Rd Sharma XII Vol 1 2018 Solutions for Class 12 Science Math Chapter 15 Mean Value Theorems are provided here with simple step-by-step explanations. These solutions for Mean Value Theorems are extremely popular among Class 12 Science students for Math Mean Value Theorems Solutions come handy for quickly completing your homework and preparing for exams. All questions and answers from the Rd Sharma XII Vol 1 2018 Book of Class 12 Science Math Chapter 15 are provided here for you for free. You will also love the ad-free experience on Meritnation’s Rd Sharma XII Vol 1 2018 Solutions. All Rd Sharma XII Vol 1 2018 Solutions for class Class 12 Science Math are prepared by experts and are 100% accurate.

#### Question 1:

Verify Lagrange's mean value theorem for the following functions on the indicated intervals. In each case find a point 'c' in the indicated interval as stated by the Lagrange's mean value theorem
(i) f(x) = x2 − 1 on [2, 3]
(ii) f(x) = x3 − 2x2x + 3 on [0, 1]
(iii) f(x) = x(x −1) on [1, 2]
(iv) f(x) = x2 − 3x + 2 on [−1, 2]
(v) f(x) = 2x2 − 3x + 1 on [1, 3]
(vi) f(x) = x2 − 2x + 4 on [1, 5]
(vii) f(x) = 2xx2 on [0, 1]
(viii) f(x) = (x − 1)(x − 2)(x − 3) on [0, 4]
(ix) $f\left(x\right)=\sqrt{25-{x}^{2}}$ on [−3, 4]
(x) f(x) = tan1 x on [0, 1]
(xi)
(xii) f(x) = x(x + 4)2 on [0, 4]
(xiii)
(xiv) f(x) = x2 + x − 1 on [0, 4]
(xv) f(x) = sin x − sin 2xx on [0, π]
(xvi) f(x) = x3 − 5x2 − 3x on [1, 3]

(i) We have

$f\left(x\right)={x}^{2}-1$

Since a polynomial function is everywhere continuous and differentiable, $f\left(x\right)$ is continuous on  and differentiable on .
Thus, both conditions of Lagrange's mean value theorem are satisfied.
So, there must exist at least one real number ​ such that
$f\text{'}\left(c\right)=\frac{f\left(3\right)-f\left(2\right)}{3-2}$

Now, $f\left(x\right)={x}^{2}-1$
$⇒f\text{'}\left(x\right)=2x$ ,  $f\left(3\right)={\left(3\right)}^{2}-1=8$ ,  $f\left(2\right)={\left(2\right)}^{2}-1=3$

∴  $f\text{'}\left(x\right)=\frac{f\left(3\right)-f\left(2\right)}{3-2}$

$⇒2x=\frac{8-3}{1}\phantom{\rule{0ex}{0ex}}⇒x=\frac{5}{2}$
Thus,  such that $f\text{'}\left(c\right)=\frac{f\left(3\right)-f\left(2\right)}{3-2}$.

Hence, Lagrange's theorem is verified.

(ii) We have,

$f\left(x\right)={x}^{3}-2{x}^{2}-x+3=0$

Since a polynomial function is everywhere continuous and differentiable.

Therefore, $f\left(x\right)$ is continuous on  and differentiable on .
Thus, both conditions of Lagrange's mean value theorem are satisfied.
So, there must exist at least one real number ​ such that
$f\text{'}\left(c\right)=\frac{f\left(1\right)-f\left(0\right)}{1-0}$

Now, $f\left(x\right)={x}^{3}-2{x}^{2}-x+3=0$
$⇒f\text{'}\left(x\right)=3{x}^{2}-4x-1$ ,   ,  $f\left(0\right)=3$

∴  $f\text{'}\left(x\right)=\frac{f\left(1\right)-f\left(0\right)}{1-0}$

Thus,  such that $f\text{'}\left(c\right)=\frac{f\left(1\right)-f\left(0\right)}{1-0}$.

Hence, Lagrange's theorem is verified.

(iii) We have,

$f\left(x\right)=x\left(x-1\right)$ which can be rewritten as $f\left(x\right)={x}^{2}-x$

Since a polynomial function is everywhere continuous and differentiable.

Therefore, $f\left(x\right)$ is continuous on  and differentiable on .
Thus, both conditions of Lagrange's mean value theorem are satisfied.
So, there must exist at least one real number ​ such that
$f\text{'}\left(c\right)=\frac{f\left(2\right)-f\left(1\right)}{2-1}$

Now, $f\left(x\right)={x}^{2}-x$

$⇒f\text{'}\left(x\right)=2x-1$ ,   ,  $f\left(1\right)=0$

∴  $f\text{'}\left(x\right)=\frac{f\left(2\right)-f\left(1\right)}{2-1}$

$⇒2x-1=\frac{2-0}{2-1}\phantom{\rule{0ex}{0ex}}⇒2x-1-2=0\phantom{\rule{0ex}{0ex}}⇒2x=3\phantom{\rule{0ex}{0ex}}⇒x=\frac{3}{2}$
Thus,  such that $f\text{'}\left(c\right)=\frac{f\left(2\right)-f\left(1\right)}{2-1}$.

Hence, Lagrange's theorem is verified.

(iv) We have,

$f\left(x\right)={x}^{2}-3x+2$

Since a polynomial function is everywhere continuous and differentiable.

Therefore, $f\left(x\right)$ is continuous on  and differentiable on .
Thus, both conditions of Lagrange's mean value theorem are satisfied.
So, there must exist at least one real number ​ such that
$f\text{'}\left(c\right)=\frac{f\left(2\right)-f\left(-1\right)}{2+1}=\frac{f\left(2\right)-f\left(-1\right)}{3}$

Now, $f\left(x\right)={x}^{2}-3x+2$
$⇒f\text{'}\left(x\right)=2x-3$ ,  $f\left(2\right)=0$ ,  $f\left(-1\right)={\left(-1\right)}^{2}-3\left(-1\right)+2=6$

∴  $f\text{'}\left(x\right)=\frac{f\left(2\right)-f\left(-1\right)}{3}$

$⇒2x-3=-2\phantom{\rule{0ex}{0ex}}⇒2x-1=0\phantom{\rule{0ex}{0ex}}⇒x=\frac{1}{2}$
Thus,  such that $f\text{'}\left(c\right)=\frac{f\left(2\right)-f\left(-1\right)}{2-\left(-1\right)}$.

Hence, Lagrange's theorem is verified.

(v) We have,

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

Since a polynomial function is everywhere continuous and differentiable.

Therefore, $f\left(x\right)$ is continuous on  and differentiable on .
Thus, both conditions of Lagrange's mean value theorem are satisfied.
So, there must exist at least one real number ​ such that
$f\text{'}\left(c\right)=\frac{f\left(3\right)-f\left(1\right)}{3-1}=\frac{f\left(3\right)-f\left(1\right)}{2}$

Now, $f\left(x\right)=2{x}^{2}-3x+1$
$⇒f\text{'}\left(x\right)=4x-3$ ,  $f\left(3\right)=10$ ,  $f\left(1\right)=2{\left(1\right)}^{2}-3\left(1\right)+1=0$

∴  $f\text{'}\left(x\right)=\frac{f\left(3\right)-f\left(1\right)}{2}$

$⇒4x-3=\frac{10-0}{2}\phantom{\rule{0ex}{0ex}}⇒4x-3-5=0\phantom{\rule{0ex}{0ex}}⇒x=2$
Thus,  such that $f\text{'}\left(c\right)=\frac{f\left(3\right)-f\left(1\right)}{3-1}$.

Hence, Lagrange's theorem is verified.

(vi) We have,

$f\left(x\right)={x}^{2}-2x+4$

Since a polynomial function is everywhere continuous and differentiable.

Therefore, $f\left(x\right)$ is continuous on  and differentiable on .
Thus, both conditions of Lagrange's mean value theorem are satisfied.
So, there must exist at least one real number ​ such that
$f\text{'}\left(c\right)=\frac{f\left(5\right)-f\left(1\right)}{5-1}=\frac{f\left(5\right)-f\left(1\right)}{4}$

Now, $f\left(x\right)={x}^{2}-2x+4$
$⇒f\text{'}\left(x\right)=2x-2$ ,  $f\left(5\right)=25-10+4=19$ ,  $f\left(1\right)=1-2+4=3$

∴  $f\text{'}\left(x\right)=\frac{f\left(5\right)-f\left(1\right)}{4}$

$⇒2x-2=\frac{19-3}{4}\phantom{\rule{0ex}{0ex}}⇒2x-2-4=0\phantom{\rule{0ex}{0ex}}⇒x=\frac{6}{2}=3$
Thus,  such that $f\text{'}\left(c\right)=\frac{f\left(5\right)-f\left(1\right)}{5-1}$.

Hence, Lagrange's theorem is verified.

(vii) We have,

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

Since a polynomial function is everywhere continuous and differentiable.

Therefore, $f\left(x\right)$ is continuous on  and differentiable on .
Thus, both conditions of Lagrange's mean value theorem are satisfied.
So, there must exist at least one real number ​ such that
$f\text{'}\left(c\right)=\frac{f\left(1\right)-f\left(0\right)}{1-0}=\frac{f\left(1\right)-f\left(0\right)}{1}$

Now, $f\left(x\right)=2x-{x}^{2}$
$⇒f\text{'}\left(x\right)=2-2x$ ,  $f\left(1\right)=2-1=1$ ,  $f\left(0\right)=0$

∴  $f\text{'}\left(x\right)=\frac{f\left(1\right)-f\left(0\right)}{1}$

$⇒2-2x=\frac{1-0}{1}\phantom{\rule{0ex}{0ex}}⇒-2x=1-2\phantom{\rule{0ex}{0ex}}⇒x=\frac{1}{2}$
Thus,  such that $f\text{'}\left(c\right)=\frac{f\left(1\right)-f\left(0\right)}{1-0}$.

