{\mathrm{lim}}f\left(x\right)=\underset{h\to 0}{\mathrm{lim}}f\left(\mathrm{\pi }-h\right)=\underset{h\to 0}{\mathrm{lim}}k\left(\mathrm{\pi }-\mathrm{h}\right)+1=k\mathrm{\pi }+1$
(RHL at x = $\mathrm{\pi }$) = $\underset{x\to {\mathrm{\pi }}^{+}}{\mathrm{lim}}f\left(x\right)=\underset{h\to 0}{\mathrm{lim}}f\left(\mathrm{\pi }+h\right)=\underset{h\to 0}{\mathrm{lim}}\mathrm{cos}\left(\mathrm{\pi }+h\right)=\mathrm{cos\pi }=-1$

If f(x) is continuous at x = $\mathrm{\pi }$, then
$\underset{x\to {\mathrm{\pi }}^{-}}{\mathrm{lim}}f\left(x\right)=\underset{x\to {\mathrm{\pi }}^{+}}{\mathrm{lim}}f\left(x\right)$
$⇒k\mathrm{\pi }+1=-1\phantom{\rule{0ex}{0ex}}⇒\mathrm{k}=\frac{-2}{\mathrm{\pi }}$

(v) Given:

We have
(LHL at x = 5) = $\underset{x\to {5}^{-}}{\mathrm{lim}}f\left(x\right)=\underset{h\to 0}{\mathrm{lim}}f\left(5-h\right)=\underset{h\to 0}{\mathrm{lim}}k\left(5-\mathrm{h}\right)+1=5k+1$
(RHL at x = 5) = $\underset{x\to {5}^{+}}{\mathrm{lim}}f\left(x\right)=\underset{h\to 0}{\mathrm{lim}}f\left(5+h\right)=\underset{h\to 0}{\mathrm{lim}}3\left(5+\mathrm{h}\right)-5=10$

If f(x) is continuous at x = 5, then
$\underset{x\to {5}^{-}}{\mathrm{lim}}f\left(x\right)=\underset{x\to {5}^{+}}{\mathrm{lim}}f\left(x\right)\phantom{\rule{0ex}{0ex}}⇒5k+1=10\phantom{\rule{0ex}{0ex}}⇒k=\frac{9}{5}$

(vi) Given:

If f(x) is continuous at x = 5, then
$\underset{x\to 5}{\mathrm{lim}}f\left(x\right)=f\left(5\right)\phantom{\rule{0ex}{0ex}}⇒\underset{x\to 5}{\mathrm{lim}}\left(x+5\right)=k\phantom{\rule{0ex}{0ex}}⇒k=5+5=10$

(vii) Given:

We have
(LHL at x = 1) = $\underset{x\to {1}^{-}}{\mathrm{lim}}f\left(x\right)=\underset{h\to 0}{\mathrm{lim}}f\left(1-h\right)=\underset{h\to 0}{\mathrm{lim}}4=4$
(RHL at x = 1) = $\underset{x\to {1}^{+}}{\mathrm{lim}}f\left(x\right)=\underset{h\to 0}{\mathrm{lim}}f\left(1+h\right)=\underset{h\to 0}{\mathrm{lim}}k{\left(1+\mathrm{h}\right)}^{2}=k$

If f(x) is continuous at x = 1, then
$\underset{x\to {1}^{-}}{\mathrm{lim}}f\left(x\right)=\underset{x\to {1}^{+}}{\mathrm{lim}}f\left(x\right)\phantom{\rule{0ex}{0ex}}⇒k=4$

(viii) Given:

We have
(LHL at x = 0) = $\underset{x\to {0}^{-}}{\mathrm{lim}}f\left(x\right)=\underset{h\to 0}{\mathrm{lim}}f\left(0-h\right)=\underset{h\to 0}{\mathrm{lim}}k\left({\left(-h\right)}^{2}+2\right)=2k$
(RHL at x = 0) = $\underset{x\to {0}^{+}}{\mathrm{lim}}f\left(x\right)=\underset{h\to 0}{\mathrm{lim}}f\left(0+h\right)=\underset{h\to 0}{\mathrm{lim}}3h+1=1$

If f(x) is continuous at x = 0, then
$\underset{x\to {0}^{-}}{\mathrm{lim}}f\left(x\right)=\underset{x\to {0}^{+}}{\mathrm{lim}}f\left(x\right)\phantom{\rule{0ex}{0ex}}⇒2k=1\phantom{\rule{0ex}{0ex}}⇒k=\frac{1}{2}$

(ix) Given:

If f(x) is continuous at x = 2, then
$\underset{x\to 2}{\mathrm{lim}}f\left(x\right)=f\left(2\right)\phantom{\rule{0ex}{0ex}}⇒\underset{x\to 2}{\mathrm{lim}}\left(x+5\right)=k\phantom{\rule{0ex}{0ex}}⇒k=2+5=7$

#### Question 37:

Find the values of a and b so that the function f given by
is continuous at x = 3 and x = 5.

Given:

We have
(LHL at x = 3) = $\underset{x\to {3}^{-}}{\mathrm{lim}}f\left(x\right)=\underset{h\to 0}{\mathrm{lim}}f\left(3-h\right)$$=\underset{h\to 0}{\mathrm{lim}}\left(1\right)=1$

(RHL at x = 3) = $\underset{x\to {3}^{+}}{\mathrm{lim}}f\left(x\right)=\underset{h\to 0}{\mathrm{lim}}f\left(3+h\right)$$=\underset{h\to 0}{\mathrm{lim}}a\left(3+h\right)+b=3a+b$

(LHL at x = 5) = $\underset{x\to {5}^{-}}{\mathrm{lim}}f\left(x\right)=\underset{h\to 0}{\mathrm{lim}}f\left(5-h\right)$$=\underset{h\to 0}{\mathrm{lim}}\left(a\left(5-h\right)+b\right)=5a+b$

(RHL at x = 5) = $\underset{x\to {5}^{+}}{\mathrm{lim}}f\left(x\right)=\underset{h\to 0}{\mathrm{lim}}f\left(5+h\right)$$=\underset{h\to 0}{\mathrm{lim}}7=7$

If f(x) is continuous at x = 3 and 5, then

∴ ​

On solving eqs. (1) and (2), we get

#### Question 38:

If . Show that f is continuous at x = 1.

Given:

We have
(LHL at x = 1) = $\underset{x\to {1}^{-}}{\mathrm{lim}}f\left(x\right)=\underset{h\to 0}{\mathrm{lim}}f\left(1-h\right)$$=\underset{h\to 0}{\mathrm{lim}}\frac{{\left(1-h\right)}^{2}}{2}=\frac{1}{2}$

(RHL at x = 1) = $\underset{x\to {1}^{+}}{\mathrm{lim}}f\left(x\right)=\underset{h\to 0}{\mathrm{lim}}f\left(1+h\right)$$=\underset{h\to 0}{\mathrm{lim}}\left[2{\left(1+h\right)}^{2}-3\left(1+h\right)+\frac{3}{2}\right]=2-3+\frac{3}{2}=\frac{1}{2}$

Also, $f\left(1\right)=\frac{{\left(1\right)}^{2}}{2}=\frac{1}{2}$

∴ ​

Hence, the given function is continuous at $x=1$.

#### Question 39:

Discuss the continuity of the f(x) at the indicated points:
(i) f(x) = | x | + | x − 1 | at x = 0, 1.
(ii) f(x) = | x − 1 | + | x + 1 | at x = −1, 1.

(i) Given: $f\left(x\right)=\left|x\right|+\left|x-1\right|$

We have
(LHL at x = 0) = $\underset{x\to {0}^{-}}{\mathrm{lim}}f\left(x\right)=\underset{h\to 0}{\mathrm{lim}}f\left(0-h\right)$$=\underset{h\to 0}{\mathrm{lim}}\left[\left|0-h\right|+\left|0-h-1\right|\right]=1$

(RHL at x = 0) = $\underset{x\to {0}^{+}}{\mathrm{lim}}f\left(x\right)=\underset{h\to 0}{\mathrm{lim}}f\left(0+h\right)$$=\underset{h\to 0}{\mathrm{lim}}\left[\left|0+h\right|+\left|0+h-1\right|\right]=1$

Also, $f\left(0\right)=\left|0\right|+\left|0-1\right|=0+1=1$

Now,

(LHL at x = 1) = $\underset{x\to {1}^{-}}{\mathrm{lim}}f\left(x\right)=\underset{h\to 0}{\mathrm{lim}}f\left(1-h\right)=\underset{h\to 0}{\mathrm{lim}}\left(\left|1-h\right|+\left|1-h-1\right|\right)=1+0=1$

(RHL at x =1) = $\underset{x\to {1}^{+}}{\mathrm{lim}}f\left(x\right)=\underset{h\to 0}{\mathrm{lim}}f\left(1+h\right)=\underset{h\to 0}{\mathrm{lim}}\left(\left|1+h\right|+\left|1+h-1\right|\right)=1+0=1$

Also, $f\left(1\right)=\left|1\right|+\left|1-1\right|=1+0=1$

∴ ​

Hence, $f\left(x\right)$ is continuous at .

(ii) Given: $f\left(x\right)=\left|x-1\right|+\left|x+1\right|$

We have
(LHL at x = −1) = $\underset{x\to -{1}^{-}}{\mathrm{lim}}f\left(x\right)=\underset{h\to 0}{\mathrm{lim}}f\left(-1-h\right)$$=\underset{h\to 0}{\mathrm{lim}}\left[\left|-1-h-1\right|+\left|-1-h+1\right|\right]=2+0=2$

(RHL at x = −1) = $\underset{x\to -{1}^{+}}{\mathrm{lim}}f\left(x\right)=\underset{h\to 0}{\mathrm{lim}}f\left(-1+h\right)$$=\underset{h\to 0}{\mathrm{lim}}\left[\left|-1+h-1\right|+\left|-1+h+1\right|\right]=2+0=2$

Also, $f\left(-1\right)=\left|-1-1\right|+\left|-1+1\right|=\left|-2\right|=2$

Now,

(LHL at x = 1) = $\underset{x\to {1}^{-}}{\mathrm{lim}}f\left(x\right)=\underset{h\to 0}{\mathrm{lim}}f\left(1-h\right)=\underset{h\to 0}{\mathrm{lim}}\left(\left|1-h-1\right|+\left|1-h+1\right|\right)=0+2=2$

(RHL at x =1) = $\underset{x\to {1}^{+}}{\mathrm{lim}}f\left(x\right)=\underset{h\to 0}{\mathrm{lim}}f\left(1+h\right)=\underset{h\to 0}{\mathrm{lim}}\left(\left|1+h-1\right|+\left|1+h+1\right|\right)=0+2=2$

Also, $f\left(1\right)=\left|1+1\right|+\left|1-1\right|=2$

∴ ​

Hence, $f\left(x\right)$ is continuous at .

#### Question 40:

Prove that is discontinuous at x = 0

The given function can be rewritten as

$⇒$

We have
(LHL at x = 0) = $\underset{x\to {0}^{-}}{\mathrm{lim}}f\left(x\right)=\underset{h\to 0}{\mathrm{lim}}f\left(0-h\right)=\underset{h\to 0}{\mathrm{lim}}f\left(-h\right)=\underset{h\to 0}{\mathrm{lim}}2=2$

(RHL at x = 0) = $\underset{x\to {0}^{+}}{\mathrm{lim}}f\left(x\right)=\underset{h\to 0}{\mathrm{lim}}f\left(0+h\right)=\underset{h\to 0}{\mathrm{lim}}f\left(h\right)=\underset{h\to 0}{\mathrm{lim}}0=0$

∴ ​$\underset{x\to {0}^{-}}{\mathrm{lim}}f\left(x\right)\ne \underset{x\to {0}^{+}}{\mathrm{lim}}f\left(x\right)$

Thus,  f(x) is discontinuous at x = 0
.

#### Question 41:

If , then what should be the value of
k so that f(x) is continuous at x = 0.

