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#### 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 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:

A function (x) = 1 + $\frac{1}{x}$ is defined on the closed interval [1, 3]. A point in the interval, where the function satisfies the mean value theorem, is ______________.

The function $f\left(x\right)=1+\frac{1}{x}$ is defined on the interval [1, 3].

f(x) is continuous on [1, 3] and differentiable on (1, 3).

So, by mean value theorem there must exist at least one real number c ∈ (1, 3) such that

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

$⇒-\frac{1}{{c}^{2}}=\frac{-\frac{2}{3}}{2}$

$⇒\frac{1}{{c}^{2}}=\frac{1}{3}$

$⇒{c}^{2}=3$

$⇒c=±\sqrt{3}$

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

Hence, a point in the interval where the given function satisfies the mean value theorem is $\sqrt{3}$.

A function (x) = 1 + $\frac{1}{x}$ is defined on the closed interval [1, 3]. A point in the interval, where the function satisfies the mean value theorem, is .

#### Question 2:

For the function f(x) = 8x2 - 7x + 5, x ∈ [-6, 6], the value of c for the lagrange's mean value theorem is __________________.

The given function is $f\left(x\right)=8{x}^{2}-7x+5$.

f(x) is a polynomial function.

We know that a polynomial function is everywhere continuous and differentiable. So, f(x) is continuous on [−6, 6] and differentiable on (−6, 6). Thus, both the conditions of Lagrange's mean value theorem are satisfied.

So, there must exist at least one real number c ∈ (−6, 6) such that

$f\text{'}\left(c\right)=\frac{f\left(6\right)-f\left(-6\right)}{6-\left(-6\right)}$

Now, $f\left(x\right)=8{x}^{2}-7x+5$

$⇒f\text{'}\left(x\right)=16x-7$

$\therefore f\text{'}\left(c\right)=\frac{f\left(6\right)-f\left(-6\right)}{6-\left(-6\right)}$

$⇒16c-7=\frac{\left[8×{\left(6\right)}^{2}-7×6+5\right]-\left[8×{\left(-6\right)}^{2}-7×\left(-6\right)+5\right]}{12}$

$⇒16c-7=\frac{-84}{12}=-7$

$⇒16c=0$

$⇒c=0$

Thus, c = 0 ∈ (−6, 6) such that $f\text{'}\left(c\right)=\frac{f\left(6\right)-f\left(-6\right)}{6-\left(-6\right)}$.

Hence, the value of c is 0.

For the function f(x) = 8x2 − 7x + 5, x ∈ [−6, 6], the value of c for the Lagrange's mean value theorem is ___0___.

#### Question 3:

If the function  f(x) = x3 – 6x2 + ax + b defined on [1, 3] satisfies Roll's theorem for c$2+\frac{1}{\sqrt{3}},$ then a = ___________, b = __________.

The given function is f(x) = x3 − 6x2 + ax + b.

It is given that f(x) defined on [1, 3] satisfies Rolle's theorem for $c=2+\frac{1}{\sqrt{3}}$.

$\therefore f\left(1\right)=f\left(3\right)$ and $f\text{'}\left(c\right)=0$

Now,

$f\left(1\right)=f\left(3\right)$

$⇒1-6+a+b=27-54+3a+b$

$⇒-5+a=-27+3a$

$⇒2a=22$

$⇒a=11$

Also,

f(x) = x3 − 6x2 + ax + b

$⇒f\text{'}\left(x\right)=3{x}^{2}-12x+a$

$\therefore f\text{'}\left(c\right)=0$

$⇒3{c}^{2}-12c+a=0$

$⇒3{c}^{2}-12c+11=0$      (a = 11)

Since both equations f(1) = f(3) and $f\text{'}\left(c\right)=3{c}^{2}-12c+11=0$ are independent of b, so b can taken any real value.

a = 11 and b ∈ R

Thus, if the function  f(x) = x3 – 6x2 + ax + b defined on [1, 3] satisfies Roll's theorem for $c=2+\frac{1}{\sqrt{3}}$, then a = 11 and b ∈ R.