Hence, Lagrange's theorem is verified.

(viii) We have,

$f\left(x\right)=\left(x-1\right)\left(x-2\right)\left(x-3\right)$ which can be rewritten as $f\left(x\right)={x}^{3}-6{x}^{2}+11x-6$

Since a polynomial function is everywhere continuous and differentiable.

Therefore, $f\left(x\right)$ is continuous on  and differentiable on .
Thus, both conditions of Lagrange's mean value theorem are satisfied.
So, there must exist at least one real number ​ such that
$f\text{'}\left(c\right)=\frac{f\left(4\right)-f\left(0\right)}{4-0}=\frac{f\left(4\right)-f\left(0\right)}{4}$

Now, $f\left(x\right)={x}^{3}-6{x}^{2}+11x-6$
$⇒f\text{'}\left(x\right)=3{x}^{2}-12x+11$ ,  $f\left(0\right)=-6$ ,  $f\left(4\right)=64-96+44-6=6$

∴  $f\text{'}\left(x\right)=\frac{f\left(4\right)-f\left(0\right)}{4-0}$

Thus,  such that $f\text{'}\left(c\right)=\frac{f\left(4\right)-f\left(0\right)}{4-0}$.

Hence, Lagrange's theorem is verified.

(ix) We have,

$f\left(x\right)=\sqrt{25-{x}^{2}}$

Here, $f\left(x\right)$ will exist,
if
$25-{x}^{2}\ge 0\phantom{\rule{0ex}{0ex}}⇒{x}^{2}\le 25\phantom{\rule{0ex}{0ex}}⇒-5\le x\le 5$

Since for each , the function $f\left(x\right)$ attains a unique definite value.
So, $f\left(x\right)$ is continuous on

Also, $f\text{'}\left(x\right)=\frac{1}{2\sqrt{25-{x}^{2}}}\left(-2x\right)=\frac{-x}{\sqrt{25-{x}^{2}}}$ exists for all

So, $f\left(x\right)$ is differentiable on .

Thus, both the conditions of lagrange's theorem are satisfied.

Consequently, there exists some  such that

$f\text{'}\left(c\right)=\frac{f\left(4\right)-f\left(-3\right)}{4+3}=\frac{f\left(4\right)-f\left(-3\right)}{7}$

Now, $f\left(x\right)=\sqrt{25-{x}^{2}}$

$f\text{'}\left(x\right)=\frac{-x}{\sqrt{25-{x}^{2}}}$ ,  $f\left(-3\right)=4$ ,  $f\left(4\right)=3$

∴  $f\text{'}\left(x\right)=\frac{f\left(4\right)-f\left(-3\right)}{4+3}$

Thus,  such that $f\text{'}\left(c\right)=\frac{f\left(4\right)-f\left(-3\right)}{4-\left(-3\right)}$.

Hence, Lagrange's theorem is verified.

(x) We have,

$f\left(x\right)={\mathrm{tan}}^{-1}x$

Clearly,  $f\left(x\right)$ is continuous on  and derivable on $\left(0,1\right)$

Thus, both the conditions of lagrange's theorem are satisfied.

Consequently, there exists some  such that

$f\text{'}\left(c\right)=\frac{f\left(1\right)-f\left(0\right)}{1-0}=\frac{f\left(1\right)-f\left(0\right)}{1}$

Now, $f\left(x\right)={\mathrm{tan}}^{-1}x$

$f\text{'}\left(x\right)=\frac{1}{1+{x}^{2}}$ ,  $f\left(1\right)=\frac{\mathrm{\pi }}{4}$ ,  $f\left(0\right)=0$

∴  $f\text{'}\left(x\right)=\frac{f\left(1\right)-f\left(0\right)}{1-0}$

Thus,  such that $f\text{'}\left(c\right)=\frac{f\left(1\right)-f\left(0\right)}{1-0}$.

Hence, Lagrange's theorem is verified.

(xi) We have,

$f\left(x\right)=x+\frac{1}{x}=\frac{{x}^{2}+1}{x}$

Clearly,  $f\left(x\right)$ is continuous on  and derivable on

Thus, both the conditions of lagrange's theorem are satisfied.

Consequently, there exists some  such that

$f\text{'}\left(c\right)=\frac{f\left(3\right)-f\left(1\right)}{3-1}=\frac{f\left(3\right)-f\left(1\right)}{2}$

Now, $f\left(x\right)=\frac{{x}^{2}+1}{x}$

$f\text{'}\left(x\right)=\frac{{x}^{2}-1}{{x}^{2}}$ ,  $f\left(1\right)=2$ ,  $f\left(3\right)=\frac{10}{3}$

∴  $f\text{'}\left(x\right)=\frac{f\left(3\right)-f\left(1\right)}{2}$

Thus,  such that $f\text{'}\left(c\right)=\frac{f\left(3\right)-f\left(1\right)}{3-1}$.

Hence, Lagrange's theorem is verified.

(xii) We have,

$f\left(x\right)=x{\left(x+4\right)}^{2}=x\left({x}^{2}+16+8x\right)={x}^{3}+8{x}^{2}+16x$

Since $f\left(x\right)$ is a polynomial function which is everywhere continuous and differentiable.

Therefore,  $f\left(x\right)$ is continuous on  and derivable on $\left(0,4\right)$

Thus, both the conditions of lagrange's theorem are satisfied.

Consequently, there exists some  such that

$f\text{'}\left(c\right)=\frac{f\left(4\right)-f\left(0\right)}{4-0}=\frac{f\left(4\right)-f\left(0\right)}{4}$

Now, $f\left(x\right)={x}^{3}+8{x}^{2}+16x$

$f\text{'}\left(x\right)=3{x}^{2}+16x+16$ ,  $f\left(4\right)=64+128+64=256$ ,  $f\left(0\right)=0$

∴  $f\text{'}\left(x\right)=\frac{f\left(4\right)-f\left(0\right)}{4-0}$

Thus,  such that $f\text{'}\left(c\right)=\frac{f\left(4\right)-f\left(0\right)}{4-0}$.

Hence, Lagrange's theorem is verified.

(xiii) We have,

$f\left(x\right)=\sqrt{{x}^{2}-4}$

Here, $f\left(x\right)$ will exist,
if

Since for each , the function $f\left(x\right)$ attains a unique definite value.
So, $f\left(x\right)$ is continuous on

Also, $f\text{'}\left(x\right)=\frac{1}{2\sqrt{{x}^{2}-4}}\left(2x\right)=\frac{x}{\sqrt{{x}^{2}-4}}$ exists for all

So, $f\left(x\right)$ is differentiable on .

Thus, both the conditions of lagrange's theorem are satisfied.

Consequently, there exists some  such that

$f\text{'}\left(c\right)=\frac{f\left(4\right)-f\left(2\right)}{4-2}=\frac{f\left(4\right)-f\left(2\right)}{2}$

Now, $f\left(x\right)=\sqrt{{x}^{2}-4}$

$f\text{'}\left(x\right)=\frac{x}{\sqrt{{x}^{2}-4}}$ ,  $f\left(4\right)=2\sqrt{3}$ ,  $f\left(2\right)=0$

∴  $f\text{'}\left(x\right)=\frac{f\left(4\right)-f\left(2\right)}{4-2}$

Thus,  such that $f\text{'}\left(c\right)=\frac{f\left(4\right)-f\left(2\right)}{4-2}$.

Hence, Lagrange's theorem is verified.

(xiv) We have,

$f\left(x\right)={x}^{2}+x-1$

Since polynomial function is everywhere continuous and differentiable.

Therefore, $f\left(x\right)$ is continuous on  and differentiable on

Thus, both the conditions of lagrange's theorem are satisfied.

Consequently, there exists some  such that

$f\text{'}\left(c\right)=\frac{f\left(4\right)-f\left(0\right)}{4-0}=\frac{f\left(4\right)-f\left(0\right)}{4}$

Now, $f\left(x\right)={x}^{2}+x-1$

$f\text{'}\left(x\right)=2x+1$ ,  $f\left(4\right)=19$ ,  $f\left(0\right)=-1$

∴  $f\text{'}\left(x\right)=\frac{f\left(4\right)-f\left(0\right)}{4-0}$

Thus,  such that $f\text{'}\left(c\right)=\frac{f\left(4\right)-f\left(0\right)}{4-0}$.

Hence, Lagrange's theorem is verified.

(xv) We have,

$f\left(x\right)=\mathrm{sin}x-\mathrm{sin}2x-x$

Since   are everywhere continuous and differentiable

Therefore, $f\left(x\right)$ is continuous on $\left[0,\mathrm{\pi }\right]$ and differentiable on

Thus, both the conditions of lagrange's theorem are satisfied.