The given function can be rewritten as

We have
(LHL at x = 0) = $\underset{x\to {0}^{-}}{\mathrm{lim}}f\left(x\right)=\underset{h\to 0}{\mathrm{lim}}f\left(0-h\right)=\underset{h\to 0}{\mathrm{lim}}f\left(-h\right)=\underset{h\to 0}{\mathrm{lim}}-2{\left(-h\right)}^{2}+k=k$

(RHL at x = 0) = $\underset{x\to {0}^{+}}{\mathrm{lim}}f\left(x\right)=\underset{h\to 0}{\mathrm{lim}}f\left(0+h\right)=\underset{h\to 0}{\mathrm{lim}}f\left(h\right)=\underset{h\to 0}{\mathrm{lim}}\left(2{h}^{2}+k\right)=k$

If $f\left(x\right)$ is continuous at $x=0$, then

k can be any real number.

#### Question 42:

For what value of λ is the function

continuous at x = 0? What about continuity at x = ± 1?

The given function f is

If f is continuous at x = 0, then

Therefore, there is no value of λ for which f(x) is continuous at x = 0.

At x = 1,

f (1) = 4x + 1 = 4 × 1 + 1 = 5

Therefore, for any values of λ, f is continuous at x = 1

At x = -1, we have

f (-1) = $\lambda \left(1+2\right)=3\lambda$

$\underset{x\to -1}{\mathrm{lim}}\lambda \left(1+2\right)=3\lambda \phantom{\rule{0ex}{0ex}}\therefore \underset{x\to -1}{\mathrm{lim}}f\left(x\right)=f\left(-1\right)$

Therefore, for any values of λf is continuous at x = -1

#### Question 43:

For what value of k is the following function continuous at x = 2?

Given:

We have
(LHL at x = 2) = $\underset{x\to {2}^{-}}{\mathrm{lim}}f\left(x\right)=\underset{h\to 0}{\mathrm{lim}}f\left(2-h\right)$$=\underset{h\to 0}{\mathrm{lim}}\left(2\left(2-h\right)+1\right)=5$

(RHL at x = 2) = $\underset{x\to {2}^{+}}{\mathrm{lim}}f\left(x\right)=\underset{h\to 0}{\mathrm{lim}}f\left(2+h\right)$$=\underset{h\to 0}{\mathrm{lim}}3\left(2+h\right)-1=5$

Also, $f\left(2\right)=k$

If f(x) is continuous at x = 2, then

Hence, for k = 5, $f\left(x\right)$ is continuous at $x=2$.

#### Question 44:

Let If f(x) is continuous at x = $\frac{\mathrm{\pi }}{2}$, find a and b.

Given:

We have
(LHL at x = $\frac{\mathrm{\pi }}{2}$) = $\underset{x\to {\frac{\mathrm{\pi }}{2}}^{-}}{\mathrm{lim}}f\left(x\right)=\underset{h\to 0}{\mathrm{lim}}f\left(\frac{\mathrm{\pi }}{2}-h\right)$

$=\underset{h\to 0}{\mathrm{lim}}\left(\frac{1-{\mathrm{sin}}^{3}\left(\frac{\mathrm{\pi }}{2}-h\right)}{3{\mathrm{cos}}^{2}\left(\frac{\mathrm{\pi }}{2}-h\right)}\right)\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}=\underset{h\to 0}{\mathrm{lim}}\left(\frac{1-{\mathrm{cos}}^{3}h}{3{\mathrm{sin}}^{2}h}\right)\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}=\frac{1}{3}\underset{h\to 0}{\mathrm{lim}}\left(\frac{\left(1-\mathrm{cos}h\right)\left(1+{\mathrm{cos}}^{2}h+\mathrm{cos}h\right)}{\left(1-\mathrm{cos}h\right)\left(1+\mathrm{cos}h\right)}\right)\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}=\frac{1}{3}\underset{h\to 0}{\mathrm{lim}}\left(\frac{\left(1+{\mathrm{cos}}^{2}h+\mathrm{cos}h\right)}{\left(1+\mathrm{cos}h\right)}\right)\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}=\frac{1}{3}\left(\frac{1+1+1}{1+1}\right)=\frac{1}{2}$

(RHL at x = $\frac{\mathrm{\pi }}{2}$) = $\underset{x\to {\frac{\mathrm{\pi }}{2}}^{+}}{\mathrm{lim}}f\left(x\right)=\underset{h\to 0}{\mathrm{lim}}f\left(\frac{\mathrm{\pi }}{2}+h\right)$
$=\underset{h\to 0}{\mathrm{lim}}\left(\frac{b\left[1-\mathrm{sin}\left(\frac{\mathrm{\pi }}{2}+h\right)\right]}{{\left[\mathrm{\pi }-2\left(\frac{\mathrm{\pi }}{2}+h\right)\right]}^{2}}\right)\phantom{\rule{0ex}{0ex}}=\underset{h\to 0}{\mathrm{lim}}\left(\frac{b\left(1-\mathrm{cos}h\right)}{{\left[-2h\right]}^{2}}\right)\phantom{\rule{0ex}{0ex}}=\underset{h\to 0}{\mathrm{lim}}\left(\frac{2b{\mathrm{sin}}^{2}\frac{h}{2}}{4{h}^{2}}\right)\phantom{\rule{0ex}{0ex}}=\underset{h\to 0}{\mathrm{lim}}\left(\frac{2b{\mathrm{sin}}^{2}\frac{h}{2}}{16\frac{{h}^{2}}{4}}\right)\phantom{\rule{0ex}{0ex}}=\frac{b}{8}\underset{h\to 0}{\mathrm{lim}}{\left(\frac{\mathrm{sin}\frac{h}{2}}{\frac{h}{2}}\right)}^{2}\phantom{\rule{0ex}{0ex}}=\frac{b}{8}×1\phantom{\rule{0ex}{0ex}}=\frac{b}{8}$

Also, $f\left(\frac{\mathrm{\pi }}{2}\right)=a$

If f(x) is continuous at x = $\frac{\mathrm{\pi }}{2}$ , then

#### Question 45:

If the functions f(x), defined below is continuous at x = 0, find the value of k.

Given:

We have
(LHL at x = 0) = $\underset{x\to {0}^{-}}{\mathrm{lim}}f\left(x\right)=\underset{h\to 0}{\mathrm{lim}}f\left(0-h\right)$

$=\underset{h\to 0}{\mathrm{lim}}\left(\frac{1-\mathrm{cos}2\left(-h\right)}{2{\left(-h\right)}^{2}}\right)\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}=\underset{h\to 0}{\mathrm{lim}}\left(\frac{1-\mathrm{cos}2h}{2{h}^{2}}\right)\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}=\frac{1}{2}\underset{h\to 0}{\mathrm{lim}}\left(\frac{2{\mathrm{sin}}^{2}h}{{h}^{2}}\right)\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}=\frac{2}{2}\underset{h\to 0}{\mathrm{lim}}\left(\frac{{\mathrm{sin}}^{2}h}{{h}^{2}}\right)\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}=\frac{2}{2}\underset{h\to 0}{\mathrm{lim}}{\left(\frac{\mathrm{sin}h}{h}\right)}^{2}\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}=1×1\phantom{\rule{0ex}{0ex}}=1$

(RHL at x = 0) = $\underset{x\to {0}^{+}}{\mathrm{lim}}f\left(x\right)=\underset{h\to 0}{\mathrm{lim}}f\left(0+h\right)=\underset{h\to 0}{\mathrm{lim}}f\left(h\right)$$=\underset{h\to 0}{\mathrm{lim}}\left(1\right)=1$
Also, $f\left(0\right)=k$

If f(x) is continuous at x = 0, then
​

$⇒1=1=k$

Hence, the required value of k is 1.

#### Question 46:

Find the relationship between 'a' and 'b' so that the function 'f' defined by

is continuous at x = 3.

Given:

We have
(LHL at x = 3) = $\underset{x\to {3}^{-}}{\mathrm{lim}}f\left(x\right)=\underset{h\to 0}{\mathrm{lim}}f\left(3-h\right)=\underset{h\to 0}{\mathrm{lim}}a\left(3-h\right)+1=3a+1$

(RHL at x = 3) = $\underset{x\to {3}^{+}}{\mathrm{lim}}f\left(x\right)=\underset{h\to 0}{\mathrm{lim}}f\left(3+h\right)=\underset{h\to 0}{\mathrm{lim}}b\left(3+h\right)+3=3b+3$

Hence, the required relationship between .

#### Question 1:

Prove that the function is everywhere continuous.

When x < 0, we have
$f\left(x\right)=\frac{\mathrm{sin}x}{x}$

We know that sin x as well as the identity function x are everywhere continuous.
So, the quotient function $\frac{\mathrm{sin}x}{x}$ is continuous at each x < 0.

When x > 0, we have $f\left(x\right)=x+1$, which is a polynomial function.
Therefore, $f\left(x\right)$ is continuous at each x > 0.

Now,
Let us consider the point x = 0.

Given:

We have
(LHL at x = 0) = $\underset{x\to {0}^{-}}{\mathrm{lim}}f\left(x\right)=\underset{h\to 0}{\mathrm{lim}}f\left(0-h\right)=\underset{h\to 0}{\mathrm{lim}}f\left(-h\right)=\underset{h\to 0}{\mathrm{lim}}\left(\frac{\mathrm{sin}\left(-h\right)}{-h}\right)=\underset{h\to 0}{\mathrm{lim}}\left(\frac{\mathrm{sin}\left(h\right)}{h}\right)=1$

(RHL at x = 0) = $\underset{x\to {0}^{+}}{\mathrm{lim}}f\left(x\right)=\underset{h\to 0}{\mathrm{lim}}f\left(0+h\right)=\underset{h\to 0}{\mathrm{lim}}f\left(h\right)=\underset{h\to 0}{\mathrm{lim}}\left(h+1\right)=1$

Also,
$f\left(0\right)=0+1=1$

$\underset{x\to {0}^{-}}{\mathrm{lim}}f\left(x\right)=\underset{x\to {0}^{+}}{\mathrm{lim}}f\left(x\right)=f\left(0\right)$

Thus, $f\left(x\right)$ is continuous at x = 0.

Hence, $f\left(x\right)$ is everywhere continuous.

#### Question 2:

Discuss the continuity of the function

Given:

We have
(LHL at x = 0) = $\underset{x\to {0}^{-}}{\mathrm{lim}}f\left(x\right)=\underset{h\to 0}{\mathrm{lim}}f\left(0-h\right)=\underset{h\to 0}{\mathrm{lim}}f\left(-h\right)=\underset{h\to 0}{\mathrm{lim}}\left(-1\right)=-1$

(RHL at x = 0) = $\underset{x\to {0}^{+}}{\mathrm{lim}}f\left(x\right)=\underset{h\to 0}{\mathrm{lim}}f\left(0+h\right)=\underset{h\to 0}{\mathrm{lim}}f\left(h\right)=\underset{h\to 0}{\mathrm{lim}}\left(1\right)=1$

$\underset{x\to {0}^{-}}{\mathrm{lim}}f\left(x\right)\ne \underset{x\to {0}^{+}}{\mathrm{lim}}f\left(x\right)$

Thus, $f\left(x\right)$ is discontinuous at x = 0.

#### Question 3:

Find the points of discontinuity, if any, of the following functions:
(i)

(ii)

(iii)

(iv)

(v)

(vi)

(vii)

(viii)

(ix)

(x)

(xi)

(xii)

(xiii)

(i)

When x $\ne$1, then
$f\left(x\right)={x}^{3}-{x}^{2}+2x-2$

We know that a polynomial function is everywhere continuous.
So, $f\left(x\right)={x}^{3}-{x}^{2}+2x-2$ is continuous at each x $\ne$1.

At x = 1, we have

(LHL at x = 1) = $\underset{x\to {1}^{-}}{\mathrm{lim}}f\left(x\right)=\underset{h\to 0}{\mathrm{lim}}f\left(1-h\right)=\underset{h\to 0}{\mathrm{lim}}\left({\left(1-h\right)}^{3}-{\left(1-h\right)}^{2}+2\left(1-h\right)-2\right)=1-1+2-2=0$

(RHL at x = 1) = $\underset{x\to {1}^{+}}{\mathrm{lim}}f\left(x\right)=\underset{h\to 0}{\mathrm{lim}}f\left(1+h\right)=\underset{h\to 0}{\mathrm{lim}}\left({\left(1+h\right)}^{3}-{\left(1+h\right)}^{2}+2\left(1+h\right)-2\right)=1-1+2-2=0$

Also, $f\left(1\right)=4$
$\underset{x\to {1}^{-}}{\mathrm{lim}}f\left(x\right)=\underset{x\to {1}^{+}}{\mathrm{lim}}f\left(x\right)\ne f\left(1\right)$

Thus, $f\left(x\right)$ is discontinuous at x = 1.

