Exercise 7.7-Additional Problems - Chapter 7 - Applications of Differential Calculus - 12th Maths Guide Samacheer Kalvi Solutions - Tamil Medium

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Additional Problems
Question 1.
Discuss the curves $y=x^4-4 x^3$ with respect to concavity and points of inflection.
Solution:
$
\begin{aligned}
& f(x)=x^4-4 x^3 \Rightarrow f^{\prime}(x)=4 x^3-12 x^2 \\
& f^{\prime}(x)=12 x^2-24 x=12 x(x-2)
\end{aligned}
$
Since $f^{\prime}(x)=0$ when $x=0$ or 2 , we divide the real line into three intervals.
$(-\infty, 0),(0,2),(2, \infty)$ and complete the following chart.

The point $(0, f(0)$ i.e., $(0,0)$ is an inflection point since the curve changes from concave upward to concave downward there. Also $(2, f(2))$ i.e., $(2,-16)$ is an inflection point since the curve changes from concave downward to concave upward there.

Question 2.
Find the intervals of concavity and the points of inflection of the following functions.
$
f(x)=2 x^3+5 x^2-4 x
$
Solution:
$
\begin{aligned}
& y=f(x)=2 x^3+5 x^2-4 x \\
& f^{\prime}(x)=6 x^2+10 x-4 \\
& f^{\prime}(x)=12 x+10 ; f^{\prime \prime}(x)=12 \\
& f^{\prime}(x)=0 \Rightarrow 12 x+10=0 \\
& 12 x=-10 \text { (or }) x=\frac{-10}{12}=\frac{-5}{6} \\
& \text { At } x=\frac{-5}{6}, f^{\prime \prime \prime}(x)=12 \neq 0
\end{aligned}
$
So, the point of inflection is at $x=\frac{-5}{6}$.
$
\text { At } \begin{aligned}
x & =\frac{-5}{6}, y=2\left(\frac{-5}{6}\right)^3+5\left(\frac{-5}{6}\right)^2-4\left(\frac{-5}{6}\right) \\
& =2\left(\frac{-125}{216}\right)+5\left(\frac{25}{36}\right)+\frac{20}{6}=\frac{-125}{108}+\frac{125}{36}+\frac{20}{6} \\
& =\frac{-125+375+360}{108} \quad=\frac{610}{108}=\frac{305}{54}
\end{aligned}
$
The point of inflection is $\left(\frac{-5}{6}, \frac{305}{54}\right)$
$
\begin{aligned}
f^{\prime \prime}(x) & =0 \Rightarrow 12 x+10=0 \\
12 x & =-10 \\
, x & =\frac{-10}{12}=\frac{-5}{6}
\end{aligned}
$
So, the intervals are $(-\infty,-5 / 6),(-5 / 6, \infty)$
Consider $\mathrm{x}$ in $(-\infty,-5 / 6)$ say $\mathrm{x}=-1$
$f^{\prime}(x)=-12+10=-2<0 \Rightarrow$ the curve convex upward in the interval $(-\infty,-5 / 6)$ Consider $x$ in $(-5 / 6, \infty)$ say $x=0$ $\mathrm{f}^{\prime}(\mathrm{x})=0+10=10>0$
$\Rightarrow$ the curve is concave upward in $(-5 / 6,-\infty)$
Thus, the curve is concave upward in $(-5 / 6, \infty)$ and convex upward in $(-\infty,-5 / 6)$
The point of inflection is $\left(\frac{-5}{6}, \frac{305}{54}\right)$

Question 3.
$
f(x)=x^4-6 x^2
$
Solution:
$
\begin{aligned}
& f(x)=x^4-6 x^2 ; f^{\prime}(x)=4 x^3-12 x ; f^{\prime \prime}(x)=12 x^2-12 \\
& f^{\prime \prime}(x)=24 x \\
& \left\{\begin{aligned}
f^{\prime \prime}(x) & =0 \Rightarrow 12 x^2=12 \\
x^2 & =1 \\
x & =\pm 1 \\
\text { At } x & =\pm 1 \\
f^{\prime \prime}(x) & =24(\pm 1)=\pm 24 \neq 0
\end{aligned}\right\} \\
&
\end{aligned}
$
So, the points of inflection are at $\mathrm{x}=\pm 1$.
$
f^{\prime}(x)=0 \Rightarrow x^2=1 \Rightarrow x=\pm 1
$
So, the intervals are $(-\infty,-1),(-1,1),(1, \infty)$
When $x \in(-\infty,-1)$, say $x=-2$
$
f^{\prime}(x)=12(-2)^2-12=48-12=36>0
$
$\Rightarrow \mathrm{f}(\mathrm{x})$ is concave upward in $(-\infty,-1)$ when $x \in(-1,1)$ say $x=0$.
$\mathrm{f}^{\prime}(\mathrm{x})=0-12=-12<0$
$\Rightarrow \mathrm{f}(\mathrm{x})$ convex upward.
$\Rightarrow \mathrm{x}=-1$ is a point of inflection. Again when $\mathrm{x} \in(-1,1)$, the curve is convex upward and when $\mathrm{x} \in(1, \infty)$ say $\mathrm{x}=$
2.
$f^{\prime \prime}(x)=(1, \infty)$ say $x=2$.
$\mathrm{f}^{\prime \prime}(\mathrm{x})=12(4)-12>0$
$\Rightarrow$ the curve is concave upward.
$\Rightarrow \mathrm{x}=1$ is a point of inflection.
Thus $\mathrm{x}=\pm 1$ are the points of inflection.
At $x=1, f(x)=1-6=-5$
At $x=-1, f(x)=(-1)^4-6(-1)^2=1-6=-5$
The curve concave upward in $(-\infty,-1) \mathrm{u}(1, \infty)$ and convex upward in $(-1,1)$ and the points of inflection are (1, -5) and $(-1,-5)$.