Consequently, there exists some  such that

$f\text{'}\left(c\right)=\frac{f\left(\mathrm{\pi }\right)-f\left(0\right)}{\mathrm{\pi }-0}=\frac{f\left(\mathrm{\pi }\right)-f\left(0\right)}{\mathrm{\pi }}$

Now, $f\left(x\right)=\mathrm{sin}x-\mathrm{sin}2x-x$

$f\text{'}\left(x\right)=\mathrm{cos}x-2\mathrm{cos}2x-1$ ,  $f\left(\mathrm{\pi }\right)=-\mathrm{\pi }$ ,  $f\left(0\right)=0$

∴  $f\text{'}\left(x\right)=\frac{f\left(\mathrm{\pi }\right)-f\left(0\right)}{\mathrm{\pi }-0}$

Thus,  such that $f\text{'}\left(c\right)=\frac{f\left(\mathrm{\pi }\right)-f\left(0\right)}{\mathrm{\pi }-0}$.

Hence, Lagrange's theorem is verified.

(xvi) We have,

$f\left(x\right)={x}^{3}-5{x}^{2}-3x$

Since polynomial function is everywhere continuous and differentiable

Therefore, $f\left(x\right)$ is continuous on  and differentiable on

Thus, both the conditions of lagrange's theorem are satisfied.

Consequently, there exists some  such that

$f\text{'}\left(c\right)=\frac{f\left(3\right)-f\left(1\right)}{3-1}=\frac{f\left(3\right)-f\left(1\right)}{2}$

Now, $f\left(x\right)={x}^{3}-5{x}^{2}-3x$

$f\text{'}\left(x\right)=3{x}^{2}-10x-3$ ,  $f\left(3\right)=-27$ ,  $f\left(1\right)=-7$

∴  $f\text{'}\left(x\right)=\frac{f\left(3\right)-f\left(1\right)}{2}$

Thus,  such that $f\text{'}\left(c\right)=\frac{f\left(3\right)-f\left(1\right)}{3-1}$.

Hence, Lagrange's theorem is verified.

#### Question 2:

Discuss the applicability of Lagrange's mean value theorem for the function
f(x) = | x | on [−1, 1]

Given:
$f\left(x\right)=\left|x\right|$

If Lagrange's theorem is applicable for the given function, then $f\left(x\right)$ is continuous on  and differentiable on .

But it is known that $f\left(x\right)=\left|x\right|$ is not differentiable at .

Thus, our supposition is wrong.
Therefore, Lagrange's theorem is not applicable for the given function.

#### Question 3:

Show that the lagrange's mean value theorem is not applicable to the function
f(x) = $\frac{1}{x}$ on [−1, 1]

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

Clearly, $f\left(x\right)$ does not exist for x = 0

Thus, the given function is discontinuous on .

Hence, Lagrange's mean value theorem is not applicable for the given function on .1x

#### Question 4:

Verify the  hypothesis and conclusion of Lagrange's man value theorem for the function
f(x) = $\frac{1}{4x-1},$ 1≤ x ≤ 4.

The given function is $f\left(x\right)=\frac{1}{4x-1}$.

Since for each , the function attains a unique definite value, $f\left(x\right)$ is continuous on .

Also, $f\text{'}\left(x\right)=\frac{-4}{{\left(4x-1\right)}^{2}}$ exists for all

Thus, both the conditions of Lagrange's mean value theorem are satisfied.

Consequently, there exists some  such that
$f\text{'}\left(c\right)=\frac{f\left(4\right)-f\left(1\right)}{4-1}=\frac{f\left(4\right)-f\left(1\right)}{3}$

Now,
$f\left(x\right)=\frac{1}{4x-1}$$⇒$$f\text{'}\left(x\right)=\frac{-4}{{\left(4x-1\right)}^{2}}$

$\therefore$ $f\text{'}\left(x\right)=\frac{f\left(4\right)-f\left(1\right)}{4-1}$
$⇒f\text{'}\left(x\right)=\frac{\frac{1}{15}-\frac{1}{3}}{4-1}=\frac{-4}{45}\phantom{\rule{0ex}{0ex}}⇒\frac{-4}{{\left(4x-1\right)}^{2}}=\frac{-4}{45}\phantom{\rule{0ex}{0ex}}⇒{\left(4x-1\right)}^{2}=45\phantom{\rule{0ex}{0ex}}⇒16{x}^{2}-8x-44=0\phantom{\rule{0ex}{0ex}}⇒4{x}^{2}-2x-11=0\phantom{\rule{0ex}{0ex}}⇒x=\frac{1}{4}\left(1±3\sqrt{5}\right)$

Thus,  such that $f\text{'}\left(c\right)=\frac{f\left(4\right)-f\left(1\right)}{4-1}$.

Hence, Lagrange's theorem is verified.

#### Question 5:

Find a point on the parabola y = (x − 4)2, where the tangent is parallel to the chord joining (4, 0) and (5, 1).

​Let:
$f\left(x\right)={\left(x-4\right)}^{2}={x}^{2}-8x+16$

The tangent to the curve is parallel to the chord joining the points  and .

Assume that the chord joins the points  and .

$\therefore$

The polynomial function is everywhere continuous and differentiable.

So, ${x}^{2}-8x+16$ is continuous on  and differentiable on .

Thus, both the conditions of Lagrange's theorem are satisfied.

Consequently, there exists  such that $f\text{'}\left(c\right)=\frac{f\left(5\right)-f\left(4\right)}{5-4}$.

Now,
$f\left(x\right)={x}^{2}-8x+16⇒$$f\text{'}\left(x\right)=2x-8$

$\therefore$$f\text{'}\left(x\right)=\frac{f\left(5\right)-f\left(4\right)}{5-4}$$⇒$$2x-8=\frac{1}{1}⇒2x=9⇒x=\frac{9}{2}$

Thus,  such that ​$f\text{'}\left(c\right)=\frac{f\left(5\right)-f\left(4\right)}{5-4}$.

Clearly,
$f\left(c\right)={\left(\frac{9}{2}-4\right)}^{2}=\frac{1}{4}$

Thus, , i.e.​  $\left(\frac{9}{2},\frac{1}{4}\right)$, is a point on the given curve where the tangent is parallel to the chord joining the points (4, 0) and (5, 1).

#### Question 6:

Find a point on the curve y = x2 + x, where the tangent is parallel to the chord joining (0, 0) and (1, 2).

​Let:
$f\left(x\right)={x}^{2}+x$

The tangent to the curve is parallel to the chord joining the points  and .

Assume that the chord joins the points  and .

$\therefore$

The polynomial function is everywhere continuous and differentiable.

So, $f\left(x\right)={x}^{2}+x$ is continuous on  and differentiable on .

Thus, both the conditions of Lagrange's theorem are satisfied.

Consequently, there exists  such that $f\text{'}\left(c\right)=\frac{f\left(1\right)-f\left(0\right)}{1-0}$.

Now,
$f\left(x\right)={x}^{2}+x$$⇒$$f\text{'}\left(x\right)=2x+1$

$\therefore$$f\text{'}\left(x\right)=\frac{f\left(1\right)-f\left(0\right)}{1-0}$$⇒$$2x+1=\frac{2-0}{1-0}⇒2x=1⇒x=\frac{1}{2}$

Thus, $c=\frac{1}{2}\in \left(0,1\right)$ such that ​$f\text{'}\left(c\right)=\frac{f\left(1\right)-f\left(0\right)}{1-0}$.

Clearly,
$f\left(c\right)={\left(\frac{1}{2}\right)}^{2}+\frac{1}{2}=\frac{3}{4}$.

Thus, , i.e.​  , is a point on the given curve where the tangent is parallel to the chord joining the points (4, 0) and (5, 1).

#### Question 7:

Find a point on the parabola y = (x − 3)2, where the tangent is parallel to the chord joining (3, 0) and (4, 1).

​Let:
$f\left(x\right)={\left(x-3\right)}^{2}={x}^{2}-6x+9$

The tangent to the curve is parallel to the chord joining the points  and .

Assume that the chord joins the points  and .

$\therefore$

The polynomial function is everywhere continuous and differentiable.

So, $f\left(x\right)={x}^{2}-6x+9$ is continuous on  and differentiable on .

Thus, both the conditions of Lagrange's theorem are satisfied.

Consequently, there exists  such that $f\text{'}\left(c\right)=\frac{f\left(4\right)-f\left(3\right)}{4-3}$.

Now,
$f\left(x\right)={x}^{2}-6x+9$$⇒$$f\text{'}\left(x\right)=2x-6$

$\therefore$$f\text{'}\left(x\right)=\frac{f\left(4\right)-f\left(3\right)}{4-3}$$⇒$$2x-6=\frac{1-0}{4-3}⇒2x=7⇒x=\frac{7}{2}$

Thus,  such that ​$f\text{'}\left(c\right)=\frac{f\left(4\right)-f\left(3\right)}{4-3}$.

Clearly,
$f\left(c\right)={\left(\frac{7}{2}-3\right)}^{2}=\frac{1}{4}$

Thus, , i.e.  , is a point on the given curve where the tangent is parallel to the chord joining the points  and .

#### Question 8:

Find the points on the curve y = x3 − 3x, where the tangent to the curve is parallel to the chord joining (1, −2) and (2, 2).

​Let:
$f\left(x\right)={x}^{3}-3x$

The tangent to the curve is parallel to the chord joining the points  and .

Assume that the chord joins the points  and .

$\therefore$

The polynomial function is everywhere continuous and differentiable.

So, $f\left(x\right)={x}^{3}-3x$ is continuous on  and differentiable on .

Thus, both the conditions of Lagrange's theorem are satisfied.