Hence, the only point of discontinuity for $f\left(x\right)$ is x = 1.

(ii)
Given:

When x $\ne$2, then
$f\left(x\right)=\frac{{x}^{4}-16}{x-2}=\frac{{x}^{4}-{2}^{4}}{x-2}=\frac{\left({x}^{2}+4\right)\left(x-2\right)\left(x+2\right)}{x-2}=\left({x}^{2}+4\right)\left(x+2\right)$

We know that a polynomial function is everywhere continuous.
Therefore, the functions are everywhere continuous.
So, the product function $\left({x}^{2}+4\right)\left(x+2\right)$ is everywhere continuous.
Thus, f(x) is continuous at every x $\ne$2.

At x = 2, we have

(LHL at x = 2) = $\underset{x\to {2}^{-}}{\mathrm{lim}}f\left(x\right)=\underset{h\to 0}{\mathrm{lim}}f\left(2-h\right)=\underset{h\to 0}{\mathrm{lim}}\left[{\left(2-h\right)}^{2}+4\right]\left(2-h+2\right)=8\left(4\right)=32$

(RHL at x = 2) = $\underset{x\to {2}^{+}}{\mathrm{lim}}f\left(x\right)=\underset{h\to 0}{\mathrm{lim}}f\left(2+h\right)=\underset{h\to 0}{\mathrm{lim}}\left[{\left(2+h\right)}^{2}+4\right]\left(2+h+2\right)=8\left(4\right)=32$

Also, $f\left(2\right)=16$

$\underset{x\to {2}^{-}}{\mathrm{lim}}f\left(x\right)=\underset{x\to {2}^{+}}{\mathrm{lim}}f\left(x\right)\ne f\left(2\right)$

Thus, $f\left(x\right)$ is discontinuous at x = 2.

Hence, the only point of discontinuity for $f\left(x\right)$ is x = 2.

(iii)

When x < 0, then
$f\left(x\right)=\frac{\mathrm{sin}x}{x}$

We know that sin x as well as the identity function x are everywhere continuous.
So, the quotient function $\frac{\mathrm{sin}x}{x}$ is continuous at each x < 0.

When x > 0, then
$f\left(x\right)=2x+3$, which is a polynomial function.
Therefore, $f\left(x\right)$ is continuous at each x > 0.

Now,
Let us consider the point x = 0.
Given:

We have
(LHL at x = 0) = $\underset{x\to {0}^{-}}{\mathrm{lim}}f\left(x\right)=\underset{h\to 0}{\mathrm{lim}}f\left(0-h\right)=\underset{h\to 0}{\mathrm{lim}}f\left(-h\right)=\underset{h\to 0}{\mathrm{lim}}\left(\frac{\mathrm{sin}\left(-h\right)}{-h}\right)=\underset{h\to 0}{\mathrm{lim}}\left(\frac{\mathrm{sin}\left(h\right)}{h}\right)=1$

(RHL at x = 0) = $\underset{x\to {0}^{+}}{\mathrm{lim}}f\left(x\right)=\underset{h\to 0}{\mathrm{lim}}f\left(0+h\right)=\underset{h\to 0}{\mathrm{lim}}f\left(h\right)=\underset{h\to 0}{\mathrm{lim}}\left(2h+3\right)=3$

$\underset{x\to {0}^{-}}{\mathrm{lim}}f\left(x\right)\ne \underset{x\to {0}^{+}}{\mathrm{lim}}f\left(x\right)$

Thus, $f\left(x\right)$ is discontinuous at x = 0.

Hence, the only point of discontinuity for $f\left(x\right)$ is x = 0.

(iv)

When x $\ne$ 0, then
$f\left(x\right)=\frac{\mathrm{sin}3x}{x}$

We know that sin 3x as well as the identity function x are everywhere continuous.
So, the quotient function $\frac{\mathrm{sin}3x}{x}$ is continuous at each x $\ne$ 0.

Let us consider the point x = 0.
Given:

We have

(LHL at x = 0) =

(RHL at x = 0) =
Also, $f\left(0\right)=4$

$\underset{x\to {0}^{-}}{\mathrm{lim}}f\left(x\right)=\underset{x\to {0}^{+}}{\mathrm{lim}}f\left(x\right)\ne f\left(0\right)$
Thus, $f\left(x\right)$ is discontinuous at x = 0.

Hence, the only point of discontinuity for $f\left(x\right)$ is x = 0.

(v)

When x $\ne$ 0, then
$f\left(x\right)=\frac{\mathrm{sin}x}{x}+\mathrm{cos}x$

We know that sin x as well as the identity function x both are everywhere continuous.
So, the quotient function $\frac{\mathrm{sin}x}{x}$ is continuous at each x $\ne$ 0.
Also, cos x is everywhere continuous.
Therefore, is continuous at each x $\ne$ 0.

Let us consider the point x = 0.
Given:

We have
(LHL at x = 0) = $\underset{x\to {0}^{-}}{\mathrm{lim}}f\left(x\right)=\underset{h\to 0}{\mathrm{lim}}f\left(0-h\right)=\underset{h\to 0}{\mathrm{lim}}f\left(-h\right)=\underset{h\to 0}{\mathrm{lim}}\left(\frac{\mathrm{sin}\left(-h\right)}{-h}+\mathrm{cos}\left(-h\right)\right)=\underset{h\to 0}{\mathrm{lim}}\left(\frac{\mathrm{sin}\left(-h\right)}{-h}\right)+\underset{h\to 0}{\mathrm{lim}}\mathrm{cos}\left(-h\right)=1+1=2$

(RHL at x = 0) = $\underset{x\to {0}^{+}}{\mathrm{lim}}f\left(x\right)=\underset{h\to 0}{\mathrm{lim}}f\left(0+h\right)=\underset{h\to 0}{\mathrm{lim}}f\left(h\right)=\underset{h\to 0}{\mathrm{lim}}\left(\frac{\mathrm{sin}\left(h\right)}{h}+\mathrm{cos}\left(h\right)\right)=\underset{h\to 0}{\mathrm{lim}}\left(\frac{\mathrm{sin}\left(h\right)}{h}\right)+\underset{h\to 0}{\mathrm{lim}}\mathrm{cos}\left(h\right)=1+1=2$
Also, $f\left(0\right)=5$

$\underset{x\to {0}^{-}}{\mathrm{lim}}f\left(x\right)=\underset{x\to {0}^{+}}{\mathrm{lim}}f\left(x\right)\ne f\left(0\right)$
Thus, $f\left(x\right)$ is discontinuous at x = 0.

Hence, the only point of discontinuity for  $f\left(x\right)$ is x = 0.

(vi)

When x $\ne$ 0, then
$f\left(x\right)=\frac{{x}^{4}+{x}^{3}+2{x}^{2}}{{\mathrm{tan}}^{-1}x}$

We know that ${x}^{4}+{x}^{3}+2{x}^{2}$ is a polynomial function which is everywhere continuous.
Also, ${\mathrm{tan}}^{-1}x$ is everywhere continuous.
So, the quotient function $\frac{{x}^{4}+{x}^{3}+2{x}^{2}}{{\mathrm{tan}}^{-1}x}$ is continuous at each x $\ne$ 0.

Let us consider the point x = 0.

Given:

We have
(LHL at x = 0) = $\underset{x\to {0}^{-}}{\mathrm{lim}}f\left(x\right)=\underset{h\to 0}{\mathrm{lim}}f\left(0-h\right)=\underset{h\to 0}{\mathrm{lim}}f\left(-h\right)=\underset{h\to 0}{\mathrm{lim}}\left(\frac{{\left(-h\right)}^{4}+{\left(-h\right)}^{3}+2{\left(-h\right)}^{2}}{{\mathrm{tan}}^{-1}\left(-h\right)}\right)=\underset{h\to 0}{\mathrm{lim}}\left(\frac{{\left(h\right)}^{3}-{\left(h\right)}^{2}+2\left(h\right)}{-\frac{{\mathrm{tan}}^{-1}\left(h\right)}{h}}\right)=\frac{0}{\left(-1\right)}=0$

(RHL at x = 0) = $\underset{x\to {0}^{+}}{\mathrm{lim}}f\left(x\right)=\underset{h\to 0}{\mathrm{lim}}f\left(0+h\right)=\underset{h\to 0}{\mathrm{lim}}f\left(h\right)=\underset{h\to 0}{\mathrm{lim}}\left(\frac{{\left(h\right)}^{4}+{\left(h\right)}^{3}+2{\left(h\right)}^{2}}{{\mathrm{tan}}^{-1}\left(h\right)}\right)=\underset{h\to 0}{\mathrm{lim}}\left(\frac{{\left(h\right)}^{3}+{\left(h\right)}^{2}+2\left(h\right)}{\frac{{\mathrm{tan}}^{-1}\left(h\right)}{h}}\right)=\frac{0}{1}=0$
Also, $f\left(0\right)=10$

$\underset{x\to {0}^{-}}{\mathrm{lim}}f\left(x\right)=\underset{x\to {0}^{+}}{\mathrm{lim}}f\left(x\right)\ne f\left(0\right)$
Thus, $f\left(x\right)$ is discontinuous at x = 0.

Hence, the only point of discontinuity for  $f\left(x\right)$ is x = 0.

(vii)
Given:

We have

It is given that $f\left(0\right)=7$

$\underset{x\to 0}{\mathrm{lim}}f\left(x\right)\ne f\left(0\right)$

Hence, the given function is discontinuous at x = 0 and continuous elsewhere.

(viii)

When x > 1, then
$f\left(x\right)=\left|x-3\right|$

Since modulus function is a continuous function, $f\left(x\right)$ is continuous for each x > 1.

When x < 1, then
$f\left(x\right)=\frac{{x}^{2}}{4}-\frac{3x}{2}+\frac{13}{4}$

Since, are continuous being polynomial functions, will also be continuous.
Also, $\frac{13}{4}$ is continuous being a polynomial function.

$⇒\frac{{x}^{2}}{4}-\frac{3x}{2}+\frac{13}{4}$ is continuous for each $x<1$.

$⇒f\left(x\right)$ is continuous for each x < 1.

At x = 1, we have
(LHL at x=1) =

(RHL at x=1) =
Also, $f\left(1\right)=\left|1-3\right|=\left|-2\right|=2$

Thus,

Hence, $f\left(x\right)$ is continuous at x= 1.

Thus, the given function is nowhere discontinuous.

(ix)

At $x\le -3$, we have
$f\left(x\right)=\left|x\right|+3$

Since modulus function and constant function are continuous, $f\left(x\right)=\left|x\right|+3$ is continuous for each $x\le -3$.

At $-3, we have
$f\left(x\right)=-2x$
Since polynomial function is continuous and constant function is continuous, $f\left(x\right)=-2x$ is continuous for each$-3.

At $x>3$, we have
$f\left(x\right)=6x+2$

Since polynomial function is continuous and constant function is continuous, $f\left(x\right)=6x+2$ is continuous for each $x>3$.

Now, we check the continuity of the function at the point $x=3$.

We have
(LHL at x=3) = $\underset{x\to {3}^{-}}{\mathrm{lim}}f\left(x\right)=\underset{h\to 0}{\mathrm{lim}}f\left(3-h\right)=\underset{h\to 0}{\mathrm{lim}}-2\left(3-h\right)=-6$

(RHL at x=3) = $\underset{x\to {3}^{+}}{\mathrm{lim}}f\left(x\right)=\underset{h\to 0}{\mathrm{lim}}f\left(3+h\right)=\underset{h\to 0}{\mathrm{lim}}6\left(3+h\right)+2=20$
$⇒\underset{x\to {3}^{-}}{\mathrm{lim}}f\left(x\right)\ne \underset{x\to {3}^{+}}{\mathrm{lim}}f\left(x\right)$

Hence, the only point of discontinuity of the given function is $x=3$

(x)
Given:

The given function f is defined at all the points of the real line.

Let c be a point on the real line.

