Exercise 7.3-Additional Problems - Chapter 7 Applications of Differential Calculus 12th Maths Guide Samacheer Kalvi Solutions - SaraNextGen [2024-2025]

Updated By SaraNextGen

On April 24, 2024, 11:35 AM

Additional Questions Solved
Question 1.
Verify Rolle's theorem for the following:
(i) $f(x)=x^3-3 x+3,0 \leq x \leq 1$
(ii) $f(x)=\tan x, 0 \leq x \leq \pi$
(iii) $f(x)=|x|,-1 \leq x \leq 1$
(iv) $f(x)=\sin ^2 x, 0 \leq x \leq \pi$
(v) $f(x)=e^x \sin x, 0 \leq x \leq \pi$
(vi) $f(x)=x(x-1)(x-2), 0 \leq x \leq 2$
Solution:
(i) $f(x)=x^3-3 x+3,0 \leq x \leq 1$. $f$ is continuous on $[0,1]$ and differentiable in $(0,1)$ $\mathrm{f}(0)=3$ and $\mathrm{f}(1)=1 \therefore \mathrm{f}(\mathrm{a}) \neq \mathrm{f}(\mathrm{b})$
$\therefore$ Rolle's theorem, does not hold, since $\mathrm{f}(\mathrm{a})=\mathrm{f}(\mathrm{b})$ is not satisfied.
Also note that $f^{\prime}(x)=3 x^2-3=0 \Rightarrow x^2=1 \Rightarrow x=\pm 1$
There exists no point $c \in(0,1)$ satisfying $f^{\prime}(c)=0$.
(ii) $f(x)=\tan x, 0 \leq x \leq \pi$.
$\mathrm{f}^{\prime}(\mathrm{x})$ is not continuous in $[0, \pi]$ as $\tan \mathrm{x}$ tends to $+\infty$ at $\mathrm{x}=\frac{\pi}{2}$,
$\therefore$ Rolle's theorem is not applicable.
(iii) $f(x)=|x|,-1 \leq x \leq 1$
$f$ is continuous in $[-1,1]$ but not differentiable in $(-1,1)$ since $f^{\prime}(0)$ does not exist.
$\therefore$ Rolle's theorem is not applicable.
(iv) $f(x)=\sin ^2 x, 0 \leq x \leq \pi$
$f$ is continuous in $[0, \pi]$ and differentiable in $(0, \pi) . f(0)=f(\pi)=0$
(i.e.,) f satisfies hypothesis of Rolle's theorem.
$f^{\prime}(x)=2 \sin x \cos x=\sin 2 x$
$f^{\prime}(c)=0 \Rightarrow \sin 2 c=0 \Rightarrow 2 c=0, \pi, 2 \pi, 3 \pi, \ldots \Rightarrow c=0, \frac{\pi}{2}, \pi, \frac{3 \pi}{2}, \ldots$
Since $c=\frac{\pi}{2} \in(0, \pi)$, the suitable $c$ of Rolle's theorem is $c=\frac{\pi}{2}$.

(v) $f(x)=e^x, \sin x, 0 \leq x \leq \pi$
$\mathrm{e}^{\mathrm{x}}$ and $\sin \mathrm{x}$ are continuous for all $\mathrm{x}$, therefore the product $\mathrm{e}^{\mathrm{x}} \sin \mathrm{x}$ is continuous in $0 \leq \mathrm{x} \leq \pi$.
$f^{\prime}(x)=e^x \sin x+e^x \cos x=e^x(\sin x+\cos x)$ exist in $0<x<\pi \Rightarrow f^{\prime}(x)$ is differentiable in $(0, \pi)$
$f(0)=e^{\circ} \sin 0=0$
$\mathrm{f}(\pi)=\mathrm{e}^\pi \sin \pi=0$
$\therefore \mathrm{f}$ satisfies hypothesis of Rolle's theorem.
Thus there exists $c \in(0, \pi)$ satisfying $f^{\prime}(c)=0 \Rightarrow e^c(\sin c+\cos c)=0$
$\Rightarrow e^c=0$ or $\sin c+\cos c=0$
$\Rightarrow \mathrm{e}^{\mathrm{c}}=0 \Rightarrow \mathrm{c}=-\infty$ which is not meaningful here.
$\Rightarrow \sin c=-\cos c \Rightarrow \frac{\sin c}{\cos c}=-1 \Rightarrow \tan c=-1=\tan \frac{3 \pi}{4}$
$\Rightarrow c=\frac{3 \pi}{4}$ is the required point.
(vi) $f(x)=x(x-1)(x-2), 0 \leq x \leq 2$
$f$ is not continuous in $[0,2]$ and differentiable in $(0,2)$
$f(0)=0=f(2)$, satisfying hypothesis of Rolle's theorem.
Now $f^{\prime}(x)=(x-1)(x-2)+x(x-2)+x(x-1)=0$
$
\Rightarrow 3 x^2-6 x+2=0 \Rightarrow x=1 \pm \frac{1}{\sqrt{3}}
$
The required $c$ in Rolle's theorem is $1 \pm \frac{1}{\sqrt{3}} \in(0,2)$
Note: There could exist more than one such ' $c$ ' appearing in the statement of Rolle's theorem.