Consequently, there exists  such that $f\text{'}\left(c\right)=\frac{f\left(2\right)-f\left(1\right)}{2-1}$.

Now,
$f\left(x\right)={x}^{3}-3x$$⇒$$f\text{'}\left(x\right)=3{x}^{2}-3$

$\therefore$$f\text{'}\left(x\right)=\frac{f\left(2\right)-f\left(1\right)}{2-1}$$⇒$$3{x}^{2}-3=\frac{2+2}{2-1}⇒3{x}^{2}=7⇒x=±\sqrt{\frac{7}{3}}$

Thus, $c=±\sqrt{\frac{7}{3}}$ such that ​$f\text{'}\left(c\right)=\frac{f\left(2\right)-f\left(1\right)}{2-1}$.

Clearly,
$f\left(\sqrt{\frac{7}{3}}\right)=\left[{\left(\frac{7}{3}\right)}^{\frac{3}{2}}-3\sqrt{\frac{7}{3}}\right]=\sqrt{\frac{7}{3}}\left[\frac{7}{3}-3\right]=\sqrt{\frac{7}{3}}\left[\frac{-2}{3}\right]=\frac{-2}{3}\sqrt{\frac{7}{3}}$ and $f\left(-\sqrt{\frac{7}{3}}\right)=\frac{2}{3}\sqrt{\frac{7}{3}}$

∴ $f\left(c\right)=\mp \frac{2}{3}\sqrt{\frac{7}{3}}$

Thus, , i.e.​  , are points on the given curve where the tangent is parallel to the chord joining the points  and .

#### Question 9:

Find a point on the curve y = x3 + 1 where the tangent is parallel to the chord joining (1, 2) and (3, 28).

​Let:
$f\left(x\right)={x}^{3}+1$

The tangent to the curve is parallel to the chord joining the points  and .

Assume that the chord joins the points  and .

$\therefore$

The polynomial function is everywhere continuous and differentiable.

So, $f\left(x\right)={x}^{3}+1$ is continuous on  and differentiable on .

Thus, both the conditions of Lagrange's theorem are satisfied.

Consequently, there exists  such that $f\text{'}\left(c\right)=\frac{f\left(3\right)-f\left(1\right)}{3-1}$.

Now,
$f\left(x\right)={x}^{3}+1$$⇒$$f\text{'}\left(x\right)=3{x}^{2}$

$\therefore$$f\text{'}\left(x\right)=\frac{f\left(3\right)-f\left(1\right)}{3-1}$$⇒$$3{x}^{2}=\frac{26}{2}⇒3{x}^{2}=13⇒x=±\sqrt{\frac{13}{3}}$

Thus, $c=\sqrt{\frac{13}{3}}$ such that ​$f\text{'}\left(c\right)=\frac{f\left(3\right)-f\left(1\right)}{3-1}$.

Clearly,
$f\left(c\right)=\left[{\left(\frac{13}{3}\right)}^{\frac{3}{2}}+1\right]$

Thus, , i.e.​  , is a point on the given curve where the tangent is parallel to the chord joining the points  and .

#### Question 10:

Let C be a curve defined parametrically as . Determine a point P on C, where the tangent to C is parallel to the chord joining the points (a, 0) and (0, a).                                                                                                                                 [CBSE 2014]

#### Question 11:

Using Lagrange's mean value theorem, prove that

(ba) sec2 a < tan b − tan a < (ba) sec2 b

where 0 < a < b < $\frac{\mathrm{\pi }}{2}$.

​Consider, the function

Clearly, $f\left(x\right)$ is continuous on  and derivable on .

Thus, both the conditions of Lagrange's theorem are satisfied.

Consequently,  such that $f\text{'}\left(c\right)=\frac{f\left(b\right)-f\left(a\right)}{b-a}$.

Now,
$f\left(x\right)=\mathrm{tan}x$ $⇒$ $f\text{'}\left(x\right)=se{c}^{2}x$

$\therefore$ $f\text{'}\left(c\right)=\frac{f\left(b\right)-f\left(a\right)}{b-a}$ $⇒$

Now,

Hence proved.

#### Question 1:

If f (x) = Ax2 + Bx + C is such that f (a) = f (b), then write the value of c in Rolle's theorem.

We have
$f\left(x\right)=A{x}^{2}+Bx+C$

Differentiating the given function with respect to x, we get
$f\text{'}\left(x\right)=2Ax+B$
$⇒f\text{'}\left(c\right)=2Ac+B$

From (1), we have

$c=\frac{a+b}{2}$

#### Question 2:

State Rolle's theorem.

Rolle's Theorem:

Let be a real valued function  defined on the closed interval $\left[a,b\right]$ such that
(i) it is continuous on the closed interval ​$\left[a,b\right]$,
(ii) it is differentiable on the open interval  and
(iii) $f\left(a\right)=f\left(b\right)$
Then, there exists a real number $c\in \left(a,b\right)$ such that $f\text{'}\left(c\right)=0$.

#### Question 3:

State Lagrange's mean value theorem.

Lagrange's Mean Value Theorem:

Let $f\left(x\right)$ be a function defined on $\left[a,b\right]$ such that
(i) it is continuous on ​$\left[a,b\right]$ and
(ii) it is differentiable on $\left(a,b\right)$.

Then, there exists a real number $c\in \left(a,b\right)$ such that $f\text{'}\left(c\right)=\frac{f\left(b\right)-f\left(a\right)}{b-a}$.

#### Question 4:

If the value of c prescribed in Rolle's theorem for the function
f (x) = 2x (x − 3)n on the interval write the value of n (a positive integer).

We have
$f\left(x\right)=2x{\left(x-3\right)}^{n}$

Differentiating the given function with respect to x, we get

$f\text{'}\left(x\right)=2\left[xn{\left(x-3\right)}^{n-1}+{\left(x-3\right)}^{n}\right]\phantom{\rule{0ex}{0ex}}⇒f\text{'}\left(x\right)=2{\left(x-3\right)}^{n}\left[\frac{xn}{\left(x-3\right)}+1\right]\phantom{\rule{0ex}{0ex}}⇒f\text{'}\left(c\right)=2{\left(c-3\right)}^{n}\left[\frac{cn}{\left(c-3\right)}+1\right]$

Given:
$f\text{'}\left(\frac{3}{4}\right)=0$

#### Question 5:

Find the value of c prescribed by Lagrange's mean value theorem for the function
$f\left(x\right)=\sqrt{{x}^{2}-4}$ defined on [2, 3].

We have

$f\left(x\right)=\sqrt{{x}^{2}-4}$

Here, $f\left(x\right)$ will exist, if

Since for each , the function $f\left(x\right)$ attains a unique definite value, $f\left(x\right)$ is continuous on .

Also, $f\text{'}\left(x\right)=\frac{1}{2\sqrt{{x}^{2}-4}}\left(2x\right)=\frac{x}{\sqrt{{x}^{2}-4}}$ exists for all .

So, $f\left(x\right)$ is differentiable on .

Thus, both the conditions of lagrange's theorem are satisfied.

Consequently, there exists  such that

$f\text{'}\left(c\right)=\frac{f\left(3\right)-f\left(2\right)}{3-2}=\frac{f\left(3\right)-f\left(2\right)}{1}$

Now,
$f\left(x\right)=\sqrt{{x}^{2}-4}$

$f\text{'}\left(x\right)=\frac{x}{\sqrt{{x}^{2}-4}}$ ,  $f\left(3\right)=\sqrt{5}$ ,  $f\left(2\right)=0$

∴  $f\text{'}\left(x\right)=\frac{f\left(3\right)-f\left(2\right)}{3-2}$

Thus,  such that $f\text{'}\left(c\right)=\frac{f\left(3\right)-f\left(2\right)}{3-2}$.

Hence, Lagrange's theorem is verified.

#### Question 1:

If the polynomial equation

n positive integer, has two different real roots α and β, then between α and β, the equation

(a) exactly one root
(b) almost one root
(c) at least one root
(d) no root

(c) at least one root

We observe that, $n{a}_{n}{x}^{n-1}+\left(n-1\right){a}_{n-1}{x}^{n-2}+...+{a}_{1}=0$ is the derivative of the
polynomial ${a}_{n}{x}^{n}+{a}_{n-1}{x}^{n-1}+{a}_{n-2}{x}^{n-2}+...+{a}_{2}{x}^{2}+{a}_{1}x+{a}_{0}=0$

Polynomial function is continuous every where in R and consequently derivative in R
Therefore, ${a}_{n}{x}^{n}+{a}_{n-1}{x}^{n-1}+{a}_{n-2}{x}^{n-2}+...+{a}_{2}{x}^{2}+{a}_{1}x+{a}_{0}$ is continuous on and derivative on .
Hence, it satisfies the both the conditions of Rolle's theorem.

By algebraic interpretation of Rolle's theorem, we know that between any two roots of a function $f\left(x\right)$, there exists at least one root of its derivative.

Hence, the equation $n{a}_{n}{x}^{n-1}+\left(n-1\right){a}_{n-1}{x}^{n-2}+...+{a}_{1}=0$ will have at least one root between .

#### Question 2:

If 4a + 2b + c = 0, then the equation 3ax2 + 2bx + c = 0 has at least one real root lying in the interval
(a) (0, 1)
(b) (1, 2)
(c) (0, 2)
(d) none of these

(c) (0, 2)

f is continuous in the closed interval [0, 2] and f is derivable in the open interval (0, 2).