Case I:

Therefore, f is continuous at all points x, such that x < 1

Case II:

If c = 1, then the left hand limit of f at x = 1 is,

The right hand limit of f at x = 1 is,

It is observed that the left and right hand limit of f at x = 1 do not coincide.

Therefore, f is not continuous at x = 1

Case III:

Therefore, f is continuous at all points x, such that x > 1

Thus, from the above observation, it can be concluded that x = 1 is the only point of discontinuity of f.

(xi) The given function is

The given function is defined at all points of the real line.

Let c be a point on the real line.

Case I:

Therefore, f is continuous at all points x, such that x < 0

Case II:

The left hand limit of f at x = 0 is,

The right hand limit of f at x = 0 is,

Therefore, f is continuous at x = 0

Case III:

Therefore, f is continuous at all points of the interval (0, 1).

Case IV:

The left hand limit of f at x = 1 is,

The right hand limit of f at x = 1 is,

It is observed that the left and right hand limits of f at x = 1 do not coincide.

Therefore, f is not continuous at x = 1

Case V:

Therefore, f is continuous at all points x, such that x > 1

Hence, f is not continuous only at x = 1

(xii)

The given function f is

It is evident that f is defined at all points of the real line.

Let c be a real number.

Case I:

Therefore, f is continuous at all points x, such that x 0

Case II:

Therefore, f is continuous at x = 0

From the above observations, it can be concluded that f is continuous at every point of the real line.

Thus, f is a continuous function.

(xiii)
The given function
f is

The given function is defined at all points of the real line.

Let c be a point on the real line.

Case I:

Therefore, f is continuous at all points x, such that x < −1

Case II:

The left hand limit of f at x = −1 is,

The right hand limit of f at x = −1 is,

Therefore, f is continuous at x = −1

Case III:

Therefore, f is continuous at all points of the interval (−1, 1).

Case IV:

The left hand limit of f at x = 1 is,

The right hand limit of f at x = 1 is,

Therefore, f is continuous at x = 2

Case V:

Therefore, f is continuous at all points x, such that x > 1

Thus, from the above observations, it can be concluded that f is continuous at all points of the real line.

#### Question 4:

In the following, determine the value(s) of constant(s) involved in the definition so that the given function is continuous:
(i)

(ii)

(iii)

(iv)

(v)

(vi)

(vii)

(viii)

(i) Given:

If $f\left(x\right)$ is continuous at x = 0, then
$\underset{x\to 0}{\mathrm{lim}}f\left(x\right)=f\left(0\right)$

(ii) Given:

If $f\left(x\right)$ is continuous at x = 2, then
$\underset{x\to {2}^{-}}{\mathrm{lim}}f\left(x\right)=\underset{x\to {2}^{+}}{\mathrm{lim}}f\left(x\right)$

$⇒\underset{h\to 0}{\mathrm{lim}}f\left(2-h\right)=\underset{h\to 0}{\mathrm{lim}}f\left(2+h\right)\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}⇒\underset{h\to 0}{\mathrm{lim}}\left(k\left(2-h\right)+5\right)=\underset{h\to 0}{\mathrm{lim}}\left(2+h-1\right)\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}⇒2k+5=1\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}⇒2k=-4\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}⇒k=-2$

(iii) Given:

If $f\left(x\right)$ is continuous at x = 0, then
$\underset{x\to {0}^{-}}{\mathrm{lim}}f\left(x\right)=\underset{x\to {0}^{+}}{\mathrm{lim}}f\left(x\right)$

Hence, there does not exist any value of k, which can make the given function continuous.

(iv) Given:

If $f\left(x\right)$ is continuous at x = 3 and 5, then

(v)

Given:

If $f\left(x\right)$ is continuous at x = −1 and 0, then

(vi)

Given:

If $f\left(x\right)$ is continuous at x = 0, then
$\underset{x\to {0}^{-}}{\mathrm{lim}}f\left(x\right)=\underset{x\to {0}^{+}}{\mathrm{lim}}f\left(x\right)$

(vii)

Given:

If $f\left(x\right)$ is continuous at x = 2 and 10,  then

(viii)

Given:

If $f\left(x\right)$ is continuous at x = $\frac{\mathrm{\pi }}{2}$, then
$\underset{x\to {\frac{\mathrm{\pi }}{2}}^{-}}{\mathrm{lim}}f\left(x\right)=f\left(\frac{\mathrm{\pi }}{2}\right)$

#### Question 5:

The function
is continuous on (0, ∞), then find the most suitable values of a and b.

Given:  f is continuous on $\left(0,\infty \right)$

∴  f is continuous at x = 1 and $\sqrt{2}$

At x = 1, we have

$\underset{x\to {1}^{+}}{\mathrm{lim}}f\left(x\right)=\underset{h\to 0}{\mathrm{lim}}f\left(1+h\right)=\underset{h\to 0}{\mathrm{lim}}\left(a\right)=a$

Also,

At x = $\sqrt{2}$, we have

f is continuous at x = 1 and $\sqrt{2}$

If a = 1, then

If a = −1, then

Hence, the most suitable values of a and b are

a = −1, b = 1  or a = 1, $b=1±\sqrt{2}$

#### Question 6:

Find the values of a and b so that the function f(x) defined by

becomes continuous on [0, π].

Given: f is continuous on .

f is continuous at x = $\frac{\mathrm{\pi }}{4}$ and $\frac{\mathrm{\pi }}{2}$

At x = $\frac{\mathrm{\pi }}{4}$, we have

At x = $\frac{\mathrm{\pi }}{2}$, we have

Since f is continuous at x = $\frac{\mathrm{\pi }}{4}$ and x = $\frac{\mathrm{\pi }}{2}$, we get

#### Question 7:

The function f(x) is defined as follows:

If f is continuous on [0, 8], find the values of a and b.

Given: f is continuous on .

f is continuous at x = 2 and x = 4

At x = 2, we have

Also,
At x = 4, we have

f is continuous at x = 2 and x = 4

On simplifying eqs. (1) and (2), we get

#### Question 8:

If for x ≠ π/4, find the value which can be assigned to f(x) at x = π/4 so that the function f(x) becomes continuous every where in [0, π/2].

When $x\ne \frac{\mathrm{\pi }}{4}$,  and  are continuous in .

Thus, the quotient function  is continuous in $\left[0,\frac{\mathrm{\pi }}{2}\right]$ for each $x\ne \frac{\mathrm{\pi }}{4}$.

So, if $f\left(x\right)$ is continuous at $x=\frac{\mathrm{\pi }}{4}$, then it will be everywhere continuous in .

Now,
Let us consider the point x = $\frac{\mathrm{\pi }}{4}$.

Given:

We have
(LHL at x = $\frac{\mathrm{\pi }}{4}$) =

(RHL at x = $\frac{\mathrm{\pi }}{4}$) =

If $f\left(x\right)$ is continuous at $x=\frac{\mathrm{\pi }}{4}$, then

∴  $f\left(\frac{\mathrm{\pi }}{4}\right)=\frac{1}{2}$

Hence, for ​$f\left(\frac{\mathrm{\pi }}{4}\right)=\frac{1}{2}$, the function $f\left(x\right)$ will be everywhere continuous in ​.

#### Question 9:

Discuss the continuity of the function

When x < 2, we have

We know that a polynomial function is everywhere continuous.
So, $f\left(x\right)$ is continuous for each x < 2.

When $x>2$, we have
$f\left(x\right)=\frac{3x}{2}$

The functions 3x and 2 are continuous being the polynomial and constant function, respectively.
Thus, the quotient function $\frac{3x}{2}$ is continuous at each x > 2.

Now,
Let us consider the point x = 2.
Given:

We have
(LHL at x = 2) =

(RHL at x = 2) =

Also,
$f\left(2\right)=\frac{3\left(2\right)}{2}=3$

∴ $\underset{x\to {2}^{-}}{\mathrm{lim}}f\left(x\right)=\underset{x\to {2}^{+}}{\mathrm{lim}}f\left(x\right)=f\left(2\right)$

Thus, $f\left(x\right)$ is continuous at x = 2.

Hence, $f\left(x\right)$ is everywhere continuous.

#### Question 10:

Discuss the continuity of f(x) = sin | x |.

This function f is defined for every real number and f can be written as the composition of two functions as,

f = h o g, where

It has to be proved first that are continuous functions.

Clearly, g is defined for all real numbers.

Let c be a real number.

Case I:

Therefore, g is continuous at all points x < 0

Case II:

Therefore, g is continuous at all points x > 0

Case III:

Therefore, g is continuous at x = 0

From the above three observations, it can be concluded that g is continuous at all points.

Now, h (x) = sin x

It is evident that h (x) = sin x is defined for every real number.

Let c be a real number.
Put
x = c + k

If x c, then k 0

h (c) = sin c

So, h is a continuous function.

is a continuous function.

#### Question 11:

Prove that

is everywhere continuous.

When x < 0, we have
$f\left(x\right)=\frac{\mathrm{sin}x}{x}$

We know that sin x as well as the identity function x are everywhere continuous.
So, the quotient function $\frac{\mathrm{sin}x}{x}$ is continuous at each x < 0.

When x > 0, we have
$f\left(x\right)=x+1$, which is a polynomial function.
Therefore, $f\left(x\right)$ is continuous at each x > 0.

Now,
Let us consider the point x = 0.
Given:

We have
(LHL at x = 0) =

(RHL at x = 0) =
Also,
$f\left(0\right)=0+1=1$

$\underset{x\to {0}^{-}}{\mathrm{lim}}f\left(x\right)=\underset{x\to {0}^{+}}{\mathrm{lim}}f\left(x\right)=f\left(0\right)$

Thus, $f\left(x\right)$ is continuous at x = 0.

Hence, $f\left(x\right)$ is everywhere continuous.

#### Question 12:

Show that the function g (x) = x − [x] is discontinuous at all integral points. Here [x] denotes the greatest integer function.

Given:

It is evident that g is defined at all integral points.

Let $n\in Z$.

Then,

The left hand limit of f at x = n is,

The right hand limit of f at x = n is,

It is observed that the left and right hand limits of f at x = n do not coincide.
i.e.

So, f is not continuous at x = n, $n\in Z$

Hence, g is discontinuous at all integral points.

#### Question 13:

Discuss the continuity of the following functions:
(i) f(x) = sin x + cos x
(ii) f(x) = sin x − cos x
(iii) f(x) = sin x cos x

It is known that if g and h are two continuous functions, then are also continuous.

It has to proved first that g (x) = sin x and h (x) = cos x are continuous functions.

Let g (x) = sin x

It is evident that g (x) = sin x is defined for every real number.

Let c be a real number. Put x = c + h

If x c, then h 0

So, g is a continuous function.

Let h (x) = cos x

It is evident that h (x) = cos x is defined for every real number.

Let c be a real number. Put x = c + h

If x c, then h 0

h (c) = cos c

So, h is a continuous function.

Therefore, it can be concluded that

(i) f (x) = g (x) + h (x) = sin x + cos x is a continuous function.

(ii) f (x) = g (x) − h (x) = sin x − cos x is a continuous function.

(iii) f (x) = g (x$×$ h (x) = sin x cos x is a continuous function.

#### Question 14:

Show that f (x) = cos x2 is a continuous function.

Given: f (x) = cos (x2)

This function f is defined for every real number and f can be written as the composition of two functions as

f = g o h, where g (x) = cos x and h (x) = x2

It has to be first proved that g (x) = cos x and h (x) = x2 are continuous functions.

It is evident that g is defined for every real number.

Let c be a real number.

Then, g (c) = cos c

So, g (x) = cos x is a continuous function.

Now,
h
(x) = x2

Clearly, h is defined for every real number.

Let k be a real number, then h (k) = k2

So, h is a continuous function.

It is known that for real valued functions g and h, such that (g o h) is defined at x = c, if g is continuous at x = c and if f is continuous at g (c), then, (f o g) is continuous at x = c.

Therefore, is a continuous function.

#### Question 15:

Show that f (x) = | cos x | is a continuous function.

The given function is

This function f is defined for every real number and f can be written as the composition of two functions as,

f = g o h, where

It has to be first proved that are continuous functions.

Clearly, g is defined for all real numbers.

Let c be a real number.

Case I:

So, g is continuous at all points x < 0.

Case II:

So, g is continuous at all points x > 0.

Case III:

So, g is continuous at x = 0

From the above three observations, it can be concluded that g is continuous at all points.

