Exercise 7.3 - 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

Ex $7.3$
Queston 1.
Explain why Rolle's theorem is not applicable to the following functions in the respective intervals.
(i) $f(x)=\left|\frac{1}{x}\right|, x \in[-1,1]$
(ii) $f(x)=\tan x, x \in[0, \pi]$
(iii) $f(x)=x-2 \log x, x \in[2,7]$
Solution:
(i) $f(x)=\left|\frac{1}{x}\right|$
$f(x)$ is not continuous at $x=0$. So Rolle's Theorem is not applicable.
(ii) $f(x)=\tan x$
$f(x)$ is not continuous at $x=\frac{\pi}{2}$. So Rolle's Theorem is not applicable..
(iii) $f(x)=x-2 \log x$
$f(x)=x-2 \log x$
$f(2)=2-2 \log 2=2-\log 4$
$f(7)=7-2 \log 7=7-\log 49$
$\mathrm{f}(2) \neq \mathrm{f}(7)$
So Rolle's theorem is not applicable.
Question 2.
Using the Rolle's theorem, determine the values of $\mathrm{x}$ at which the tangent is parallel to the $\mathrm{x}$-axis for the following functions:
(i) $f(x)=x^2-x, x \in[0,1]$
(ii) $f(x)=\frac{x^2-2 x}{x+2}, x \in[-1,6]$
(iii) $f(x)=\sqrt{x}-\frac{x}{2}, x \in[0,9]$
Solution:
Tangent is parallel to $x$ axis. So $\frac{d y}{d x}=0$
(i) $f(x)=x^2-x$
$
\begin{aligned}
& f^{\prime}(x)=2 x-1 \\
& f^{\prime}(x)=0 \Rightarrow 2 x-1=0 \Rightarrow x=\frac{1}{2} \in[0,1]
\end{aligned}
$

(ii) $f(x)=\frac{x^2-2 x}{x+2}$
$
\begin{aligned}
f^{\prime}(x) & =\frac{(x+2)(2 x-2)-\left(x^2-2 x\right)(1)}{(x+2)^2}=\frac{x^2+4 x-4}{(x+2)^2} \\
f^{\prime}(x) & =\frac{x^2+4 x-4}{(x+2)^2} \\
f^{\prime}(x) & =0 \Rightarrow x^2+4 x-4=0 \\
x & =\frac{-4 \pm \sqrt{16+16}}{2}=\frac{-4 \pm 4 \sqrt{2}}{2}=-2 \pm 2 \sqrt{2}
\end{aligned}
$
But $x \in[-1,6]$
$
\therefore x=-2 \pm 2 \sqrt{2}
$
(iii) $f(x)=\sqrt{x}-\frac{x}{3}$
$
\begin{aligned}
f^{\prime}(x) & =\frac{1}{2 \sqrt{x}}-\frac{1}{3} \\
f^{\prime}(x) & =0 \Rightarrow \frac{1}{2 \sqrt{x}}-\frac{1}{3}=0 \\
\Rightarrow \frac{1}{2 \sqrt{x}} & =\frac{1}{3} \Rightarrow 2 \sqrt{x}=3 \Rightarrow \sqrt{x}=\frac{3}{2} \Rightarrow x=\frac{9}{4} \\
x & =\frac{9}{4} \in[0,9]
\end{aligned}
$
Question 3.
Explain why Lagrange's mean value theorem is not applicable to the following functions in the respective intervals :
(i) $f(x)=\frac{x+1}{x}, x \in[-1,2]$
(ii) $f(x)=|3 x+1|, x \in[-1,3]$
Solution:
(i) $f(x)=\frac{x+1}{x}$