Also, f(0) = f(2)

By Rolle's Theorem,

#### Question 3:

For the function f (x) = x + $\frac{1}{x}$, x ∈ [1, 3], the value of c for the Lagrange's mean value theorem
is
(a) 1
(b) $\sqrt{3}$
(c) 2
(d) none of these

(b)$\sqrt{3}$

We have
$f\left(x\right)=x+\frac{1}{x}=\frac{{x}^{2}+1}{x}$

Clearly,  $f\left(x\right)$ is continuous on  and derivable on .

Thus, both the conditions of Lagrange's theorem are satisfied.

Consequently, there exists  such that

$f\text{'}\left(c\right)=\frac{f\left(3\right)-f\left(1\right)}{3-1}=\frac{f\left(3\right)-f\left(1\right)}{2}$

Now, $f\left(x\right)=\frac{{x}^{2}+1}{x}$

$f\text{'}\left(x\right)=\frac{{x}^{2}-1}{{x}^{2}}$ ,  $f\left(1\right)=2$ ,  $f\left(3\right)=\frac{10}{3}$

∴  $f\text{'}\left(x\right)=\frac{f\left(3\right)-f\left(1\right)}{2}$

Thus,  such that $f\text{'}\left(c\right)=\frac{f\left(3\right)-f\left(1\right)}{3-1}$.

#### Question 4:

If from Lagrange's mean value theorem, we have

(a) a < x1b
(b) ax1 < b
(c) a < x1 < b
(d) ax1b

(c) a < x1 < b

In the Lagrange's mean value theorem,  such that $f\text{'}\left(c\right)=\frac{f\left(b\right)-f\left(a\right)}{b-a}$.

So, if there is ${x}_{1}$ such that $f\text{'}\left({x}_{1}\right)=\frac{f\left(b\right)-f\left(a\right)}{b-a}$, then .
$⇒a<{x}_{1}

#### Question 5:

Rolle's theorem is applicable in case of ϕ (x) = asin x, a > a in
(a) any interval
(b) the interval [0, π]
(c) the interval (0, π/2)
(d) none of these

(b) the interval [0, π]152

The given function is $\varphi \left(x\right)={a}^{\mathrm{sin}x}$, where > 0.

Differentiating the given function with respect to x, we get

∴

Also, the given function is derivable and hence continuous on the interval .

Hence, the Rolle's theorem is applicable on the given function in the interval ​.

#### Question 6:

The value of c in Rolle's theorem when
f (x) = 2x3 − 5x2 − 4x + 3, x ∈ [1/3, 3] is

(a) 2

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

(c) −2

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

(a) 2

Given:
$f\left(x\right)=2{x}^{3}-5{x}^{2}-4x+3$

Differentiating the given function with respect to x, we get

Thus, $c=2\in \left(\frac{1}{3},3\right)$ for which Rolle's theorem holds.

Hence, the required value of c is 2.

#### Question 7:

When the tangent to the curve y = x log x is parallel to the chord joining the points (1, 0) and (e, e), the value of x is

(a) e1/1−e

(b) e(e−1)(2e−1)

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

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

(a) e1/1−e

Given:
$y=f\left(x\right)=x\mathrm{log}x$

Differentiating the given function with respect to x,  we get

$f\text{'}\left(x\right)=1+\mathrm{log}x$

$⇒$ Slope of the tangent to the curve = $1+\mathrm{log}x$

Also,
Slope of the chord joining the points , (m) = $\frac{e}{e-1}$

The tangent to the curve is parallel  to the chord joining the points .

#### Question 8:

The value of c in Rolle's theorem for the function $f\left(x\right)=\frac{x\left(x+1\right)}{{e}^{x}}$ defined on [−1, 0] is
(a) 0.5

(b) $\frac{1+\sqrt{5}}{2}$

(c) $\frac{1-\sqrt{5}}{2}$

(d) −0.5

(c) $\frac{1-\sqrt{5}}{2}$152

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

Differentiating the given function with respect to x, we get

Hence, the required value of c is $\frac{1-\sqrt{5}}{2}$.

#### Question 9:

The value of c in Lagrange's mean value theorem for the function f (x) = x (x − 2) when x ∈ [1, 2] is
(a) 1
(b) 1/2
(c) 2/3
(d) 3/2

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

We have
f (x) = x (x − 2)

It can be rewritten as $f\left(x\right)={x}^{2}-2x$.

We know that a polynomial function is everywhere continuous and differentiable.

Since $f\left(x\right)$ is a polynomial , it is continuous on  and differentiable on .

Thus, $f\left(x\right)$ satisfies both the conditions of Lagrange's theorem on .

So, there must exist at least one real number  such that

$f\text{'}\left(c\right)=\frac{f\left(2\right)-f\left(1\right)}{2-1}=\frac{f\left(2\right)-f\left(1\right)}{1}$

Now, $f\left(x\right)={x}^{2}-2x$
$⇒f\text{'}\left(x\right)=2x-2$,
and

$⇒f\text{'}\left(x\right)=\frac{0+1}{1}\phantom{\rule{0ex}{0ex}}⇒2x-2=1\phantom{\rule{0ex}{0ex}}⇒x=\frac{3}{2}$

∴

#### Question 10:

The value of c in Rolle's theorem for the function f (x) = x3 − 3x in the interval [0, $\sqrt{3}$] is
(a) 1
(b) −1
(c) 3/2
(d) 1/3

(a) 1

The given function is $f\left(x\right)={x}^{3}-3x$.
This is a polynomial function, which is continuous and derivable in R.
Therefore, the function is continuous on [0, $\sqrt{3}$] and derivable on (0, $\sqrt{3}$ ).

Differentiating the given function with respect to x, we get

Thus,  for which Rolle's theorem holds.

Hence, the required value of c is 1.

#### Question 11:

If f (x) = ex sin x in [0, π], then c in Rolle's theorem is

(a) π/6

(b) π/4

(c) π/2

(d) 3π/4

(d) 3π/4

The given function is $f\left(x\right)={e}^{x}\mathrm{sin}x$.

Differentiating the given function with respect to x, we get

Thus, $c=\frac{3\mathrm{\pi }}{4}\in \left(0,\mathrm{\pi }\right)$ for which Rolle's theorem holds.

Hence, the required value of c is 3π/4.

#### Question 1:

Discuss the applicability of Rolle's theorem for the following functions on the indicated intervals
(i) f(x) = 3 + (x − 2)2/3 on [1, 3]

(ii) f(x) = [x] for −1 ≤ x ≤ 1, where [x] denotes the greatest integer not exceeding x

(iii) f(x) = sin$\frac{1}{x}$for −1 ≤ x ≤ 1

(iv) f(x) = 2x2 − 5x + 3 on [1, 3]

(v) f(x) = x2/3 on [−1, 1]

(vi) $f\left(x\right)=\left\{\begin{array}{ll}-4x+5,& 0\le x\le 1\\ 2x-3,& 1

(i) The given function is $f\left(x\right)=3+{\left(x-2\right)}^{\frac{2}{3}}$.

Differentiating with respect to x, we get

$f\text{'}\left(x\right)=\frac{2}{3}{\left(x-2\right)}^{\frac{2}{3}-1}\phantom{\rule{0ex}{0ex}}⇒f\text{'}\left(x\right)=\frac{2}{3}{\left(x-2\right)}^{\frac{-1}{3}}\phantom{\rule{0ex}{0ex}}⇒f\text{'}\left(x\right)=\frac{2}{3{\left(x-2\right)}^{\frac{1}{3}}}$

Clearly, we observe that for x=2$f\text{'}\left(x\right)$ does not exist.

Therefore, $f\left(x\right)$ is not derivable on .

Hence, Rolle's theorem is not applicable for the given function.

(ii) The given function is $f\left(x\right)=\left[x\right]$.
The domain of is given to be .

Let  such that is not an integer.
Then,
$\underset{x\to c}{\mathrm{lim}}f\left(x\right)=f\left(c\right)$

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

Now, let $c=0$.

Then,

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

Thus, is discontinuous at = 0.

Therefore, $f\left(x\right)$ is not continuous in .

Rolle's theorem is not applicable for the given function.

(iii) The given function is $f\left(x\right)=\mathrm{sin}\frac{1}{x}$.
The domain of is given to be .

It is known that $\underset{x\to 0}{\mathrm{lim}}\mathrm{sin}\frac{1}{x}$ does not exist.

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

Hence, Rolle's theorem is not applicable for the given function.

(iv) The given function is $f\left(x\right)=2{x}^{2}-5x+3$ on .
The domain of is given to be .
It is a polynomial function.
Thus, it is everywhere derivable and hence continuous.

But

Hence, Rolle's theorem is not applicable for the given function.

(v) The given function is $f\left(x\right)={x}^{\frac{2}{3}}$ on .
The domain of is given to be .

Differentiating $f\left(x\right)$ with respect to x, we get

$f\text{'}\left(x\right)=\frac{2}{3}{x}^{-\frac{1}{3}}$
We observe that at $x=0$$f\text{'}\left(x\right)$ is not defined.
Hence, Rolle's theorem is not applicable for the given function.