Now, h (x) = cos x

It is evident that h (x) = cos x is defined for every real number.

Let c be a real number.
Put
x = c + h

If x c, then h 0

h (c) = cos c

So, h (x) = cos x is a continuous function.

It is known that for real valued functions g and h,such that (g o h) is defined at x = c, if g is continuous at x = c and if f is continuous at g (c), then (f o g) is continuous at x = c.

Therefore, is a continuous function.

#### Question 16:

Find all the points of discontinuity of f defined by f (x) = | x | − | x + 1 |.

Given:

The two functions, g and h, are defined as

Then, f = g h

The continuity of g and h is examined first.

Clearly, g is defined for all real numbers.

Let c be a real number.

Case I:

So, g is continuous at all points x < 0.

Case II:

So, g is continuous at all points x > 0.

Case III:

So, g is continuous at x = 0

From the above three observations, it can be concluded that g is continuous at all points.

Clearly, h is defined for every real number.

Let c be a real number.

Case I:

So, h is continuous at all points x < −1.

Case II:

So, h is continuous at all points x > −1.

Case III:

So, h is continuous at x = −1

From the above three observations, it can be concluded that h is continuous at all points of the real line.

So, g and h are continuous functions.

Thus,
f = gh is also a continuous function.

Therefore, f has no point of discontinuity.

#### Question 17:

Determine if is a continuous function?

The given function f is

It is evident that f is defined at all points of the real line.

Let c be a real number.

Case I:

So, f is continuous at all points x 0

Case II:

So, f is continuous at x = 0

From the above observations, it can be concluded that f is continuous at every point of the real line.

Thus, f is a continuous function.

#### Question 18:

Given the function $f\left(x\right)=\frac{1}{x+2}$. Find the points of discontinuity of the function f(f(x)).

$f\left[f\left(x\right)\right]=\frac{1}{\frac{1}{x+2}+2}=\frac{x+2}{2x+5}$

Hence, the function is discontinuous at

#### Question 19:

Find all point of discontinuity of the function

Hence, the function is discontinuous at

#### Question 1:

Define continuity of a function at a point.

Continuity at a point:

A function $f\left(x\right)$ is said to be continuous at a point xa of its domain, iff $\underset{x\to \mathrm{a}}{\mathrm{lim}}f\left(x\right)=f\left(a\right)$.

#### Question 2:

What happens to a function f (x) at x = a, if $\underset{x\to a}{\mathrm{lim}}$ f (x) = f (a)?

If $f\left(x\right)$ is a function defined in its domain such that $\underset{x\to \mathrm{a}}{\mathrm{lim}}f\left(x\right)=f\left(a\right)$, then $f\left(x\right)$ becomes continuous at $x=a$.

#### Question 3:

Find f (0), so that $f\left(x\right)=\frac{x}{1-\sqrt{1-x}}$ becomes continuous at x = 0.

If $f\left(x\right)$is continuous at x = 0, then $\underset{x\to 0}{\mathrm{lim}}f\left(x\right)=f\left(0\right)$          ...(1)

Given: $f\left(x\right)=\frac{x}{1-\sqrt{1-x}}$

$⇒f\left(0\right)=2$

So, for $f\left(0\right)=2$, the function f(x) becomes continuous at x = 0.

#### Question 4:

If is continuous at x = 0, then write the value of k.

If $f\left(x\right)$ is continuous at $x=0$, then
$\underset{x\to 0}{\mathrm{lim}}f\left(x\right)=f\left(0\right)$

#### Question 5:

If the function is continuous at x = 0, find f (0).

Given:  is continuous at $x=0$.
$\underset{x\to 0}{\mathrm{lim}}f\left(x\right)=f\left(0\right)$

#### Question 6:

If is continuous at x = 4, find k.

Given:

If $f\left(x\right)$ is continuous at $x=4$, then
$\underset{x\to 4}{\mathrm{lim}}f\left(x\right)=f\left(4\right)$

$⇒\underset{x\to 4}{\mathrm{lim}}\left(\frac{{x}^{2}-16}{x-4}\right)=k$

$⇒\underset{x\to 4}{\mathrm{lim}}\frac{\left(x+4\right)\left(x-4\right)}{\left(x-4\right)}=k\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}⇒\underset{x\to 4}{\mathrm{lim}}\left(x+4\right)=k\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}⇒k=8$

#### Question 7:

Determine whether is continuous at x = 0 or not.

Given:

We have

Hence, $f\left(x\right)$ is continuous at $x=0$.

#### Question 8:

If is continuous at x = 0, find k.

Given:
If $f\left(x\right)$ is continuous at $x=0$, then
$\underset{x\to 0}{\mathrm{lim}}f\left(x\right)=f\left(0\right)$

#### Question 9:

If is continuous at x = 0, write the value of k.

Given,

If $f\left(x\right)$ is continuous at $x=0$, then

$\underset{x\to 0}{\mathrm{lim}}f\left(x\right)=f\left(0\right)$

$⇒\underset{x\to 0}{\mathrm{lim}}\left(\frac{{\mathrm{sin}}^{-1}x}{x}\right)=f\left(0\right)$

#### Question 10:

Write the value of b for which is continuous at x = 1.

Given:

If $f\left(x\right)$ is continuous at $x=1$, then
$\underset{x\to {1}^{-}}{\mathrm{lim}}f\left(x\right)=\underset{x\to {1}^{+}}{\mathrm{lim}}f\left(x\right)=f\left(1\right)$                   ...(1)

Now,
$\underset{x\to {1}^{-}}{\mathrm{lim}}f\left(x\right)=\underset{h\to 0}{\mathrm{lim}}f\left(1-h\right)=\underset{h\to 0}{\mathrm{lim}}5\left(1-h\right)-4=5-4=1$

$\underset{x\to {1}^{+}}{\mathrm{lim}}f\left(x\right)=\underset{h\to 0}{\mathrm{lim}}f\left(1+h\right)=\underset{h\to 0}{\mathrm{lim}}4{\left(1+h\right)}^{2}+3b\left(1+h\right)=4+3b$

Also, $f\left(1\right)=5\left(1\right)-4=1$

$⇒1=4+3b\phantom{\rule{0ex}{0ex}}⇒-3=3b\phantom{\rule{0ex}{0ex}}⇒b=-1$

Thus, for $b=-1$, the function $f\left(x\right)$ is continuous at $x=1$.

#### Question 11:

Determine the value of the constant 'k' so that function f is continuous at $x=$0.

Since the function is continuous at x = 0, therefore,

$\underset{x\to {0}^{-}}{\mathrm{lim}}f\left(x\right)=\underset{x\to {0}^{+}}{\mathrm{lim}}f\left(x\right)=f\left(0\right)\phantom{\rule{0ex}{0ex}}⇒\underset{x\to 0}{\mathrm{lim}}\frac{-kx}{x}=\underset{x\to 0}{\mathrm{lim}}3=3\phantom{\rule{0ex}{0ex}}⇒-k=3\phantom{\rule{0ex}{0ex}}⇒k=-3$

#### Question 12:

Find the value of k for which the function $f\left(x\right)=\left\{\begin{array}{l}\begin{array}{cc}\frac{{x}^{2}+3x-10}{x-2},& x\ne 2\\ k,& x=2\end{array}\end{array}\right\$is continuous at x = 2

Given,
$f\left(x\right)=\left\{\begin{array}{l}\begin{array}{cc}\frac{{x}^{2}+3x-10}{x-2},& x\ne 2\\ k,& x=2\end{array}\end{array}\right\$
$\underset{x\to {2}^{-}}{\mathrm{lim}}\left(\frac{{x}^{2}+3x-10}{x-2}\right)=\underset{x\to {2}^{-}}{\mathrm{lim}}\left(x+5\right)=7\phantom{\rule{0ex}{0ex}}f\left(2\right)=k\phantom{\rule{0ex}{0ex}}\underset{x\to {2}^{+}}{\mathrm{lim}}\left(\frac{{x}^{2}+3x-10}{x-2}\right)=\underset{x\to {2}^{+}}{\mathrm{lim}}\left(x+5\right)=7$
If $f\left(x\right)$ is continuous at $x=2$, then
$\underset{x\to {2}^{-}}{\mathrm{lim}}f\left(x\right)=f\left(2\right)=\underset{x\to {2}^{+}}{\mathrm{lim}}f\left(x\right)$
$⇒k=7$

#### Question 1:

The function $f\left(x\right)=\frac{4-{x}^{2}}{4x-{x}^{3}}$
(a) discontinuous at only one point
(b) discontinuous exactly at two points
(c) discontinuous exactly at three points
(d) none of these

(C) discontinuous exactly at three points

Given: $f\left(x\right)=\frac{4-{x}^{2}}{4x-{x}^{3}}$
$⇒f\left(x\right)=\frac{4-{x}^{2}}{x\left(4-{x}^{2}\right)}$

Clearly, $f\left(x\right)$ is defined and continuous for all real numbers except .

Therefore, $f\left(x\right)$ is discontinuous exactly at three points.

#### Question 2:

If f (x) = | xa | ϕ (x), where ϕ (x) is continuous function, then
(a) f' (a+) = ϕ (a)
(b) f' (a) = −ϕ (a)
(c) f' (a+) = f' (a)
(d) none of these

(a)  $f\text{'}\left({a}^{+}\right)=\varphi \left(a\right)$
(b)  $f\text{'}\left({a}^{-}\right)=-\varphi \left(a\right)$

Here,

Also,

#### Question 3:

If , then at x = 1
(a) f (x) is continuous and f' (1+) = log10 e
(b) f (x) is continuous and f' (1+) = log10 e
(c) f (x) is continuous and f' (1) = log10 e
(d) f (x) is continuous and f' (1) = −log10 e

(a) f (x) is continuous and $f\text{'}$ (1+) = ${\mathrm{log}}_{10}e$
(d) f (x) is continuous and $f\text{'}$ (1) = ${\mathrm{log}}_{10}e$

Given:

Also,

#### Question 4:

If is continuous at x = 0, then k equals
(a) $16\sqrt{2}$ log 2 log 3
(b) $16\sqrt{2}$ ln 6
(c) $16\sqrt{2}$ ln 2 ln 3
(d) none of these

Given:
If $f\left(x\right)$ is continuous at $x=0$, then
$\underset{x\to 0}{\mathrm{lim}}f\left(x\right)=f\left(0\right)$

#### Question 5:

If f (x) defined by then f (x) is continuous for all
(a) x
(b) x except at x = 0
(c) x except at x = 1
(d) x except at x = 0 and x = 1.

(d) x except at x = 0 and x = 1.

Given:

So,

$\underset{x\to {0}^{-}}{\mathrm{lim}}f\left(x\right)=\underset{h\to 0}{\mathrm{lim}}f\left(-h\right)=1$

Also,

$\underset{x\to {0}^{+}}{\mathrm{lim}}f\left(x\right)=\underset{h\to 0}{\mathrm{lim}}f\left(h\right)=-1$

$⇒\underset{x\to {0}^{+}}{\mathrm{lim}}f\left(x\right)\ne \underset{x\to {0}^{-}}{\mathrm{lim}}f\left(x\right)$

Thus, $f\left(x\right)$ is discontinuous at $x=0$.

Now,

$\underset{x\to {1}^{-}}{\mathrm{lim}}f\left(x\right)=\underset{h\to 0}{\mathrm{lim}}f\left(1-h\right)=-1$

$\underset{x\to {1}^{+}}{\mathrm{lim}}f\left(x\right)=\underset{h\to 0}{\mathrm{lim}}f\left(1+h\right)=1$

$⇒\underset{x\to {1}^{+}}{\mathrm{lim}}f\left(x\right)\ne \underset{x\to {1}^{-}}{\mathrm{lim}}f\left(x\right)$

So, $f\left(x\right)$ is discontinuous at $x=1$.

Hence, $f\left(x\right)$ is continuous for all except at $x=0$ and x = 1.

