Exercise 7.10 - 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 $7.10$
Choose the correct or the most suitable answer.
Question 1.
The volume of a sphere is increasing in volume at the rate of $3 \pi \mathrm{cm}^3 / \mathrm{sec}$. The rate of change of its radius when radius is $\frac{1}{2} \mathrm{~cm} \ldots .$.
(a) $3 \mathrm{~cm} / \mathrm{s}$
(b) $2 \mathrm{~cm} / \mathrm{s}$
(c) $1 \mathrm{~cm} / \mathrm{s}$
(d) $\frac{1}{2} \mathrm{~cm} / \mathrm{s}$
Solution:
(a) $3 \mathrm{~cm} / \mathrm{s}$
Hint:
$
\text { Hint: } \begin{aligned}
v & =\frac{4}{3} \pi r^3 \\
\cdot, \frac{d v}{d t} & =\frac{4}{3} \pi\left(3 r^2 \frac{d r}{d t}\right)
\end{aligned}
$
Given $\frac{d v}{d t}=3 \pi$ and $r=\frac{1}{2}$ we get
$
\begin{aligned}
3 \pi & =\frac{4}{3} \pi(3)\left(\frac{1}{4}\right) \frac{d r}{d t} \\
\frac{d r}{d t} & =3
\end{aligned}
$
Question 2.
A balloon rises straight up at $10 \mathrm{~m} / \mathrm{s}$. An observer is $40 \mathrm{~m}$ away from the spot where the balloon left the ground. Find the rate of change of the balloon's angle of elevation in radian per second when the balloon is 30 metres above the ground.
(1) $\frac{3}{25} \mathrm{radians} / \mathrm{sec}$
(2) $\frac{4}{25}$ radians/sec
(3) $\frac{1}{5}$ radians $/ \mathrm{sec}$
(4) $\frac{1}{3}$ radians $/ \mathrm{sec}$
Solution:
(b) $\frac{4}{25} \mathrm{radian} / \mathrm{sec}$

Hint:
(1) $\frac{3}{25}$ radians/sec
(2) $\frac{4}{25}$ radians/sec
(3) $\frac{1}{5}$ radians/sec
(4) $\frac{1}{3}$ radians $/ \mathrm{sec}$
$
\text { (i.e.,) } \begin{aligned}
10 & =40\left(\frac{50}{40}\right)^2 \frac{d \theta}{d t} \\
\therefore \frac{d \theta}{d t} & =\frac{10 \times 40}{2500}=\frac{4}{25} \text { radians } / \mathrm{sec}
\end{aligned}
$

Question 3.
The position of a particle moving along a horizontal line of any time $t$ is given by $s(t)=80 t-16 t^2$. The time at which the particle is at rest is .......
(1) $t=0$
(2) $t=\frac{1}{3}$
(3) $t=1$
(4) $t=3$
Solution:
(2) $t=\frac{1}{3}$
Hint:
$
\begin{aligned}
s(t) & =3 t^2-2 t-8 \\
v & =\frac{d s}{d t}=6 t-2 \\
v & =0 \Rightarrow 6 t-2=0 \Rightarrow t=2 / 6=\frac{1}{3}
\end{aligned}
$

Question 4.
A stone is thrown up vertically. The height it reaches at time $t$ seconds is given by $x=80 t-16 t^2$. The stone reaches the maximum height in time $t$ seconds is given by .....
(a) 2
(b) $2.5$
(c) 3
(d) $3.5$
Solution:
(b) $2.5$
Hint:
$
\begin{aligned}
x & =80 t-16 t^2 \\
\frac{d x}{d t} & =80-32 t \\
\frac{d x}{d t} & =0 \Rightarrow 80=32 t \\
t & =\frac{80}{32}=\frac{5}{2}=2.5
\end{aligned}
$