(vi) The given function is

At x = 0, we have

$\underset{x\to {1}^{-}}{\mathrm{lim}}f\left(x\right)=\underset{h\to 0}{\mathrm{lim}}f\left(1-h\right)=\underset{h\to 0}{\mathrm{lim}}\left[-4\left(1-h\right)+5\right]=1$
And
$\underset{x\to {1}^{+}}{\mathrm{lim}}f\left(x\right)=\underset{h\to 0}{\mathrm{lim}}f\left(1+h\right)=\underset{h\to 0}{\mathrm{lim}}\left[2\left(1+h\right)-3\right]=-1$

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

Thus, $f\left(x\right)$ is discontinuous at $x=1$.
Hence, Rolle's theorem is not applicable for the given function.

#### Question 2:

Verify Rolle's theorem for each of the following functions on the indicated intervals
(i) f(x) = x2 − 8x + 12 on [2, 6]

(ii) f(x) = x2 − 4x + 3 on [1, 3]

(iii) f(x) = (x − 1) (x − 2)2 on [1, 2]

(iv) f(x) = x(x − 1)2 on [0, 1]

(v) f(x) = (x2 − 1) (x − 2) on [−1, 2]

(vi) f(x) = x(x − 4)2 on the interval [0, 4]

(vii)
f(x) = x(x −2)2 on the interval [0, 2]

(viii)
f(x) = x2 + 5x + 6 on the interval [−3, −2]

(i) Given:
$f\left(x\right)={x}^{2}-8x+12$

We know that a polynomial function is everywhere derivable and hence continuous.

So, being a polynomial function, $f\left(x\right)$ is continuous and derivable on

Also,

Thus, all the conditions of Rolle's theorem are satisfied.

Now, we have to show that there exists  such that $f\text{'}\left(c\right)=0$.

We have

Thus, .

Hence, Rolle's theorem is verified.

(ii) Given:

We know that a polynomial function is everywhere derivable and hence continuous.

So, being a polynomial function, $f\left(x\right)$ is continuous and derivable on

Also,

Thus, all the conditions of Rolle's theorem are satisfied.

Now, we have to show that there exists  such that $f\text{'}\left(c\right)=0$.

We have

Thus, .

Hence, Rolle's theorem is verified.

(iii) Given:
$f\left(x\right)=\left(x-1\right){\left(x-2\right)}^{2}$
i.e. $f\left(x\right)={x}^{3}+4x-4{x}^{2}-{x}^{2}-4+4x$
i.e. $f\left(x\right)={x}^{3}-5{x}^{2}+8x-4$

We know that a polynomial function is everywhere derivable and hence continuous.

So, being a polynomial function, $f\left(x\right)$ is continuous and derivable on

Also,

Thus, all the conditions of Rolle's theorem are satisfied.

Now, we have to show that there exists  such that $f\text{'}\left(c\right)=0$.

We have

Thus, .

Hence, Rolle's theorem is verified.

(iv) Given:
$f\left(x\right)=x{\left(x-1\right)}^{2}$
$⇒f\left(x\right)=x\left({x}^{2}-2x+1\right)$

We know that a polynomial function is everywhere derivable and hence continuous.

So, $f\left(x\right)$ being a polynomial function is continuous and derivable on

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

Thus, all the conditions of Rolle's theorem are satisfied.

Now, we have to show that there exists  such that $f\text{'}\left(c\right)=0$.

We have

Thus, .

Hence, Rolle's theorem is verified.

(v) Given:
$f\left(x\right)=\left({x}^{2}-1\right)\left(x-2\right)$
i.e. $f\left(x\right)={x}^{3}-2{x}^{2}-x+2$

We know that a polynomial function is everywhere derivable and hence continuous.

So, being a polynomial function, $f\left(x\right)$ is continuous and derivable on

Also,
$f\left(-1\right)=f\left(2\right)=0$

Thus, all the conditions of Rolle's theorem are satisfied.

Now, we have to show that there exists  such that $f\text{'}\left(c\right)=0$.

We have

Thus, .

Hence, Rolle's theorem is verified.

(vi) Given function is
$f\left(x\right)=x{\left(x-4\right)}^{2}$, which can be rewritten as $f\left(x\right)={x}^{3}-8{x}^{2}+16x$.

We know that a polynomial function is everywhere derivable and hence continuous.

So, being a polynomial function, $f\left(x\right)$ is continuous and derivable on

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

Thus, all the conditions of Rolle's theorem are satisfied.

Now, we have to show that there exists  such that $f\text{'}\left(c\right)=0$.

We have

Thus, .

Hence, Rolle's theorem is verified.

(vii) The given function is
$f\left(x\right)=x{\left(x-2\right)}^{2}$, which can be rewritten as $f\left(x\right)={x}^{3}-4{x}^{2}+4x$.

We know that a polynomial function is everywhere derivable and hence continuous.

So, being a polynomial function, $f\left(x\right)$ is continuous and derivable on

Also,
$f\left(0\right)=f\left(2\right)=0$

Thus, all the conditions of Rolle's theorem are satisfied.

Now, we have to show that there exists  such that $f\text{'}\left(c\right)=0$.

We have

Thus, .

Hence, Rolle's theorem is verified.

(viii) Given function is
$f\left(x\right)={x}^{2}+5x+6$.

We know that a polynomial function is everywhere derivable and hence continuous.

So, being a polynomial function, $f\left(x\right)$ is continuous and derivable on
Also,

Thus, all the conditions of the Rolle's theorem are satisfied.

Now, we have to show that there exists  such that $f\text{'}\left(c\right)=0$.

We have

Thus, .

Hence, Rolle's theorem is verified.

#### Question 3:

Verify Rolle's theorem for each of the following functions on the indicated intervals
(i) f(x) = cos 2 (x − π/4) on [0, π/2]

(ii) f(x) = sin 2x on [0, π/2]

(iii) f(x) = cos 2x on [−π/4, π/4]

(iv) f(x) = ex sin x on [0, π]

(v) f(x) = ex cos x on [−π/2, π/2]

(vi) f(x) = cos 2x on [0, π]

(vii) f(x) = on 0 ≤ x ≤ π

(viii) f(x) = sin 3x on [0, π]

(ix) f(x) = ${{e}^{1-x}}^{2}$ on [−1, 1]

(x) f(x) = log (x2 + 2) − log 3 on [−1, 1]

(xi) f(x) = sin x + cos x on [0, π/2]

(xii) f(x) = 2 sin x + sin 2x on [0, π]

(xiii)

(xiv)

(xv) f(x) = 4sin x on [0, π]

(xvi) f(x) = x2 − 5x + 4 on [1, 4]

(xvii) f(x) = sin4 x + cos4 x on

(xviii) f(x) = sin x − sin 2x on [0, π]

(i) The given function is $f\left(x\right)=\mathrm{cos}2\left(x-\frac{\mathrm{\pi }}{4}\right)=\mathrm{cos}\left(2x-\frac{\mathrm{\pi }}{2}\right)=\mathrm{sin}2x$.

Since $\mathrm{sin}2x$ is everywhere continuous and differentiable.

Therefore, $\mathrm{sin}2x$ is continuous on  and differentiable on .

Also,
$f\left(\frac{\mathrm{\pi }}{2}\right)=f\left(0\right)=0$

Thus, $f\left(x\right)$ satisfies all the conditions of Rolle's theorem.

Now, we have to show that there exists  such that $f\text{'}\left(c\right)=0$.

We have
$f\left(x\right)=\mathrm{sin}2x\phantom{\rule{0ex}{0ex}}⇒f\text{'}\left(x\right)=2\mathrm{cos}2x$

$\therefore f\text{'}\left(x\right)=0\phantom{\rule{0ex}{0ex}}⇒2\mathrm{cos}2x=0\phantom{\rule{0ex}{0ex}}⇒\mathrm{cos}2x=0\phantom{\rule{0ex}{0ex}}⇒x=\frac{\mathrm{\pi }}{4}$

Thus,  such that $f\text{'}\left(c\right)=0$.

Hence, Rolle's theorem is verified.

(ii) The given function is $f\left(x\right)=\mathrm{sin}2x$.

Since $\mathrm{sin}2x$ is everywhere continuous and differentiable.

Therefore, $\mathrm{sin}2x$ is continuous on  and differentiable on .

Also,
$f\left(\frac{\mathrm{\pi }}{2}\right)=f\left(0\right)=0$

Thus, $f\left(x\right)$ satisfies all the conditions of Rolle's theorem.

Now, we have to show that there exists  such that $f\text{'}\left(c\right)=0$.

We have
$f\left(x\right)=\mathrm{sin}2x\phantom{\rule{0ex}{0ex}}⇒f\text{'}\left(x\right)=2\mathrm{cos}2x$

$\therefore f\text{'}\left(x\right)=0\phantom{\rule{0ex}{0ex}}⇒2\mathrm{cos}2x=0\phantom{\rule{0ex}{0ex}}⇒\mathrm{cos}2x=0\phantom{\rule{0ex}{0ex}}⇒x=\frac{\mathrm{\pi }}{4}$

Thus,  such that $f\text{'}\left(c\right)=0$.

​Hence, Rolle's theorem is verified.

(iii)

The given function is $f\left(x\right)=\mathrm{cos}2x$.

Since $\mathrm{cos}2x$ is everywhere continuous and differentiable, $\mathrm{cos}2x$ is continuous on  and differentiable on .

Also,
$f\left(\frac{\mathrm{\pi }}{4}\right)=f\left(\frac{-\mathrm{\pi }}{4}\right)=0$

Thus, $f\left(x\right)$ satisfies all the conditions of Rolle's theorem.

Now, we have to show that there exists  such that $f\text{'}\left(c\right)=0$.