The function is not continuous at $x=0$. So Lagrange's mean value theorem is not applicable in the given interval.
(ii) $f(x)=|3 x+1|, x \in[-1,3]$
The function is not differentiable at $x=\frac{-1}{3}$. So Lagrange's mean value theorem is not applicable in the given interval.
Question 4.
Using the Lagrange's mean value theorem determine the values of $x$ at which the tangent is parallel to the secant line at the end points of the given interval:
(i) $f(x)=x^3-3 x+2, x \in[-2,2]$
(ii) $f(x)=(x-2)(x-7), x \in[3,11]$
Solution:
(i) $f(x)=x^3-3 x+2$
Here $\mathrm{a}=-2, \mathrm{~b}=2$
$
\begin{aligned}
f(a) & =f(-2)=0 \\
f(b) & =f(2)=4 \\
f^{\prime}(c) & =\frac{f(b)-f(a)}{b-a}=\frac{4-0}{2+2}=\frac{4}{4}=1 \\
f(x) & =x^3-3 x+2 \\
f^{\prime}(x) & =3 x^2-3 \\
f^{\prime}(c) & =3 c^2-3
\end{aligned}
$
from (1) and (2) $\quad 3 c^2-3=1 \Rightarrow 3 c^2=4$
$
c^2=\frac{4}{3} \Rightarrow c=\pm \frac{2}{\sqrt{3}} \in[-2,2]
$
(ii) $f(x)=(x-2)(x-7)$
$
\begin{aligned}
f(x) & =x^2-9 x+14 \\
f^{\prime}(x) & =2 x-9 \\
f^{\prime}(c) & =2 c-9
\end{aligned}
$
Here $a=3, b=11$

$
\begin{aligned}
f(a) & =f(3)=-4 \\
f(b) & =f(11)=36 \\
\frac{f(b)-f(a)}{b-a} & =\frac{36+4}{11-3}=5 \\
\text { But } f^{\prime}(c) & =\frac{f(b)-f(a)}{b-a} \Rightarrow 2 c-9=5 \\
2 c & =5+9=14 \\
c & =7 \in[3,11]
\end{aligned}
$
Question 5.
Show that the value in the conclusion of the mean value theorem for
(i) $f(x)=\frac{1}{x}$ on a closed interval of positive numbers $[a, b]$ is $\sqrt{a b}$
(ii) $f(x)=\mathbf{A} x^2+\mathbf{B} x+\mathbf{C}$ on any interval $[a, b]$ is $\frac{a+b}{2}$
Solution:
$
\begin{aligned}
f(x) & =\frac{1}{x} x \in[a, b] \\
f(a) & =\frac{1}{a} ; f(b)=\frac{1}{b} \\
f^{\prime}(c) & =\frac{f(b)-f(a)}{b-a}=\frac{1}{b}-\frac{1}{a} \\
& =\frac{a-b}{(b-a) a b}=\frac{-1}{a b}
\end{aligned}
$

(i)
Now $f(x)=\frac{1}{x} \Rightarrow f^{\prime}(x)=-\frac{1}{x^2}$
So, $f^{\prime}(c)=\frac{-1}{c^2}$
from (1) and (2) $\frac{-1}{x^2}=\frac{-1}{a b} \Rightarrow x^2=a b \Rightarrow x=\pm \sqrt{a b}$ but $a$ and $b$ are positive integers
So, $x=\sqrt{a b}$

(ii)
$
\begin{aligned}
f(x) & =\mathrm{A} x^2+\mathrm{B} x+\mathrm{C}, x \in[a, b] \\
f^{\prime}(x) & =2 \mathrm{~A} x+\mathrm{B} \\
f^{\prime}(k) & =2 \mathrm{~A} k+\mathrm{B} \\
f(x) & =\mathrm{A} x^2+\mathrm{B} x+\mathrm{C}
\end{aligned}
$
So, $f(a)=\mathrm{A} a^2+\mathrm{B} a+\mathrm{C}$
and $f(b)=\mathrm{A} b^2+\mathrm{B} b+\mathrm{C}$
So, $\frac{f(b)-f(a)}{b-a}=\frac{\mathrm{A} b^2+\mathrm{B} b+\mathrm{C}-\left(\mathrm{A} a^2+\mathrm{B} a+\mathrm{C}\right)}{b-a}$ $=\frac{\mathrm{A} b^2+\mathrm{B} b+\mathrm{C}-\mathrm{A} a^2-\mathrm{B} a-\mathrm{C}}{b-a}$ $=\frac{\mathrm{A}(b+a)(b-a)+\mathrm{B}(b-a)}{b-a}$ $=\mathrm{A}(a+b)+\mathrm{B}$
$
\begin{aligned}
\text { But, } f^{\prime}(k) & =\frac{f(b)-f(a)}{b-a} \\
(1) & =(2) \\
\Rightarrow 2 \mathrm{~A} k+\mathrm{B} & =\mathrm{A}(a+b)+\mathrm{B} \\
2 \mathrm{~A} k & =\mathrm{A}(a+b) \\
\Rightarrow k & =\frac{a+b}{2}
\end{aligned}
$