Question 5.
Find the point on the curve $6 y=x^3+2$ at which $y$-coordinate changes 8 times as fast as $x$-coordinate is
(a) $(4,11)$
(b) $(4,-11)$
(c) $(-4,11)$
(d) $(-4,-11)$
Solution:
(a) $(4,11)$
Hint:
$
\begin{aligned}
6 y & =x^3+2 \\
6 \frac{d y}{d x} & =3 x^2 \\
\text { (i.e.,) } 6(8) & =3 x^2 \\
\Rightarrow x^2 & =\frac{48}{3}=16 \\
\Rightarrow x & =\pm 4
\end{aligned}
$

Question 6.
The abscissa of the point on the curve $\mathrm{f}(\mathrm{x})=\sqrt{8-2 x}$ at which the slope of the tangent is $-0.25$ ?
(a) $-8$
(b) $-4$
(c) $-2$
(d) 0
Solution:
(b) $-4$
Hint:

$
f(x)=\sqrt{8-2 x}
$
$
\begin{aligned}
f^{\prime}(x) & =\frac{1}{2 \sqrt{8-2 x}}(-2)=-\frac{1}{4} \\
\Rightarrow \quad \frac{-1}{\sqrt{8-2 x}} & =\frac{-1}{4} \\
\Rightarrow \quad \sqrt{8-2 x} & =4 \Rightarrow 8-2 x=16 \\
2 x & =8-16=-8 \\
x & =-4
\end{aligned}
$

Question 7.
The slope of the line normal to the curve $f(x)=2 \cos 4 x$ at $x=\frac{\pi}{2}$ is .......
(a) $-4 \sqrt{3}$
(b) $-4$
(c) $\frac{\sqrt{3}}{12}$
(d) $4 \sqrt{3}$
Solution:
(c) $\frac{\sqrt{3}}{12}$
Hint:
$\begin{aligned} f(x) & =2 \cos 4 x \\ f^{\prime}(x) & =2[-\sin 4 x](4) \\ & =-8 \sin 4 x \\ \text { at } f^{\prime}(x) \text { at } x=\frac{\pi}{12} & =-8 \sin \frac{\pi}{3} \\ & =-8 \frac{\sqrt{3}}{2}=-4 \sqrt{3}=m \\ \text { So slope of the normal } & =\frac{-1}{m}=\frac{-1}{-4 \sqrt{3}} \\ & =\frac{1}{4 \sqrt{3}} \times \frac{\sqrt{3}}{\sqrt{3}}=\frac{\sqrt{3}}{12}\end{aligned}$

Question 8.
The tangent to the curve $y^2-x y+9=0$ is vertical when
(a) $y=0$
(b) $y=\pm \sqrt{3}$
(c) $y=\frac{1}{2}$
(d) $y=\pm \sqrt{3}$
Solution:
(b) $y=\pm \sqrt{3}$
Hint:
$
y^2-x y+9=0
$
Differentiating w.r. to $y$
$
\begin{aligned}
2 y-x(1)-y \frac{d x}{d t} & =0 \\
\therefore \frac{d x}{d t} & =\frac{2 y-x}{y}
\end{aligned}
$
The tangent is vertical
$
\begin{aligned}
& \Rightarrow x=c \\
& \text { so } \frac{d x}{d t}=0 \text { (i.e.,) } \frac{2 y-x}{y}=0 \\
& \Rightarrow x=2 y
\end{aligned}
$
Substituting $x=2 y$ in the curve
$
\begin{aligned}
y^2-4 y^2+9=0 & \Rightarrow y^2=3 \\
& \Rightarrow y=\pm \sqrt{3}
\end{aligned}
$

Question 9.
Angle between $\mathrm{y}^2=\mathrm{x}$ and $\mathrm{x}^2=\mathrm{y}$ at the origin is

(a) $\tan ^{-1} \frac{3}{4}$
(b) $\tan ^{-1}\left(\frac{4}{3}\right)$
(c) $\frac{\pi}{2}$
(d) $\frac{\pi}{4}$