If the function  f(x) = x3 – 6x2 + ax + b defined on [1, 3] satisfies Roll's theorem for c$2+\frac{1}{\sqrt{3}},$ then a = ___11___, b = ___R___.

#### Question 4:

It is given that for the function  f(x) = x3 - 6x2 + ax + b on [1, 3], Rolle's theorem holds with c$2+\frac{1}{\sqrt{3}}.$ If f(1) = f(3) = 0, then a =_______, b =________.

The given function is f(x) = x3 − 6x2 + ax + b.

It is given that Rolle's theorem holds for f(x) defined on [1, 3] with $c=2+\frac{1}{\sqrt{3}}$.

$f\left(1\right)=f\left(3\right)=0$     (Given)

$\therefore f\left(1\right)=0$

$⇒1-6+a+b=0$

$⇒a+b=5$        .....(1)

Also,

$f\left(3\right)=0$

$⇒27-54+3a+b=0$

$⇒3a+b=27$     .....(2)

Solving (1) and (2), we get

a = 11 and b = −6

It can be verified that for a = 11 and b = −6, $f\text{'}\left(2+\frac{1}{\sqrt{3}}\right)=0$.

Thus, the values of a and b are 11 and −6, respectively.

It is given that for the function  f(x) = x3 − 6x2 + ax + b on [1, 3], Rolle's theorem holds with c$2+\frac{1}{\sqrt{3}}.$ If f(1) = f(3) = 0, then a = ___11___, b =___−6___.

#### Question 5:

For the function f(x) = logex, ∈ [1, 2], the value of c for the lagrange's mean value theorem is _______________.

The given function is f(x) = logex.

Now, f(x) = logex is differentiable and so continuous for all x > 0. So, f(x) is continuous on [1, 2] and differentiable on (1, 2). Thus, both the conditions of Lagrange's mean value theorem are satisfied.

So, there must exist at least one real number c ∈ (1, 2) such that

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

f(x) = logex

$⇒f\text{'}\left(x\right)=\frac{1}{x}$

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

$⇒\frac{1}{c}=\frac{{\mathrm{log}}_{e}2-{\mathrm{log}}_{e}1}{2-1}$

Thus, $c={\mathrm{log}}_{2}e\in \left(1,2\right)$ such that $f\text{'}\left(c\right)=\frac{f\left(2\right)-f\left(1\right)}{2-1}$.

Hence, the value of c is ${\mathrm{log}}_{2}e$.

For the function f(x) = logex, ∈ [1, 2], the value of c for the Lagrange's mean value theorem is .

#### Question 6:

The value of c in Rolle's theorem for the function  f(x) = x3 - 3x in the interval [0, $\sqrt{3}$] is _______________.

The given function is f(x) = x3 − 3x.

f(x) is a polynomial function. We know that a polynomial function is everywhere continuous and differentiable.

So, f(x) is continuous on $\left[0,\sqrt{3}\right]$ and differentiable on $\left(0,\sqrt{3}\right)$.

Also, f(0) = 0 and $f\left(\sqrt{3}\right)={\left(\sqrt{3}\right)}^{3}-3\sqrt{3}=3\sqrt{3}-3\sqrt{3}=0$

$\therefore f\left(0\right)=f\left(\sqrt{3}\right)$

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

So, there exist a real number c$\left(0,\sqrt{3}\right)$ such that $f\text{'}\left(c\right)=0$.

f(x) = x3 − 3x

$⇒f\text{'}\left(x\right)=3{x}^{2}-3$

$\therefore f\text{'}\left(c\right)=0$

$⇒3{c}^{2}-3=0$

$⇒{c}^{2}=1$

$⇒c=±1$

Thus, c = 1 ∈ $\left(0,\sqrt{3}\right)$ such that $f\text{'}\left(c\right)=0$.

Hence, the value of c is 1.

The value of c in Rolle's theorem for the function  f(x) = x3 − 3x in the interval $\left[0,\sqrt{3}\right]$ is ___1___.

#### 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:

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 4:

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 5:

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 6:

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

It is given thatis 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 thatdoes not vanish anywhere.

Hence proved.

#### Question 7:

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 8:

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 = $\frac{4}{3}$