We have
$f\left(x\right)=\mathrm{cos}2x\phantom{\rule{0ex}{0ex}}⇒f\text{'}\left(x\right)=-2\mathrm{sin}2x$

Since  such that $f\text{'}\left(c\right)=0$.

​Hence, Rolle's theorem is verified.

(iv)

The given function is $f\left(x\right)={e}^{x}\mathrm{sin}x$.

Since  are everywhere continuous and differentiable.

Therefore, being a product of these two, $f\left(x\right)$ is continuous on  and differentiable on .

Also,
$f\left(\mathrm{\pi }\right)=f\left(0\right)=0$

Thus, $f\left(x\right)$ satisfies all the conditions of Rolle's theorem.

Now, we have to show that there exists  such that $f\text{'}\left(c\right)=0$.

We have
$f\left(x\right)={e}^{x}\mathrm{sin}x\phantom{\rule{0ex}{0ex}}⇒f\text{'}\left(x\right)={e}^{x}\left(\mathrm{sin}x+\mathrm{cos}x\right)$

Since  such that $f\text{'}\left(c\right)=0$.

​Hence, Rolle's theorem is verified.

(v)

The given function is $f\left(x\right)={e}^{x}\mathrm{cos}x$.

Since  are everywhere continuous and differentiable, $f\left(x\right)$ being a product of these two is continuous on  and differentiable on .

Also,
$f\left(\frac{-\mathrm{\pi }}{2}\right)=f\left(\frac{\mathrm{\pi }}{2}\right)=0$

Thus, $f\left(x\right)$ satisfies all the conditions of Rolle's theorem.

Now, we have to show that there exists  such that $f\text{'}\left(c\right)=0$.

We have
$f\left(x\right)={e}^{x}\mathrm{cos}x\phantom{\rule{0ex}{0ex}}⇒f\text{'}\left(x\right)={e}^{x}\left(\mathrm{cos}x-\mathrm{sin}x\right)$

Since  such that $f\text{'}\left(c\right)=0$.

​Hence, Rolle's theorem is verified.

(vi)

The given function is$f\left(x\right)=\mathrm{cos}2x$.

Since $\mathrm{cos}2x$ is everywhere continuous and differentiable.

Therefore, $f\left(x\right)$  is continuous on  and differentiable on .

Also,
$f\left(\mathrm{\pi }\right)=f\left(0\right)=1$

Thus, $f\left(x\right)$ satisfies all the conditions of Rolle's theorem.

Now, we have to show that there exists  such that $f\text{'}\left(c\right)=0$.

We have
$f\left(x\right)=\mathrm{cos}2x\phantom{\rule{0ex}{0ex}}⇒f\text{'}\left(x\right)=-2\mathrm{sin}2x$

Thus,  such that $f\text{'}\left(c\right)=0$.

​Hence, Rolle's theorem is verified.

(vii)

The given function is $f\left(x\right)=\frac{\mathrm{sin}x}{{e}^{x}}$.

Since  are everywhere continuous and differentiable, being the quotient of these two, $f\left(x\right)$ is continuous on  and differentiable on .

Also,
$f\left(\mathrm{\pi }\right)=f\left(0\right)=0$

Thus, $f\left(x\right)$ satisfies all the conditions of Rolle's theorem.

Now, we have to show that there exists  such that $f\text{'}\left(c\right)=0$.

We have
$f\left(x\right)=\frac{\mathrm{sin}x}{{e}^{x}}\phantom{\rule{0ex}{0ex}}⇒f\text{'}\left(x\right)=\frac{\mathrm{cos}x-\mathrm{sin}x}{{e}^{x}}$

Thus,  such that $f\text{'}\left(c\right)=0$.

​Hence, Rolle's theorem is verified.

(viii)

The given function is$f\left(x\right)=\mathrm{sin}3x$.

Since $\mathrm{sin}3x$ is everywhere continuous and differentiable, $\mathrm{sin}3x$ is continuous on  and differentiable on $\left(0,\mathrm{\pi }\right)$.

Also,
$f\left(\mathrm{\pi }\right)=f\left(0\right)=0$

Thus, $f\left(x\right)$ satisfies all the conditions of Rolle's theorem.

Now, we have to show that there exists  such that $f\text{'}\left(c\right)=0$.

We have
$f\left(x\right)=\mathrm{sin}3x\phantom{\rule{0ex}{0ex}}⇒f\text{'}\left(x\right)=3\mathrm{cos}3x$

Thus,  such that $f\text{'}\left(c\right)=0$.

​Hence, Rolle's theorem is verified.

(ix)

The given function is$f\left(x\right)={e}^{1-{x}^{2}}$.

Since exponential function  is everywhere continuous and differentiable, ${e}^{1-{x}^{2}}$ is continuous on  and differentiable on .

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

Thus, $f\left(x\right)$ satisfies all the conditions of Rolle's theorem.

Now, we have to show that there exists  such that $f\text{'}\left(c\right)=0$.

We have
$f\left(x\right)={e}^{1-{x}^{2}}\phantom{\rule{0ex}{0ex}}⇒f\text{'}\left(x\right)=-2x{e}^{1-{x}^{2}}$

$\therefore f\text{'}\left(x\right)=0\phantom{\rule{0ex}{0ex}}⇒-2x{e}^{1-{x}^{2}}=0\phantom{\rule{0ex}{0ex}}⇒x=0$

Thus,  such that $f\text{'}\left(c\right)=0$.

​Hence, Rolle's theorem is verified.

(x)

The given function is $f\left(x\right)=\mathrm{log}\left({x}^{2}+2\right)-\mathrm{log}3$, which can be rewritten as $f\left(x\right)=\mathrm{log}\left(\frac{{x}^{2}+2}{3}\right)$.

Since logarithmic function is differentiable and so continuous in its domain, $f\left(x\right)=\mathrm{log}\left(\frac{{x}^{2}+2}{3}\right)$ is continuous on  and differentiable on .

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

Thus, $f\left(x\right)$ satisfies all the conditions of Rolle's theorem.

Now, we have to show that there exists  such that $f\text{'}\left(c\right)=0$.

We have
$f\left(x\right)=\mathrm{log}\left(\frac{{x}^{2}+2}{3}\right)\phantom{\rule{0ex}{0ex}}⇒f\text{'}\left(x\right)=\frac{3\left(2x\right)}{{x}^{2}+2}=\frac{6x}{{x}^{2}+2}$

$\therefore f\text{'}\left(x\right)=0\phantom{\rule{0ex}{0ex}}⇒\frac{6x}{{x}^{2}+2}=0\phantom{\rule{0ex}{0ex}}⇒x=0$

Thus,  such that $f\text{'}\left(c\right)=0$.

​Hence, Rolle's theorem is verified.

(xi)

The given function is.
Since  are everywhere continuous and differentiable,  is continuous on  and differentiable on .

Also,
$f\left(\frac{\mathrm{\pi }}{2}\right)=f\left(0\right)=1$

Thus, $f\left(x\right)$ satisfies all the conditions of Rolle's theorem.

Now, we have to show that there exists  such that $f\text{'}\left(c\right)=0$.

We have
$f\left(x\right)=\mathrm{sin}x+\mathrm{cos}x\phantom{\rule{0ex}{0ex}}⇒f\text{'}\left(x\right)=\mathrm{cos}x-\mathrm{sin}x$

$\therefore f\text{'}\left(x\right)=0\phantom{\rule{0ex}{0ex}}⇒\mathrm{cos}x-\mathrm{sin}x=0\phantom{\rule{0ex}{0ex}}⇒\mathrm{tan}x=1\phantom{\rule{0ex}{0ex}}⇒x=\frac{\mathrm{\pi }}{4}$

Thus,  such that $f\text{'}\left(c\right)=0$.

​Hence, Rolle's theorem is verified.

(xii)
The given function is.

Since  are everywhere continuous and differentiable, $f\left(x\right)$  is continuous on  and differentiable on .

Also,
$f\left(\mathrm{\pi }\right)=f\left(0\right)=0$

Thus, $f\left(x\right)$ satisfies all the conditions of Rolle's theorem.

Now, we have to show that there exists  such that $f\text{'}\left(c\right)=0$.

We have
$f\left(x\right)=2\mathrm{sin}x+\mathrm{sin}2x\phantom{\rule{0ex}{0ex}}⇒f\text{'}\left(x\right)=2\mathrm{cos}x+2\mathrm{cos}2x$

Thus,  such that $f\text{'}\left(c\right)=0$.

​Hence, Rolle's theorem is verified.

(xiii)

The given function is$f\left(x\right)=\frac{x}{2}-\mathrm{sin}\frac{\mathrm{\pi x}}{6}$.

Since  are everywhere continuous and differentiable, $f\left(x\right)$  is continuous on  and differentiable on .

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

Thus, $f\left(x\right)$ satisfies all the conditions of Rolle's theorem.

Now, we have to show that there exists  such that $f\text{'}\left(c\right)=0$.

We have
$f\left(x\right)=\frac{x}{2}-\mathrm{sin}\frac{\mathrm{\pi x}}{6}\phantom{\rule{0ex}{0ex}}⇒f\text{'}\left(x\right)=\frac{1}{2}-\frac{\mathrm{\pi }}{6}\mathrm{cos}\frac{\mathrm{\pi x}}{6}$

Thus,  such that $f\text{'}\left(c\right)=0$.

​Hence, Rolle's theorem is verified.

(xiv)

The given function is$f\left(x\right)=\frac{6x}{\mathrm{\pi }}-4{\mathrm{sin}}^{2}x$.

