Exercise 7.6-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 Problems
Find the absolute maximum and absolute minimum values of $f$ on the given interval.
Question 1.
$
f(x)=1-2 x-x^2,[-4,1]
$
Solution:
$
\begin{aligned}
& f(x)=1-2 x-x^2 \\
& f^{\prime}(x)=-2-2 x \\
& f^{\prime}(x)=0 \Rightarrow 2 x=-2 \Rightarrow x=-1 ; \text { The points are }-4,-1,1 \\
& \text { At } x=-4, f(x)=1-2(-4)-(-4)^2=1+8-16=-7 \\
& \text { At } x=-1, f(x)=1-2(-1)-(-1)^2=1+2-1=2 \\
& \text { At } x=1, f(x)=1-2(1)-(1)^2=1-2-1=-2
\end{aligned}
$
Therefore, the absolute minimum is $-7$ and the absolute maximum is 2 .
Question 2.
$
f(x)=x^3-12 x+1,[-3,5]
$
Solution:
$
\begin{aligned}
& f(x)=x^3-12 x+1 \\
& f^{\prime}(x)=3 x^2-12 \\
& f^{\prime}(x)=0 \Rightarrow 3 x^2-12=0 \\
& 3 x^2=12 \Rightarrow x^2=4 \Rightarrow x=\pm 2
\end{aligned}
$
The $\mathrm{x}$ values are $-3,-2,2,5$
$\mathrm{f}(\mathrm{x})($ at $\mathrm{x}=-3)=(-3)^3-12(-3)+1=-27+36+1=10$
$\mathrm{f}(\mathrm{x})(\mathrm{at} \mathrm{x}=-2)=(-2)^3-(12)(-2)+1=-8+24+1=17$
$\mathrm{f}(\mathrm{x})(\mathrm{at} \mathrm{x}=2)=2^3-12(2)+1=8-24+1=-15$
$f(x)($ at $x=5)=53-12(5)+1=125-60+1=66$
From the above four values $10,17,-15$ and 66 , we see that absolute maximum is 66 and the absolute minimum is $-15$.

Question 3.
$
f(x)=\frac{x}{x+1},[1,2]
$
Solution:
$
\begin{aligned}
f(x) & =\frac{x}{x+1} \\
f^{\prime}(x) & =\frac{(x+1)(1)-x(1)}{(x+1)^2}=\frac{x+1-x}{(x+1)^2} \\
f^{\prime}(x) & =0 \Rightarrow \frac{1}{(x+1)^2}=0
\end{aligned}
$
by which we are not getting any ' $x$ ' value.
Now
$
\begin{aligned}
& f(x)(\text { at } x=1)=\frac{1}{1+1}=\frac{1}{2} \\
& f(x)(\text { at } x=2)=\frac{2}{2+1}=\frac{2}{3}
\end{aligned}
$
So, the absolute maximum is $\frac{2}{3}$ and the absolute minimum is $\frac{1}{2}$
Question 4.
$
f(x)=\sin x+\cos x,[0, \pi / 3]
$

$
\begin{aligned}
& \mathrm{f}^{\prime}(\mathrm{x})=\sin \mathrm{x}+\cos \mathrm{x} \\
& \mathrm{f}^{\prime}(\mathrm{x})=\cos \mathrm{x}-\sin \mathrm{x} \\
& \mathrm{f}^{\prime}(\mathrm{x})=0 \Rightarrow \cos \mathrm{x}-\sin \mathrm{x}=0 \\
& \cos x=\sin x \Rightarrow \tan x=1 \quad \text { (or) } x=\frac{\pi}{4} \\
& \text { At } x=0, f(x)=\sin 0+\cos 0=1 \\
& \text { At } x=\frac{\pi}{4}, f(x)=\sin \frac{\pi}{4}+\cos \frac{\pi}{4}=\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{2}}=\frac{2}{\sqrt{2}}=\sqrt{2} \\
& \text { At } x=\frac{\pi}{3}, f(x)=\sin \frac{\pi}{3}+\cos \frac{\pi}{3}=\frac{\sqrt{3}}{2}+\frac{1}{2}=\frac{\sqrt{3+1}}{2}
\end{aligned}
$
From the values $1, \sqrt{2}$ and $\frac{\sqrt{3}+1}{2}$, the absolute maximum is $\sqrt{2}$ and the absolute minimum is 1 .