#### Question 6:

If
is continuous at x = π/2, then k =
(a) $-\frac{1}{16}$

(b) $-\frac{1}{32}$

(c) $-\frac{1}{64}$

(d) $-\frac{1}{28}$

If $f\left(x\right)$ is continuous at $x=\frac{\mathrm{\pi }}{2}$, then

#### Question 7:

If f (x) = (x + 1)cot x be continuous at x = 0, then f (0) is equal to
(a) 0
(b) 1/e
(c) e
(d) none of these

(c) e

Suppose $f\left(x\right)$ is continuous at $x=0.$

Given: $f\left(x\right)={\left(x+1\right)}^{\mathrm{cot}x}$

#### Question 8:

If
and f (x) is continuous at x = 0, then the value of k is
(a) ab
(b) a + b
(c) log a + log b
(d) none of these

Given:

If f(x) is continuous at x = 0, then
$\underset{x\to 0}{\mathrm{lim}}f\left(x\right)=f\left(0\right)$

$⇒\underset{x\to 0}{\mathrm{lim}}\left(\frac{\mathrm{log}\left(1+ax\right)-\mathrm{log}\left(1-bx\right)}{x}\right)=k$

#### Question 9:

The function
(a) is continuous at x = 0
(b) is not continuous at x = 0
(c) is not continuous at x = 0, but can be made continuous at x = 0
(d) none of these

(b) is not continuous at x = 0

Given:

We have
$\underset{x\to 0}{\mathrm{lim}}f\left(x\right)=\underset{x\to 0}{\mathrm{lim}}\left(\frac{{e}^{\frac{1}{x}}-1}{{e}^{\frac{1}{x}}+1}\right)$

If ${e}^{\frac{1}{x}}=t$, then

$\underset{x\to 0}{\mathrm{lim}}f\left(x\right)=\underset{t\to \infty }{\mathrm{lim}}\left(\frac{t-1}{t+1}\right)=\underset{t\to \infty }{\mathrm{lim}}\left(\frac{1-\frac{1}{t}}{1+\frac{1}{t}}\right)=\frac{1-0}{1+0}=1$
Also, $f\left(0\right)=0$

Hence, $f\left(x\right)$ is discontinuous at $x=0$.

#### Question 10:

Let
Then, f (x) is continuous at x = 4 when
(a) a = 0, b = 0
(b) a = 1, b = 1
(c) a = −1, b = 1
(d) a = 1, b = −1.

(d) a = 1, b = −1.

Given:

We have
(LHL at x = 4) = $\underset{x\to {4}^{-}}{\mathrm{lim}}f\left(x\right)=\underset{h\to 0}{\mathrm{lim}}f\left(4-h\right)$

$=\underset{h\to 0}{\mathrm{lim}}\left(\frac{4-h-4}{\left|4-h-4\right|}+a\right)=\underset{h\to 0}{\mathrm{lim}}\left(\frac{-h}{\left|-h\right|}+a\right)=a-1$

(RHL at x = 4) = $\underset{x\to {4}^{+}}{\mathrm{lim}}f\left(x\right)=\underset{h\to 0}{\mathrm{lim}}f\left(4+h\right)$

$=\underset{h\to 0}{\mathrm{lim}}\left(\frac{4+h-4}{\left|4+h-4\right|}+b\right)=\underset{h\to 0}{\mathrm{lim}}\left(\frac{h}{\left|h\right|}+b\right)=b+1$

Also,
$f\left(4\right)=a+b$

If f(x) is continuous at x = 4, then

$\underset{x\to {4}^{-}}{\mathrm{lim}}f\left(x\right)=\underset{x\to {4}^{+}}{\mathrm{lim}}f\left(x\right)=f\left(4\right)$

$⇒a-1=b+1=a+b$

#### Question 11:

If the function is continuous at x = 0, then the value of k is
(a) 0
(b) 1
(c) −1
(d) e.

(b)
Given:

If$f\left(x\right)$ is continuous at $x=0$, then
$\underset{x\to 0}{\mathrm{lim}}f\left(x\right)=f\left(0\right)$

#### Question 12:

Let f (x) = | x | + | x − 1|, then
(a) f (x) is continuous at x = 0, as well as at x = 1
(b) f (x) is continuous at x = 0, but not at x = 1
(c) f (x) is continuous at x = 1, but not at x = 0
(d) none of these

(a) f (x) is continuous at x = 0, as well as at x = 1

Since modulus function is everywhere continuous ,  are also everywhere continuous.

Also,
It is known that if f and g are continuous functions, then g will also be continuous.

Thus, ​ ​$\left|x\right|+\left|x-1\right|$ is everywhere continuous.

Hence, $f\left(x\right)$ is continuous at .

#### Question 13:

Let . Then, f (x) is continuous on the set
(a) R
(b) R − {1}
(c) R − {2}
(d) R − {1, 2}

(d) R − {1, 2}

Given:

So,

$\underset{x\to {1}^{-}}{\mathrm{lim}}f\left(x\right)=\underset{h\to 0}{\mathrm{lim}}f\left(1-h\right)=\underset{h\to 0}{\mathrm{lim}}\left(1-h+1\right)\left(1-h+2\right)=2×3=6$

$\underset{x\to {1}^{+}}{\mathrm{lim}}f\left(x\right)=\underset{h\to 0}{\mathrm{lim}}f\left(1+h\right)=-\underset{h\to 0}{\mathrm{lim}}\left(1+h+1\right)\left(1+h+2\right)=-2×3=-6$

Also,

$\underset{x\to {2}^{-}}{\mathrm{lim}}f\left(x\right)=\underset{h\to 0}{\mathrm{lim}}f\left(2-h\right)=-\underset{h\to 0}{\mathrm{lim}}\left(2-h+1\right)\left(2-h+2\right)=-12$

$\underset{x\to {2}^{+}}{\mathrm{lim}}f\left(x\right)=\underset{h\to 0}{\mathrm{lim}}f\left(2+h\right)=\underset{h\to 0}{\mathrm{lim}}\left(2+h+1\right)\left(2+h+2\right)=12$

Thus,

Therefore, the only points of discontinuities of the function $f\left(x\right)$ are  .

Hence, the given function is continuous on the set  R − {1, 2}
.
.

#### Question 14:

If is continuous at x = 0, then
(a) a = $-\frac{3}{2}$, b = 0, c = $\frac{1}{2}$
(b) a = $-\frac{3}{2}$, b = 1, c = $-\frac{1}{2}$
(c) a = $-\frac{3}{2}$, bR − {0}, c = $\frac{1}{2}$
(d) none of these

(c) a = $\frac{-3}{2}$, bR − {0}, c$\frac{1}{2}$

The given function can be rewritten as

We have
(LHL at x = 0) =

(RHL at x = 0) = $\underset{x\to {0}^{+}}{\mathrm{lim}}f\left(x\right)=\underset{h\to 0}{\mathrm{lim}}f\left(0+h\right)=\underset{h\to 0}{\mathrm{lim}}f\left(h\right)$

=$\underset{h\to 0}{\mathrm{lim}}\left(\frac{\sqrt{1+bh}-1}{bh}\right)=\underset{h\to 0}{\mathrm{lim}}\left(\frac{bh}{bh\left(\sqrt{1+bh}+1\right)}\right)=\underset{h\to 0}{\mathrm{lim}}\left(\frac{1}{\left(\sqrt{1+bh}+1\right)}\right)=\frac{1}{2}$

Also, $f\left(0\right)=c$

If $f\left(x\right)$ is continuous at x = 0, then
​$\underset{x\to {0}^{-}}{\mathrm{lim}}f\left(x\right)=\underset{x\to {0}^{+}}{\mathrm{lim}}f\left(x\right)=f\left(0\right)$

$⇒a=\frac{-3}{2}$

Now, $\frac{\sqrt{1+bx}-1}{bx}$ exists only if $bx\ne 0⇒b\ne 0$.

Thus, $b\in R-\left\{0\right\}$.

#### Question 15:

If is continuous at $x=\frac{\mathrm{\pi }}{2}$, then
(a) m = 1, n = 0

(b) $m=\frac{n\mathrm{\pi }}{2}+1$

(c) $n=\frac{m\mathrm{\pi }}{2}$

(d) $m=n=\frac{\mathrm{\pi }}{2}$

(c) $n=\frac{\mathrm{m}\mathrm{\pi }}{2}$n=mπ2

Here,

$f\left(\frac{\mathrm{\pi }}{2}\right)=\frac{\mathrm{m\pi }}{2}+1$

We have
(LHL at $x=\frac{\mathrm{\pi }}{2}$) =

(RHL at $x=\frac{\mathrm{\pi }}{2}$) =

Thus,
If $f\left(x\right)$ is continuous at $x=\frac{\mathrm{\pi }}{2}$, then

$⇒\frac{\mathrm{m\pi }}{2}+1=n+1\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}⇒\frac{\mathrm{m\pi }}{2}=n$

#### Question 16:

The value of f (0), so that the function
$f\left(x\right)=\frac{\sqrt{{a}^{2}-ax+{x}^{2}}-\sqrt{{a}^{2}+ax+{x}^{2}}}{\sqrt{a+x}-\sqrt{a-x}}$ becomes continuous for all x, given by
(a) a3/2
(b) a1/2
(c) −a1/2
(d) −a3/2

(c) $-{a}^{\frac{1}{2}}$

Given: $f\left(x\right)=\frac{\sqrt{{a}^{2}-ax+{x}^{2}}-\sqrt{{a}^{2}+ax+{x}^{2}}}{\sqrt{a+x}-\sqrt{a-x}}$

If $f\left(x\right)$ is continuous for all x, then it will be continuous at x = 0 as well.

So, if $f\left(x\right)$ is continuous at x = 0, then
$\underset{x\to 0}{\mathrm{lim}}f\left(x\right)=f\left(0\right)$

#### Question 17:

The function

(a) is discontinuous at finitely many points
(b) is continuous everywhere
(c) is discontinuous only at $x=±\frac{1}{n}$, nZ − {0} and x = 0
(d) none of these

Given:

Case 1:

Here,
$f\left(x\right)=1$, which is the constant function
So, $f\left(x\right)$ is continuous for all

Case 2:

Here,
, which is also a constant function.

So, $f\left(x\right)$ is continuous for all

Case 3: Consider the points x = -1 and x = 1.

We have

Similarly,  f(x) is discontinuous at x = 1.

Case 4: Consider the point x = 0.

We have
$\underset{x\to {0}^{-}}{\mathrm{lim}}f\left(x\right)=\underset{h\to 0}{\mathrm{lim}}f\left(\frac{1}{n}-h\right)=\underset{h\to 0}{\mathrm{lim}}f\left(\frac{1}{n}-h\right)={\left(\frac{1}{n-1}\right)}^{2}$

$\underset{x\to {0}^{+}}{\mathrm{lim}}f\left(x\right)=\underset{h\to 0}{\mathrm{lim}}f\left(\frac{1}{n}+h\right)=\underset{h\to 0}{\mathrm{lim}}f\left(\frac{1}{n}+h\right)={\left(\frac{1}{n}\right)}^{2}$

$\underset{x\to {0}^{+}}{\mathrm{lim}}f\left(x\right)\ne \underset{x\to {0}^{-}}{\mathrm{lim}}f\left(x\right)$

Thus, $f\left(x\right)$ is discontinuous at $x=0$.

At = 0, we have
$\underset{x\to {0}^{-}}{\mathrm{lim}}f\left(x\right)\ne 0=f\left(0\right)$

So, $f\left(x\right)$ is discontinuous at $x=0$.

Case 5: Consider the point

We have
$\underset{x\to {\frac{1}{n}}^{-}}{\mathrm{lim}}f\left(x\right)=\underset{h\to 0}{\mathrm{lim}}f\left(\frac{1}{n}-h\right)=\underset{h\to 0}{\mathrm{lim}}f\left(\frac{1}{n}-h\right)={\left(\frac{1}{n-1}\right)}^{2}$

$\underset{x\to {\frac{1}{n}}^{+}}{\mathrm{lim}}f\left(x\right)=\underset{h\to 0}{\mathrm{lim}}f\left(\frac{1}{n}+h\right)=\underset{h\to 0}{\mathrm{lim}}f\left(\frac{1}{n}+h\right)={\left(\frac{1}{n}\right)}^{2}$

$\underset{x\to {\frac{1}{\mathrm{n}}}^{+}}{\mathrm{lim}}f\left(x\right)\ne \underset{x\to {\frac{1}{\mathrm{n}}}^{-}}{\mathrm{lim}}f\left(x\right)$

Hence, $f\left(x\right)$ is discontinuous only at  $x=±\frac{1}{n}$.