Since  are everywhere continuous and differentiable, $f\left(x\right)$  is continuous on  and differentiable on .

Also,
$f\left(\frac{\mathrm{\pi }}{6}\right)=f\left(0\right)=0$

Thus, $f\left(x\right)$ satisfies all the conditions of Rolle's theorem.

Now, we have to show that there exists  such that $f\text{'}\left(c\right)=0$.

We have

Thus,  such that $f\text{'}\left(c\right)=0$.

​Hence, Rolle's theorem is verified.

(xv)

The given function is$f\left(x\right)={4}^{\mathrm{sin}x}$.

Since sine function and exponential function are everywhere continuous and differentiable, $f\left(x\right)$  is continuous on  and differentiable on .

Also,
$f\left(\mathrm{\pi }\right)=f\left(0\right)=1$

Thus, $f\left(x\right)$ satisfies all the conditions of Rolle's theorem.

Now, we have to show that there exists  such that $f\text{'}\left(c\right)=0$.

We have
$f\left(x\right)={4}^{\mathrm{sin}x}\phantom{\rule{0ex}{0ex}}⇒f\text{'}\left(x\right)={4}^{\mathrm{sin}x}\left(\mathrm{cos}x\right)\mathrm{log}4$

Thus,  such that $f\text{'}\left(c\right)=0$.

​Hence, Rolle's theorem is verified.

(xvi)

According to Rolle’s theorem, if(x) is a real valued function defined on [ab] such that it is continuous on [ab], it is differentiable on (ab) and f(a) = f(b), then there exists a real number c ∈(ab) such that f(c) = 0.

Now, f(x) is defined for all x ∈[1, 4].
At each point of [1, 4], the limit of f(x) is equal to the value of the function. Therefore, f(x) is continuous on [1, 4].

Also, exists for all x ∈(1, 4).

So, f(x) is differentiable on (1, 4).

Also,
f(1) = f(4) = 0

Thus, all the three conditions of Rolle’s theorem are satisfied.

Now, we have to show that there exists c ∈(1, 4) such that .

We have $\therefore f\text{'}\left(x\right)=0\phantom{\rule{0ex}{0ex}}⇒2x-5=0\phantom{\rule{0ex}{0ex}}⇒x=\frac{5}{2}$

[Since ∈(1, 4) such that ]

Hence, Rolle’s theorem is verified.

(xvii)

The given function is .
Since  are everywhere continuous and differentiable,  is continuous on  and differentiable on .

Also,
$f\left(\frac{\mathrm{\pi }}{2}\right)=f\left(0\right)=1$

Thus, $f\left(x\right)$ satisfies all the conditions of Rolle's theorem.

Now, we have to show that there exists  such that $f\text{'}\left(c\right)=0$.

We have
$f\left(x\right)={\mathrm{sin}}^{4}x+{\mathrm{cos}}^{4}x\phantom{\rule{0ex}{0ex}}⇒f\text{'}\left(x\right)=4{\mathrm{sin}}^{3}x\mathrm{cos}x-4{\mathrm{cos}}^{3}x\mathrm{sin}x$

Thus,  such that $f\text{'}\left(c\right)=0$.

​Hence, Rolle's theorem is verified.

(xviii)

The given function is .

Since  are everywhere continuous and differentiable, $f\left(x\right)$  is continuous on  and differentiable on .

Also,
$f\left(\mathrm{\pi }\right)=f\left(0\right)=0$

Thus, $f\left(x\right)$ satisfies all the conditions of Rolle's theorem.

Now, we have to show that there exists  such that $f\text{'}\left(c\right)=0$.

We have
$f\left(x\right)=\mathrm{sin}x-\mathrm{sin}2x\phantom{\rule{0ex}{0ex}}⇒f\text{'}\left(x\right)=\mathrm{cos}x-2\mathrm{cos}2x$

Thus,   such that $f\text{'}\left(c\right)=0$.

​Hence, Rolle's theorem is verified.​

#### Question 7:

Using Rolle's theorem, find points on the curve y = 16 − x2, x ∈ [−1, 1], where tangent is parallel to x-axis.

The equation of the curve is
$y=16-{x}^{2}$.          ...(1)

Let P$\left({x}_{1},{y}_{1}\right)$ be a point on it where the tangent is parallel to the x-axis.

Then,
${\left(\frac{dy}{dx}\right)}_{P}=0$       ...(2)

Differentiating (1) with respect to x, we get

lies on the curve $y=16-{x}^{2}$.

$\therefore$ ${y}_{1}=16-{{x}_{1}}^{2}$

When ${x}_{1}=0$,
${y}_{1}=16$

Hence,  is the required point.

#### Question 8:

At what points on the following curves, is the tangent parallel to x-axis?
(i) y = x2 on [−2, 2]
(ii) y = ${e}^{1-{x}^{2}}$ on [−1, 1]
(iii) y = 12 (x + 1) (x − 2) on [−1, 2].

(i) Let $f\left(x\right)={x}^{2}$
Since $f\left(x\right)$ is a polynomial function, it is continuous on  and differentiable on .

Also, $f\left(2\right)=f\left(-2\right)=4$

Thus, all the conditions of Rolle's theorem are satisfied.

Consequently, there exists at least one point c for which $f\text{'}\left(c\right)=0$.

But $f\text{'}\left(c\right)=0⇒2c=0⇒c=0$

By the geometrical interpretation of Rolle's theorem,  is the point on $y={x}^{2}$, where the tangent is parallel to the x-axis.

(ii) Let $f\left(x\right)={e}^{1-{x}^{2}}$
Since $f\left(x\right)$ is an exponential function, which is continuous and derivable on its domain, $f\left(x\right)$ is continuous on  and differentiable on .

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

Thus, all the conditions of Rolle's theorem are satisfied.

Consequently, there exists at least one point c for which $f\text{'}\left(c\right)=0$.

But

By the geometrical interpretation of Rolle's theorem,  is the point on $y={e}^{1-{x}^{2}}$ where the tangent is parallel to the x-axis.

(iii) Let $f\left(x\right)=12\left(x+1\right)\left(x-2\right)$    ...(1)

$⇒$$f\left(x\right)=12\left({x}^{2}-x-2\right)$
$⇒$$f\left(x\right)=12{x}^{2}-12x-24$

Since $f\left(x\right)$ is a polynomial function, $f\left(x\right)$ is continuous on  and differentiable on .

Also, $f\left(2\right)=f\left(-1\right)=0$

Thus, all the conditions of Rolle's theorem are satisfied.

Consequently, there exists at least one point c for which $f\text{'}\left(c\right)=0$.

But $f\text{'}\left(c\right)=0⇒24c-12=0⇒c=\frac{1}{2}$

(using (1))

By the geometrical interpretation of Rolle's theorem, $\left(\frac{1}{2},-27\right)$ is the point on $y=12\left(x+1\right)\left(x-2\right)$​ where the tangent is parallel to the x-axis.

#### Question 9:

If f : [−5, 5] → R is differentiable and if f' (x) doesnot vanish anywhere, then prove that f (−5) ± f (5).

It is given that is a differentiable function.
Every differentiable function is a continuous function. Thus,
(a) f is continuous in [−5, 5].
(b) is differentiable in (−5, 5).
Therefore, by the Mean Value Theorem, there exists c ∈ (−5, 5) such that It is also given that does not vanish anywhere. Hence proved.

#### Question 10:

Examine if Rolle's theorem is applicable to any one of the following functions.
(i) f (x) = [x] for x ∈ [5, 9]
(ii) f (x) = [x] for x ∈ [−2, 2]
Can you say something about the converse of Rolle's Theorem from these functions?

By Rolle’s theorem, for a function , if

(a) f is continuous on [ab],

(b) f is differentiable on (ab) and

(c) (a) = f (b),

then there exists some c ∈ (ab) such that .

Therefore, Rolle’s theorem is not applicable to those functions that do not satisfy any of the three conditions of the hypothesis.

(i) It is evident that the given function f (x) is not continuous at every integral point.

In particular, f(x) is not continuous at = 5 and = 9.

Thus, f (x) is not continuous on [5, 9]. The differentiability of f on (5, 9) is checked in the following way.

Let be an integer such that n ∈ (5, 9). Since the left and the right hand limits of f at x = n are not equal, f is not differentiable at x = n.

Thus, is not differentiable on (5, 9).

It is observed that f does not satisfy all the conditions of the hypothesis of Rolle’s theorem.

Hence, Rolle’s theorem is not applicable on .

(ii) It is evident that the given function f (x) is not continuous at every integral point.

In particular, f(x) is not continuous at = −2 and = 2.

Thus, f (x) is not continuous on [−2, 2]. The differentiability of f on (−2, 2) is checked in the following way.

Let be an integer such that n ∈ (−2, 2). Since the left and the right hand limits of f at x = n are not equal, f is not differentiable at x = n.

Thus, is not differentiable on (−2, 2).

It is observed that f does not satisfy all the conditions of the hypothesis of Rolle’s theorem.

Hence, Rolle’s theorem is not applicable on .

#### Question 11:

It is given that the Rolle's theorem holds for the function f(x) = x3 + bx2 + cx, x $\in$ [1, 2] at the point x = $\frac{4}{3}$. Find the values of b and c.

As, the Rolle's theorem holds for the function f(x) = x3 + bx2 + cxx $\in$ [1, 2] at the point x = $\frac{4}{3}$