Question 5.
$
f(x)=x-2 \cos x,[-\pi, \pi]
$
Solution:
$
\begin{aligned}
& \mathrm{f}(\mathrm{x})=\mathrm{x}-2 \cos \mathrm{x} \\
& \mathrm{f}^{\prime}(\mathrm{x})=1+2 \sin \mathrm{x} \\
& \mathrm{f}^{\prime}(\mathrm{x})=0 \Rightarrow 1+2 \sin \mathrm{x}=0 \\
& 2 \sin x=-1 \Rightarrow \sin x=-\frac{1}{2} \Rightarrow x=\frac{-\pi}{6}, \pi-\frac{\pi}{6}=\frac{5 \pi}{6} \\
& \therefore \text { The points are }-\pi, \frac{-\pi}{6}, \frac{5 \pi}{6}, \pi \\
& \text { At } x=-\pi, f(x)=-\pi-2(-1)=-\pi+2 \\
& \text { At } x=-\frac{\pi}{6}, f(x)=\frac{-\pi}{6}-2 \cos \left(\frac{\pi}{6}\right)=\frac{-\pi}{6}-2\left(\frac{\sqrt{3}}{2}\right)=\frac{-\pi}{6}-\sqrt{3} \\
& \text { At } x=\frac{5 \pi}{6}, f(x)=\frac{5 \pi}{6}-2 \cos \frac{5 \pi}{6}=\frac{5 \pi}{6}-2\left[-\cos \frac{\pi}{6}\right]=\frac{5 \pi}{6}+\sqrt{3} \\
& \text { At } x=\pi, \quad f(x)=\pi-2(-1)=\pi+2
\end{aligned}
$
From the above values, absolute maximum is $\pi+2$ and the absolute minimum is $\frac{-\pi}{6}-\sqrt{3}$. From the above values, absolute maximum is $\pi+2$ and the absolute minimum is $\frac{-\pi}{6}-\sqrt{3}$
From the local maximum and minimum values of the following functions.
Question 6.
$
2 x^3+5 x^2-4 x
$
Solution:

$\begin{aligned}
& y=f(x)=2 x^3+5 x^2-4 x \\
& \frac{d y}{d x}=6 x^2+10 x-4 ; \frac{d^2 y}{d x^2}=12 x+10 \\
& \frac{d y}{d x}=0 \Rightarrow 6 x^2+10 x-4=0 \\
& \therefore \quad \div \text { by } 2) \quad 3 x^2+5 x-2=0 \\
& 3 x^2+6 x-x-2=0 \\
& 3 x(x+2)-(x+2)=0 \\
& \text { At } x=\frac{1}{3}, \frac{d^2 y}{d x^2}=12\left(\frac{1}{3}\right)+10=14>0 \\
& \Rightarrow x=\frac{1}{3} \text { is a minimum point. } \\
& y \text { at } x=\frac{1}{3}=2\left(\frac{1}{27}\right)+5\left(\frac{1}{9}\right)-4\left(\frac{1}{3}\right)=\frac{2}{27}+\frac{5}{9}-\frac{4}{3} \\
& =\frac{2+15-36}{27}=\frac{-19}{27} \\
& \frac{d^2 y}{d x^2}(\text { at } x=-2)=12(-2)+10=-24+10=-14<0 \\
& \Rightarrow x=-2 \text { is a maximum point. } \\
& \therefore y(\text { at } x=-2)=2(-8)+5(4)-4(-2)=-16+20+8=12 \\
&
\end{aligned}$

Question 7.
$t+\cos t$
Solution:
$
\begin{aligned}
& y=t+\cos t ; \frac{d y}{d x}=1-\sin t \\
& \frac{d^2 y}{d x^2}=-\cos t \\
& \frac{d y}{d t}=0 \Rightarrow 1-\sin t=0 \\
& \sin t=1 \Rightarrow t=\frac{\pi}{2}
\end{aligned}
$
At $t=\frac{\pi}{2}, \frac{d^2 y}{d t^2}=-\cos \frac{\pi}{2}=0$; Since $\frac{d^2 y}{d x^2}=0$ at $t=\frac{\pi}{2}$
$\mathrm{t}=\frac{\pi}{2}$ is neither a maximum point nor a minimum point. So there is no maximum or minimum.