Solution:
(c) $\frac{\pi}{2}$
Hint:
The angle between the parabolas is the angle between the axes $=\frac{\pi}{2}$

Question 10.
What is the value of the limit $\lim _{x \rightarrow 0}\left(\cot x-\frac{1}{x}\right)$ ?
(a) 0
(b) 1
(c) 2
(d) $\leq$
Solution:
(a) 0
Hint:
$
\begin{aligned}
\lim _{x \rightarrow 0} \cot x-\frac{1}{x} & =\frac{\cos x}{\sin x}-\frac{1}{x} \\
& =\lim _{x \rightarrow 0} \frac{x \cos x-\sin x}{x \sin x}=\frac{0}{0} \text { form }
\end{aligned}
$
Applying L.H. rule
$
\text { - } \lim _{x \rightarrow 0} \frac{x(-\sin x)+\cos x(1)-\cos x}{x \cos x+\sin x}=\frac{-x \sin x}{x \cos x+\sin x}=\frac{0}{0}
$
Again applying L.H. Rule
$
\lim _{x \rightarrow 0}-\left[\frac{x \cos x+\sin x(1)}{x(-\sin x)+\cos x+\cos x}\right]=\frac{0}{2}=0
$

Question 11.
The function $\sin ^4 x+\cos ^4 x$ is increasing in the interval
(a) $\left[\frac{5 \pi}{8}, \frac{3 \pi}{4}\right]$
(b) $\left[\frac{\pi}{2}, \frac{5 \pi}{8}\right]$
(c) $\left[\frac{\pi}{4}, \frac{\pi}{2}\right]$
(d) $\left[0, \frac{\pi}{4}\right]$
Solution:
(c) $\left[\frac{\pi}{4}, \frac{\pi}{2}\right]$
Hint:
$
\begin{aligned}
& f(x)= \sin ^4 x+\cos ^4 x \\
& f^{\prime}(x)= 4 \sin ^3 x \cos x+4 \cos ^3 x(-\sin x) \\
&= 4 \sin x \cos x\left(\sin ^2 x-\cos ^2 x\right) \\
& f^{\prime}(x)= 0 \Rightarrow \sin x=0 \text { (or) } \cos x=0 \text { (or) } \sin ^2 x-\cos ^2 x=0 \\
& x=0 \text { (or) } x=\frac{\pi}{2} \text { (or) } x=\frac{\pi}{4}
\end{aligned}
$
In $\left[0, \frac{\pi}{4}\right] f^{\prime}(x)$ is -ve so $f(x)$ is decreasing.
In $\left[\frac{\pi}{4}, \frac{\pi}{2}\right], f^{\prime}(x)$ is +ve so $f(x)$ is increasing.

Question 12.
The number given by the Rolle's theorem for the function $x^3-3 x^2, x \in[0,3]$ is
(a) 1
(b) $\sqrt{2}$
(c) $\frac{3}{2}$
(d) 2
Solution:
(d) 2

Question 13.
The number given by the Mean value theorem for the function $\frac{1}{x}, x \in[1,9]$ is $\ldots$.
(a) 2
(b) $2.5$
(c) 3
(d) $3.5$
Solution:
(c) 3
Hint:
$
\begin{aligned}
& f(x)=\frac{1}{x} \\
& f^{\prime}(x)=\frac{-1}{x^2} \\
& b=9
\end{aligned}
$
Here $a=1, b=9$
So $\frac{f(b)-f(a)}{b-a}=\frac{\frac{1}{9}-1}{9-1}=\frac{\frac{-8}{9}}{8}=\frac{-1}{9}$
from (1) and (2)
$
-\frac{1}{x^2}=-\frac{1}{9} \Rightarrow x^2=9 \Rightarrow x=\pm 3
$

but $x=3 \in[1,9]$

Question 14.
The minimum value of the function $|3-x|+9$ is
(a) 0
(b) 3
(c) 6
(d) 9
Solution:
(d) 9
Hint:
$f(x)=|3-x|+9$
Minimum value of $|3-x|=0$
Minimum value of $|3-x|+9=0+9=9$