Question 2.
Suppose that $f(0)=-3$ and $f^{\prime}(x) \leq 5$ for all values of $x$, how large can $f(2)$ possibly be?
Solution:
Since by hypothesis $\mathrm{f}$ is differentiable, $\mathrm{f}$ is continuous everywhere. We can apply Lagrange's Law of the mean on the interval $[0,2]$, There exist atleast one ' $c$ ' $\in(0,2)$ such that
$\mathrm{f}(2)-\mathrm{f}(0)=\mathrm{f}^{\prime}(\mathrm{c})(2-0)$
$f(2)=f(0)+2 f^{\prime}(c)=-3+2 f^{\prime}(c)$
Given that $\mathrm{f}^{\prime}(\mathrm{x}) \leq 5$ for all $\mathrm{x}$.
In particular we know that $\mathrm{f}^{\prime}(\mathrm{c}) \leq 5$. Multiplying both sides of the inequality by 2 , we have $2 f^{\prime}(\mathrm{c}) \leq 10$
$f(2)=-3+2 f^{\prime}(\mathrm{c})<-3+10=7$
i.e., the largest possible value of $\mathrm{f}(2)$ is 7 .

Question 3.
Using Rolle's theorem find the points on the curves $=x^2+1,-2 \leq x \leq 2$ where the tangent is parallel to $\mathrm{X}$-axis
Solution:
$
\begin{aligned}
& y=x^2+1 \Rightarrow \frac{d y}{d x}=2 x \\
& \mathrm{a}=-2, \\
& b=2 \\
& \mathrm{f}(\mathrm{x})=\mathrm{x}^2+1 \\
& \mathrm{f}(\mathrm{a})=\mathrm{f}(-2)=4+1=5 \\
& \mathrm{f}(\mathrm{b})=\mathrm{f}(2)=4+1=5 \\
& \text { So, } \mathrm{f}(\mathrm{a})=\mathrm{f}(\mathrm{b}) \\
& \left.\mathrm{f}^{(} \mathrm{x}\right)=2 \mathrm{x} \\
& \mathrm{f}^{\prime}(\mathrm{x})=0 \Rightarrow 2 \mathrm{x}=0
\end{aligned}
$
$\mathrm{x}=0$ where $0 \in(-2,2)$
at $\mathrm{x}=0, \mathrm{y}=0+1=1$
So, the point is $(0,1)$ at $(0,1)$ the tangent drawn is parallel to $\mathrm{X}$-axis

Question 4.
Find ' $C$ ' of Lagrange's mean value theorem for the function $f(x)=2 x^3+x^2-x-1,[0,2]$

Solution:
$
\begin{aligned}
& f(x)=2 x^3+x^2-x-1 \\
& a=0 \\
& b=2 \\
& f(a)=f(0)=-1 \\
& f(b)=f(2)=2(8)=4-2-1=16+4-2-1=17
\end{aligned}
$
By Lagrange's mean value theorem, we get a constant $\mathrm{c} \in(\mathrm{a}, \mathrm{b})$ such that
$
\begin{aligned}
f^{\prime}(c) & =\frac{f(b)-f(a)}{b-a} \\
\text { i.e., } f^{\prime}(c) & =\frac{17-(-1)}{2-0}=\frac{18}{2}=9 \\
f(x) & =2 x^3+x^2-x-1 \\
f^{\prime}(x) & =6 x^2+2 x-1 \\
f^{\prime}(c) & =6 c^2+2 c-1
\end{aligned}
$
From (1) and (2),
$
\begin{aligned}
6 c^2+2 c-1 & =9 \\
\Rightarrow 6 c^2+2 c-1-9 & =0 \text { i.e., } 6 c^2+2 c-10=0 \\
(\div \text { by } 2) \Rightarrow 3 c^2+c-5 & =0 \\
c & =\frac{-1 \pm \sqrt{1-4(3)(-5)}}{2(3)} \\
\text { i.e., } c & =\frac{-1 \pm \sqrt{1+60}}{6} ; \text { i.e., } c=\frac{-1 \pm \sqrt{61}}{6} \\
\text { Here, } c & =\frac{-1+\sqrt{61}}{6} \text { where } \frac{-1+\sqrt{61}}{6} \in(0,2)
\end{aligned}
$
Thus, Lagrange's mean value theorem is true with $c=\frac{-1+\sqrt{61}}{6}$

Question 5.
Find ' $C$ ' of Lagrange's mean value theorem for the function $f(x)=x^3+x^2-3 x$ in $[1,3]$

Solution:
$
f(x)=x^3+x^2-3 x
$
$
\begin{aligned}
& \begin{array}{l}
\mathrm{a}=1, \\
\mathrm{~b}=3
\end{array} \\
& \mathrm{f}(\mathrm{a})=\mathrm{f}(1)=1-5-3=-7 \\
& \mathrm{f}(\mathrm{b})=\mathrm{f}(3)=27-5(9)-3(3) \\
& =27-45-9=-27 \\
& f^{\prime}(c)=\frac{f(b)-f(a)}{b-a}=\frac{-27-(-7)}{3-1} \\
& \quad=\frac{-27+7}{2}=\frac{-20}{2}=-10 \\
& \quad=x^3-5 x^2-3 x \\
& \mathrm{f}^{\prime}(\mathrm{x})=3 \mathrm{x}^2-10 \mathrm{x}-3 \\
& \mathrm{f}^{\prime}(\mathrm{x})=3 \mathrm{c}^2-10 \mathrm{c}-3 \\
& \text { From }(1) \text { and }(2), \\
& 3 \mathrm{c}^2-10 \mathrm{c}-3=-10 \\
& 3 \mathrm{c}^2-10 \mathrm{c}-3+10=0 \\
& 3 \mathrm{c}^2-10 \mathrm{c}+7=0 \quad 0 \Rightarrow c=\frac{7}{3} \text { (or) } 1 \\
& 3 \mathrm{c}^2-3 \mathrm{c}-7 \mathrm{c}+7=0 \\
& (3 c-7)(c-1)=\frac{7}{3} \in[1,3]
\end{aligned}
$
So, Lagrange's mean value theorem is true with $\mathrm{c}=\frac{7}{3}$