Question 6.
A race car driver is racing at 20th $\mathrm{km}$. If his speed never exceeds $150 \mathrm{~km} / \mathrm{hr}$, what is the maximum distance he can cover in the next two hours.
Solution:
By Langrange's Mean Volume theorem, $f^{\prime}(c)=\frac{f(b)-f(a)}{b-a}$
Here the interval is $[0,2]$ and $f(0)=20, f(2)=$ ?
f(b) $\left.-f(a) \leq(b-a) f^{\prime} c\right)$
here $f(a)=20$
$\Rightarrow \mathrm{f}(\mathrm{b})-20 \leq 150(2-0)$
$\Rightarrow \mathrm{f}(\mathrm{b}) \leq 320$
(i.e) $f(2)=320 \mathrm{~km}$.
Question 7.
Suppose that for a function $f(x), f^{\prime}(x) \leq 1$ for all $1 \leq x \leq 4$. Show that $f(4)-f(1) \leq 3$.
Solution:
$
\begin{aligned}
& f^{\prime}(x) \leq 1 \text { for } 1 \leq x \leq 4 \\
& \quad \Rightarrow \frac{f(4)-f(1)}{4-1} \leq 1 \\
& \left.\therefore \therefore f^{\prime}(c)=\frac{f(b)-f(a)}{b-a}\right) \Rightarrow f(4)-f(1) \leq 3
\end{aligned}
$
Question 8.
Does there exist a differentiable function $f(x)$ such that $f(0)=-1, f(2)=4$ and $f^{\prime}(x) \leq 2$ for all $x$. Justify your answer.
Solution:
$f(0)=-1, f(2)=4, f(x) \leq 2$
We know $f^{\prime}(x)=\frac{f(b)-f(a)}{b-a}$
Here $\mathrm{a}=0, \mathrm{~b}=2$
$
\begin{aligned}
f^{\prime}(x) & =\frac{f(2)-f(0)}{2-0}=\frac{4-(-1)}{2-0} \\
& =\frac{4+1}{2}=\frac{5}{2}=2.5 \notin[0,2]
\end{aligned}
$

So this is not possible
Question 9.
Show that there lies a point on the curve $f(x)=x(x+3) e^{\frac{\pi}{2}},-3 \leq x \leq 0$ where tangent drawn is parallel to the $\mathrm{x}$-axis.
Solution:
$
\begin{array}{r}
f(x)=x(x+3) e^{\frac{\pi}{2}} x \in[-3,0] \\
f(x)=\left(x^2+3 x\right) e^{\frac{\pi}{2}}
\end{array}
$
The tangent is parallel to $x$ axis
$
\begin{aligned}
\Rightarrow \frac{d y}{d x} & =0(\text { i.e }) \Rightarrow f(x)=0 \\
\text { Now } f^{\prime}(x) & =(2 x+3) e^{\frac{\pi}{2}} \\
f^{\prime}(x) & =0 \Rightarrow 2 x+3=0 \Rightarrow x=-\frac{3}{2} \in[-3,0]
\end{aligned}
$
$\Rightarrow$ There lies a point in $[-3,0]$, where tangent is parallel to $\mathrm{x}$ axis.
Question 10.
Using mean value theorem prove that for, $\mathrm{a}>0, \mathrm{~b}>0,\left|\mathrm{e}^{-\mathrm{a}}-\mathrm{e}^{-\mathrm{b}}\right|<|\mathrm{a}-\mathrm{b}|$
Solution:
$
\begin{aligned}
\text { Let } f(x) & =e^{-x} \\
f^{\prime}(x) & =-e^{-x} \\
\text { Now } f^{\prime}(c) & =\frac{f(b)-f(a)}{b-a}=\left|\frac{e^{-b}-e^{-a}}{b-a}\right| \leq\left|-e^{-c}\right| \leq 1 \Rightarrow \frac{\left|e^{-a}-e^{-b}\right|}{|a-b|} \leq 1 \\
& \Rightarrow\left|e^{-a}-e^{-b}\right|<|a-b|
\end{aligned}
$