#### Question 18:

The value of f (0), so that the function
is continuous, is given by
(a) $\frac{2}{3}$
(b) 6
(c) 2
(d) 4

(c) 2

For f(x) to be continuous at x = 0, we must have

#### Question 19:

The value of f (0) so that the function
$f\left(x\right)=\frac{2-{\left(256-7x\right)}^{1/8}}{{\left(5x+32\right)}^{1/5}-2},$ x ≠ 0 is continuous everywhere, is given by
(a) −1
(b) 1
(c) 26
(d) none of these

(d) none of these

Given:  $f\left(x\right)=\frac{2-{\left(256-7x\right)}^{\frac{1}{8}}}{{\left(5x+32\right)}^{\frac{1}{5}}-2}$

For $f\left(x\right)$ to be continuous at x = 0, we must have
$\underset{\mathrm{x}\to 0}{\mathrm{lim}}f\left(x\right)=f\left(0\right)$
$⇒f\left(0\right)=\underset{x\to 0}{\mathrm{lim}}f\left(x\right)=\underset{x\to 0}{\mathrm{lim}}\frac{2-{\left(256-7x\right)}^{\frac{1}{8}}}{{\left(5x+32\right)}^{\frac{1}{5}}-2}\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}⇒f\left(0\right)=\underset{x\to 0}{\mathrm{lim}}\frac{{256}^{\frac{1}{8}}-{\left(256-7x\right)}^{\frac{1}{8}}}{{\left(5x+32\right)}^{\frac{1}{5}}-{32}^{\frac{1}{5}}}\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}=-\underset{x\to 0}{\mathrm{lim}}\frac{\frac{\left[{\left(256-7x\right)}^{\frac{1}{8}}-{256}^{\frac{1}{8}}\right]}{x}}{\frac{\left[{\left(5x+32\right)}^{\frac{1}{5}}-{32}^{\frac{1}{5}}\right]}{x}}\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}=\frac{-7}{5}\underset{x\to 0}{\mathrm{lim}}\frac{\frac{\left[{\left(256-7x\right)}^{\frac{1}{8}}-{256}^{\frac{1}{8}}\right]}{7x}}{\frac{\left[{\left(5x+32\right)}^{\frac{1}{5}}-{32}^{\frac{1}{5}}\right]}{5x}}\phantom{\rule{0ex}{0ex}}=\frac{7}{5}\underset{x\to 0}{\mathrm{lim}}\frac{\frac{\left[{\left(256-7x\right)}^{\frac{1}{8}}-{256}^{\frac{1}{8}}\right]}{\left(256-7x\right)-256}}{\frac{\left[{\left(5x+32\right)}^{\frac{1}{5}}-{32}^{\frac{1}{5}}\right]}{5x+32-32}}\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}=\frac{7}{5}×\frac{\frac{1}{8}×{\left(256\right)}^{-\frac{7}{8}}}{\frac{1}{5}×{\left(32\right)}^{\frac{-4}{5}}}\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}=\frac{7}{5}×\frac{\frac{1}{8}×{2}^{4}}{\frac{1}{5}×{2}^{7}}\phantom{\rule{0ex}{0ex}}=\frac{7}{64}\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}$

#### Question 20:

is continuous in the interval [−1, 1], then p is equal to
(a) −1
(b) −1/2
(c) 1/2
(d) 1

(b) $-\frac{1}{2}$
Given:

If $f\left(x\right)$ is continuous at x = 0,  then
$\underset{x\to {0}^{-}}{\mathrm{lim}}f\left(x\right)=\underset{x\to {0}^{+}}{\mathrm{lim}}f\left(x\right)$

#### Question 21:

The function
is continuous for 0 ≤ x < ∞, then the most suitable values of a and b are
(a) a = 1, b = −1
(b) a = −1, b = 1 + $\sqrt{2}$
(c) a = −1, b = 1
(d) none of these

(c) a = -1, b = 1

Given: $f\left(x\right)$ is continuous for 0 ≤ x < ∞.

This means that $f\left(x\right)$ is continuous for
.

Now,

If $f\left(x\right)$ is continuous at x = 1, then
$\underset{x\to {1}^{-}}{\mathrm{lim}}f\left(x\right)=f\left(1\right)\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}⇒\underset{h\to 0}{\mathrm{lim}}f\left(1-h\right)=a\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}⇒\frac{{\left(1-h\right)}^{2}}{a}=a\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}⇒\frac{1}{a}=a\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}⇒{a}^{2}=1\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}⇒a=±1$

If $f\left(x\right)$ is continuous at x = $\sqrt{2}$, then​

$\underset{x\to {\sqrt{2}}^{-}}{\mathrm{lim}}f\left(x\right)=f\left(\sqrt{2}\right)\phantom{\rule{0ex}{0ex}}⇒\underset{h\to 0}{\mathrm{lim}}f\left(\sqrt{2}-h\right)=\frac{2{b}^{2}-4b}{2}\phantom{\rule{0ex}{0ex}}⇒\underset{h\to 0}{\mathrm{lim}}a={b}^{2}-2b\phantom{\rule{0ex}{0ex}}⇒a={b}^{2}-2b\phantom{\rule{0ex}{0ex}}⇒{b}^{2}-2b-a=0$

∴ For a = 1, we have

${b}^{2}-2b-1=0\phantom{\rule{0ex}{0ex}}⇒b=\frac{2±\sqrt{4-4\left(-1\right)}}{2}=1±\sqrt{2}$

Also,
For a = −1, we have

${b}^{2}-2b+1=0\phantom{\rule{0ex}{0ex}}⇒{\left(b-1\right)}^{2}=0\phantom{\rule{0ex}{0ex}}⇒b=1$

Thus,

#### Question 22:

If when x ≠ π/2 and f (π/2) = λ, then f (x) will be continuous function at x = π/2, where λ =
(a) 1/8
(b) 1/4
(c) 1/2
(d) none of these

(a) $\frac{1}{8}$

If $f\left(x\right)$ is continuous at $x=\frac{\mathrm{\pi }}{2}$, then
$\underset{\mathrm{x}\to \frac{\mathrm{\pi }}{2}}{\mathrm{lim}}f\left(x\right)=f\left(\frac{\mathrm{\pi }}{2}\right)$

...(1)

Suppose $\left(\frac{\mathrm{\pi }}{2}-x\right)=t$, then

#### Question 23:

The value of a for which the function
may be continuous at x = 0 is
(a) 1
(b) 2
(c) 3
(d) none of these

(d) none of these

For f(x) to be continuous at $x=0$, we must have
$\underset{\mathrm{x}\to 0}{\mathrm{lim}}f\left(x\right)=f\left(0\right)$

#### Question 24:

The function f (x) = tan x is discontinuous on the set
(a) {n π : nZ}
(b) {2n π : nZ}
(c)
(d)

When, it is not defined at the integral points. $\left[n\in Z\right]$

Hence, $f\left(x\right)$ is discontinuous on the set .

#### Question 25:

The function is continuous at x = 0, then k =
(a) 3
(b) 6
(c) 9
(d) 12

(b) 6

Given:

If $f\left(x\right)$ is continuous at $x=0$, then
$\underset{x\to 0}{\mathrm{lim}}f\left(x\right)=f\left(0\right)$

#### Question 26:

If the function

is continuous at each point of its domain, then the value of f (0) is
(a) 2

(b) $\frac{1}{3}$

(c) $-\frac{1}{3}$

(d) $\frac{2}{3}$

(b) $\frac{1}{3}$

Given:  $f\left(x\right)=\frac{2x-{\mathrm{sin}}^{-1}x}{2x+{\mathrm{tan}}^{-1}x}$

If f(x) is continuous at x = 0, then

$\underset{x\to 0}{\mathrm{lim}}f\left(x\right)=f\left(0\right)\phantom{\rule{0ex}{0ex}}⇒\underset{x\to 0}{\mathrm{lim}}\frac{2x-{\mathrm{sin}}^{-1}x}{2x+{\mathrm{tan}}^{-1}x}=f\left(0\right)\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}⇒\underset{x\to 0}{\mathrm{lim}}\frac{x\left(2-\frac{{\mathrm{sin}}^{-1}x}{x}\right)}{x\left(2+\frac{{\mathrm{tan}}^{-1}x}{x}\right)}=f\left(0\right)\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}⇒\underset{x\to 0}{\mathrm{lim}}\frac{\left(2-\frac{{\mathrm{sin}}^{-1}x}{x}\right)}{\left(2+\frac{{\mathrm{tan}}^{-1}x}{x}\right)}=f\left(0\right)\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}⇒\frac{2-\underset{x\to 0}{\mathrm{lim}}\left(\frac{{\mathrm{sin}}^{-1}x}{x}\right)}{2+\underset{x\to 0}{\mathrm{lim}}\left(\frac{{\mathrm{tan}}^{-1}x}{x}\right)}=f\left(0\right)\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}⇒\frac{2-1}{2+1}=f\left(0\right)\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}⇒f\left(0\right)=\frac{1}{3}$

#### Question 27:

The value of b for which the function
is continuous at every point of its domain, is
(a) −1

(b) 0

(c) $\frac{13}{3}$

(d) 1

(a) −1

Given: $f\left(x\right)$ is continuous at every point of its domain. So, it is continuous at $x=1$.

$⇒\underset{x\to {1}^{+}}{\mathrm{lim}}f\left(x\right)=f\left(1\right)\phantom{\rule{0ex}{0ex}}⇒\underset{h\to 0}{\mathrm{lim}}f\left(1+h\right)=f\left(1\right)\phantom{\rule{0ex}{0ex}}⇒\underset{h\to 0}{\mathrm{lim}}\left(4{\left(1+h\right)}^{2}+3b\left(1+h\right)\right)=5\left(1\right)-4\phantom{\rule{0ex}{0ex}}⇒4+3b=1\phantom{\rule{0ex}{0ex}}⇒-3=3b\phantom{\rule{0ex}{0ex}}⇒b=-1$

#### Question 28:

If $f\left(x\right)=\frac{1}{1-x}$, then the set of points discontinuity of the function f (f(f(x))) is
(a) {1}
(b) {0, 1}
(c) {−1, 1}
(d) none of these

(b) {0, 1}

Given: $f\left(x\right)=\frac{1}{1-x}$

Clearly, $f:R-\left\{1\right\}\to R$

Now,
$f\left(f\left(x\right)\right)=f\left(\frac{1}{1-x}\right)=\left(\frac{1}{1-\left(\frac{1}{1-x}\right)}\right)=\left(\frac{1-x}{-x}\right)=\left(\frac{x-1}{x}\right)$
$fof:$

Now,

$f\left(f\left(f\left(x\right)\right)\right)=f\left(\frac{x-1}{x}\right)=\left(\frac{1}{1-\left(\frac{x-1}{x}\right)}\right)=x$

$fofof:$

Thus, ​$f\left(f\left(f\left(x\right)\right)\right)$ is not defined at .

Hence, ​$f\left(f\left(f\left(x\right)\right)\right)$ is discontinuous at {0, 1}.

#### Question 29:

Let The value which should be assigned to f (x) at $x=\frac{\mathrm{\pi }}{4},$ so that it is continuous everywhere is
(a) 1
(b) 1/2
(c) 2
(d) none of these

(b) $\frac{1}{2}$

If $f\left(x\right)$ is continuous at $x=\frac{\mathrm{\pi }}{4}$, then
$\underset{\mathrm{x}\to \frac{\mathrm{\pi }}{4}}{\mathrm{lim}}f\left(x\right)=f\left(\frac{\mathrm{\pi }}{4}\right)$

If $\frac{\mathrm{\pi }}{4}-x=y$, then .

#### Question 30:

The function $f\left(x\right)=\frac{{x}^{3}+{x}^{2}-16x+20}{x-2}$ is not defined for x = 2. In order to make f (x) continuous at x = 2, Here f (2) should be defined as
(a) 0
(b) 1
(c) 2
(d) 3

Here,

So, the given function can be rewritten as

$f\left(x\right)=\frac{{\left(x-2\right)}^{2}\left(x+5\right)}{x-2}$

$⇒f\left(x\right)=\left(x-2\right)\left(x+5\right)$

If $f\left(x\right)$ is continuous at $x=2$, then
$\underset{x\to 2}{\mathrm{lim}}f\left(x\right)=f\left(2\right)$

$⇒\underset{x\to 2}{\mathrm{lim}}\left(x-2\right)\left(x+5\right)=f\left(2\right)\phantom{\rule{0ex}{0ex}}⇒f\left(2\right)=0$

Hence, in order to make $f\left(x\right)$ continuous at  should be defined as 0.