Question 4.
$
f(\theta)=\sin 2 \theta \text { in }(\theta, \pi)
$
Solution:
$
f^{\prime}(\theta)=\sin 2 \theta
$
$f^{\prime}(\theta)=(\cos 2 \theta)(2)=2 \cos 2 \theta$
$f^{\prime \prime}(\theta)=2[-\sin 2 \theta](2)=-4 \sin 2 \theta$

$
\begin{aligned}
f^{\prime}(\theta)=0 & \Rightarrow-4 \sin 2 \theta=0 \\
2 \theta & =0 \text { (or) } 2 \theta=\pi ; \text { So, } \theta=\frac{\pi}{2}
\end{aligned}
$
So, the intervals are $(0, \pi / 2),(\pi / 2, \pi)$ when $\theta$ is in $(0, \pi / 2)$, say $\theta=\pi / 4$.
$
f^{\prime \prime}(\theta)=-4 \sin 2\left(\frac{\pi}{4}\right)=-4 \sin \left(\frac{\pi}{2}\right)=-4<0
$
$\Rightarrow f(\theta)$ is convex upward in $\left(0, \frac{\pi}{2}\right)$
When $\theta$ is in the interval $\left(\frac{\pi}{2}, \pi\right)$ say $\theta=\frac{3 \pi}{4}$
$
f^{\prime \prime}(\theta)=-4 \sin 2\left(\frac{3 \pi}{4}\right)=-4 \sin \left(\frac{3 \pi}{2}\right)=-4(-1)=+4>0
$
$\Rightarrow f(\theta)$ is concave upward in the interval $\left(\frac{\pi}{2}, \pi\right)$
So, at $\theta=\frac{\pi}{2}$ we get a point of inflection. At $\theta=\frac{\pi}{2}, f(\theta)=\sin 2\left(\frac{\pi}{2}\right)=0$
So, the curve is concave upward in $\left(\frac{\pi}{2}, \pi\right)$ and concave downward in $\left(0, \frac{\pi}{2}\right)$.
The point of inflection is $\left(\frac{\pi}{2}, 0\right)$.

Question 5.
$
y=12 x^2-2 x^3-x^4
$
Solution:
$
\begin{aligned}
y & =12 x^2-2 x^3-x^4 \\
\frac{d y}{d x} & =24 x-6 x^2-4 x^3 \\
\frac{d^2 y}{d x^2} & =24-12 x-12 x^2 \\
\frac{d^2 y}{d x^2} & =0 \Rightarrow 24-12 x-12 x^2=0 \\
12\left(2-x-x^2\right) & =0 \Rightarrow x^2+x-2=0 \\
(x+2)(x-1) & =0 \\
x & =-2 \text { or } 1
\end{aligned}
$
So the intervals are $(-\infty,-2),(-2,1)(1, \infty)$
When $\mathrm{x} \in(-\infty,-2)$ say $\mathrm{x}=-3$
$
\frac{d^2 y}{d x^2}=12(2+3-9)=12(-4)>0
$
The curve is concave upward.
When $\mathrm{x} \in(-2,1)$ say $\mathrm{x}=0$.

$
\begin{aligned}
& \frac{d^2 y}{d x^2}=12(2-0-0)=24>0 \\
& \Rightarrow \text { The curve is concave upward. } \\
& x=-2 \text { is a point of inflection. } \\
& \text { When } x \in(1, \infty) \text { say } x=2 \text {. } \\
& \qquad \frac{d^2 y}{d x^2}=12(2-2-4)=12(-4)<0
\end{aligned}
$
$\Rightarrow$ The curve is convex upward. So, $x=1$ is a point of inflection. At $x=-2, y=12(4)-2(-8)-(16)=48+16-16=48$ At $x=1, y=12(1)-2(1)-1=12-2-1=9$
So, the points of inflection are $(1,9)$ and $(-2,48)$