#### Question 31:

If is continuous at x = 0, then a equals
(a) $\frac{1}{2}$

(b) $\frac{1}{3}$

(c) $\frac{1}{4}$

(d) $\frac{1}{6}$

(a) $\frac{1}{2}$

Given:
We have

(LHL at x = 0) =

(RHL at x = 0) =

#### Question 32:

If
then the value of (a, b) for which f (x) cannot be continuous at x = 1, is
(a) (2, 2)
(b) (3, 1)
(c) (4, 0)
(d) (5, 2)

(d) (5, 2)

If f(x) is continuous at x = 1, then
$\underset{\mathrm{x}\to {1}^{-}}{\mathrm{lim}}f\left(x\right)=f\left(1\right)$

Thus, the possible values of (a, b) can be . But .

Hence, for $f\left(x\right)$ cannot be continuous at = 1.

Disclaimer: The question in the book has some error. The solution here is created according to the question given in the book.

#### Question 33:

If the function f (x) defined by
is continuous at x = 0, then k =
(a) 1
(b) 5
(c) −1
(d) none of these

Given:

If f(x) is continuous at x = 0, then​$\underset{x\to 0}{\mathrm{lim}}f\left(x\right)=f\left(0\right)$.

$⇒\underset{x\to 0}{\mathrm{lim}}\left(\frac{\mathrm{log}\left(1+3x\right)-\mathrm{log}\left(1-2x\right)}{x}\right)=k$

#### Question 34:

If
then the value of a so that f (x) may be continuous at x = 0, is
(a) 25
(b) 50
(c) −25
(d) none of these

(b) 50

If $f\left(x\right)$ is continuous at $x=0$, then
$\underset{\mathrm{x}\to {0}^{-}}{\mathrm{lim}}f\left(x\right)=f\left(0\right)$

$⇒\underset{h\to 0}{\mathrm{lim}}f\left(-h\right)=f\left(0\right)$

#### Question 35:

If then the value of the function at x = 0, so that the function is continuous at x = 0, is
(a) 0
(b) −1
(c) 1
(d) indeterminate

(a) 0

Given:

Here,

If f(x) is continuous at x = 0, then$\underset{x\to 0}{\mathrm{lim}}f\left(x\right)=f\left(0\right)$.

$⇒f\left(0\right)=0$

#### Question 36:

The value of k which makes
continuous at x = 0, is
(a) 8
(b) 1
(c) −1
(d) none of these

(d) none of these

If $f\left(x\right)$ is continuous at $x=0$, then

#### Question 37:

The values of the constants a, b and c for which the function
may be continuous at x = 0, are

(a)

(b)

(c)

(d) none of these

If $f\left(x\right)$ is continuous at $x=0$, then

Also,

#### Question 38:

The points of discontinuity of the function

(a) x = 1, $x=\frac{5}{2}$

(b) $x=\frac{5}{2}$

(c) $x=1,\frac{5}{2},4$

(d) x = 0, 4

=5

If $0\le x\le 1$, then  $f\left(x\right)=2\sqrt{x}$.

Since $f\left(x\right)=2\sqrt{x}$ is a polynomial function, it is continuous.
Thus, $f\left(x\right)$ is continuous for every $0\le x\le 1$.

If $1, then  $f\left(x\right)=4-2x$. Since $2x$ is a polynomial function and 4 is a constant function, both of them are continuous. So, their difference will also be continuous.
Thus, $f\left(x\right)$ is continuous for every $1.

If $\frac{5}{2}\le x\le 4$, then  $f\left(x\right)=2x-7$. Since $2x$ is a polynomial function and 7 is continuous function, their difference will also be continuous.
Thus, $f\left(x\right)$ is continuous for every $\frac{5}{2}\le x\le 4$.

Now,
Consider the point $x=1$. Here,

$\underset{\mathrm{x}\to {1}^{-}}{\mathrm{lim}}f\left(x\right)=\underset{h\to 0}{\mathrm{lim}}f\left(1-h\right)=\underset{h\to 0}{\mathrm{lim}}\left(2\left(\sqrt{1-h}\right)\right)=2$

$\underset{\mathrm{x}\to {1}^{+}}{\mathrm{lim}}f\left(x\right)=\underset{h\to 0}{\mathrm{lim}}f\left(1+h\right)=\underset{h\to 0}{\mathrm{lim}}\left(4-2\left(1+h\right)\right)=2$

Also, $f\left(1\right)=2\sqrt{1}=2$

$⇒\underset{\mathrm{x}\to {1}^{-}}{\mathrm{lim}}f\left(x\right)=\underset{\mathrm{x}\to {1}^{+}}{\mathrm{lim}}f\left(x\right)=f\left(1\right)$

Thus, .

Now,
Consider the point $x=\frac{5}{2}$. Here,

$\underset{\mathrm{x}\to {\frac{5}{2}}^{-}}{\mathrm{lim}}f\left(x\right)=\underset{h\to 0}{\mathrm{lim}}f\left(\frac{5}{2}-h\right)=\underset{h\to 0}{\mathrm{lim}}\left(4-2\left(\frac{5}{2}-h\right)\right)=-1$

$\underset{\mathrm{x}\to {\frac{5}{2}}^{+}}{\mathrm{lim}}f\left(x\right)=\underset{h\to 0}{\mathrm{lim}}f\left(\frac{5}{2}+h\right)=\underset{h\to 0}{\mathrm{lim}}\left(2\left(\frac{5}{2}-h\right)-7\right)=-2$

Thus, .

#### Question 39:

If . Then, f (x) is continuous at $x=\frac{\mathrm{\pi }}{2}$, if
(a) $a=\frac{1}{3},$ b = 2

(b)

(c)

(d) none of these

(b)

Given:

We have
(LHL at x = $\frac{\mathrm{\pi }}{2}$) = $\underset{x\to {\frac{\mathrm{\pi }}{2}}^{-}}{\mathrm{lim}}f\left(x\right)=\underset{h\to 0}{\mathrm{lim}}f\left(\frac{\mathrm{\pi }}{2}-h\right)$

(RHL at x = $\frac{\mathrm{\pi }}{2}$) = $\underset{x\to {\frac{\mathrm{\pi }}{2}}^{+}}{\mathrm{lim}}f\left(x\right)=\underset{h\to 0}{\mathrm{lim}}f\left(\frac{\mathrm{\pi }}{2}+h\right)$

Also, $f\left(\frac{\mathrm{\pi }}{2}\right)=a$

If f(x) is continuous at x = $\frac{\mathrm{\pi }}{2}$, then

​

#### Question 40:

The points of discontinuity of the function

(a) x = 1
(b) x = 3
(c) x = 1, 3
(d) none of these

(b) x = 3

If $x\le 1$, then $f\left(x\right)=\frac{1}{5}\left(2{x}^{2}+3\right)$.
Since $2{x}^{2}+3$ is a polynomial function and $\frac{1}{5}$ is a constant function, both of them are continuous. So, their product will also be continuous.
Thus, $f\left(x\right)$ is continuous at $x\le 1$.

If $1, then $f\left(x\right)=6-5x$.

Since $5x$ is a polynomial function and $6$ is a constant function, both of them are continuous. So, their difference will also be continuous.
Thus, $f\left(x\right)$ is continuous for every $1.

If $x\ge 3$, then $f\left(x\right)=x-3$.
Since $x-3$ is a polynomial function, it is continuous. So, $f\left(x\right)$ is continuous for every $x\ge 3$.

Now,
Consider the point $x=1$. Here,

$\underset{\mathrm{x}\to {1}^{-}}{\mathrm{lim}}f\left(x\right)=\underset{h\to 0}{\mathrm{lim}}f\left(1-h\right)=\underset{h\to 0}{\mathrm{lim}}\left(\frac{1}{5}\left[2{\left(1-h\right)}^{2}+3\right]\right)=1$

$\underset{\mathrm{x}\to {1}^{+}}{\mathrm{lim}}f\left(x\right)=\underset{h\to 0}{\mathrm{lim}}f\left(1+h\right)=\underset{h\to 0}{\mathrm{lim}}\left(6-5\left(1+h\right)\right)=1$

Also,
$f\left(1\right)=\frac{1}{5}\left(2{\left(1\right)}^{2}+3\right)=1$

Thus,
$\underset{\mathrm{x}\to {1}^{-}}{\mathrm{lim}}f\left(x\right)=\underset{\mathrm{x}\to {1}^{+}}{\mathrm{lim}}f\left(x\right)=f\left(1\right)$

Hence, $f\left(x\right)$ is continuous at $x=1$.

Now,
Consider the point $x=3$. Here,

$\underset{\mathrm{x}\to {3}^{-}}{\mathrm{lim}}f\left(x\right)=\underset{h\to 0}{\mathrm{lim}}f\left(3-h\right)=\underset{h\to 0}{\mathrm{lim}}\left(6-5\left(3-h\right)\right)=-9$

$\underset{\mathrm{x}\to {3}^{+}}{\mathrm{lim}}f\left(x\right)=\underset{h\to 0}{\mathrm{lim}}f\left(3+h\right)=\underset{h\to 0}{\mathrm{lim}}\left(\left(3+h\right)-3\right)=0$

Also,
$f\left(1\right)=\frac{1}{5}\left(2{\left(1\right)}^{2}+3\right)=1$

Thus,
$\underset{\mathrm{x}\to {3}^{-}}{\mathrm{lim}}f\left(x\right)\ne \underset{\mathrm{x}\to {3}^{+}}{\mathrm{lim}}f\left(x\right)$

Hence, $f\left(x\right)$ is discontinuous at $x=3$.

So, the only point of discontinuity of $f\left(x\right)$ is $x=3$.

#### Question 41:

The value of a for which the function
is continuous at every point of its domain, is

(a) $\frac{13}{3}$
(b) 1
(c) 0
(d) −1

(d) −1

Given:

If $f\left(x\right)$ is continuous in its domain, then it will be continuous at $x=1$.

Now,
$\underset{x\to {1}^{-}}{\mathrm{lim}}f\left(x\right)=\underset{h\to 0}{\mathrm{lim}}f\left(1-h\right)=\underset{h\to 0}{\mathrm{lim}}\left[5\left(1-h\right)-4\right]=5-4=1\phantom{\rule{0ex}{0ex}}\underset{x\to {1}^{+}}{\mathrm{lim}}f\left(x\right)=\underset{h\to 0}{\mathrm{lim}}f\left(1+h\right)=\underset{h\to 0}{\mathrm{lim}}\left[4{\left(1+h\right)}^{2}+3a\left(1+h\right)\right]=4+3a$

Since f(x) is continuous at x = 1,
$\underset{x\to {1}^{-}}{\mathrm{lim}}f\left(x\right)=\underset{x\to {1}^{+}}{\mathrm{lim}}f\left(x\right)$

$⇒4+3a=1\phantom{\rule{0ex}{0ex}}⇒3a=-3\phantom{\rule{0ex}{0ex}}⇒a=-1$

#### Question 42:

If is continuous at x = π/2, then k is equal to
(a) 0
(b) $\frac{1}{2}$
(c) 1
(d) −1

(a) 0

Given:

If f(x) is continuous at $x=\frac{\mathrm{\pi }}{2}$, then
$\underset{x\mathit{\to }\frac{\mathrm{\pi }}{2}}{\mathrm{lim}}f\left(x\right)=f\left(\frac{\mathrm{\pi }}{2}\right)$

Now,
$\frac{\mathrm{\pi }}{2}-x=y$
$⇒\mathrm{\pi }-2x=2y$

Also,

$⇒\underset{y\mathit{\to }0}{\mathrm{lim}}\frac{\mathrm{sin}\left(\mathrm{cos}\left(\frac{\mathrm{\pi }}{2}-y\right)\right)-\mathrm{cos}\left(\frac{\mathrm{\pi }}{2}-y\right)}{4{y}^{2}}=k$

View NCERT Solutions for all chapters of Class 15