Question 15.
The maximum slope of the tangent to the curve $y=e^x \sin x, x \in[0,2 n]$ is at
(a) $x=\frac{\pi}{4}$
(b) $x=\frac{\pi}{2}$
(c) $x=\pi$
(d) $x=\frac{3 \pi}{2}$
Solution:
(b) $x=\frac{\pi}{2}$
Hint:
$
\begin{aligned}
y & =e^x \sin x \\
\frac{d y}{d x} & =e^x(\cos x)+\sin x\left(e^x\right) \\
& =e^x(\sin x+\cos x)=m \text { (say) }
\end{aligned}
$
Now $\frac{d m}{d x}=e^x[\cos x-\sin x]+[\sin x+\cos x] e^x$
$
\begin{aligned}
& =e^x[\cos x-\sin x+\sin x+\cos x] \\
& =2 \cos x e^x \\
\frac{d m}{d x} & =0 \Rightarrow \cos x=0 \Rightarrow x=\frac{\pi}{2}
\end{aligned}
$

Question 16.
The maximum value of the function $x^2 e^{-2 x}, x>0$ is
(a) $\frac{1}{e}$
(b) $\frac{1}{2 e}$
(c) $\frac{1}{e^2}$
(d) $\frac{4}{e^4}$
Solution:
(c) $\frac{1}{e^2}$
Hint:
$
\begin{aligned}
y & =x^2 e^{-2 x} \\
\frac{d y}{d x} & =x^2\left(e^{-2 x}\right)(-2)+e^{-2 x}(2 x) \\
& =e^{-2 x}\left[-2 x^2+2 x\right] \\
y^{\prime} & =0 \Rightarrow-2 x^2+2 x=0 \\
& \Rightarrow 2 x(1-x)=0 \Rightarrow x=0
\end{aligned}
$
at $x=0, \quad y=0$
at $x=1, y=\frac{1}{e^2}$

Question 17.
One of the closest points on the curve $x^2-y^2=4$ to the point $(6,0)$ is
(a) $(2,0)$
(b) $(\sqrt{5}, 1)$
(c) $(3, \sqrt{5})$
(d) $(\sqrt{13},-\sqrt{3})$
Solution:
(c) $(3, \sqrt{5})$
Hint:
Given $x^2-y^2=4$, point $(6,0)$
Any point on the curve is $\left(x, \pm \sqrt{x^2-4}\right)$
Distance between the points is $\sqrt{(x-6)^2+x^2-4}$
Substituting all the given options we get minimum distance point $(3, \sqrt{5})$

Question 18.
The maximum product of two positive numbers, when their sum of the squares is 200, is
(a) 100
(b) $25 \sqrt{7}$
(c) 28
(d) $24 \sqrt{14}$
Solution:
(a) 100
Hint: Let the positive numbers be $x, y$
Here $x^2+y^2=200 \Rightarrow y^2=200-x^2 \Rightarrow y=\sqrt{200-x^2}$
Now $f=x y=x \sqrt{200-x^2}$
$
\begin{aligned}
& =\sqrt{200 x^2-x^4} \\
\frac{d f}{d x} & =\frac{1}{2 \sqrt{200 x^2-x^4}}\left(400 x-4 x^3\right) \\
\frac{d f}{d x} & =0 \Rightarrow 400 x-4 x^3=0 \\
& \Rightarrow 4 x\left(100-x^2\right)=0 \\
& \Rightarrow x=0 \text { or } \pm 10
\end{aligned}
$
Here $x=10 \Rightarrow y=\sqrt{200-100}=10$
Product is $x y=(10)(10)=100$

Question 19.
The curves $=a x^4+b x^2$ with $a b>0$
(a) has no horizontal tangent
(b) is concave up
(c) is concave down
(d) has no points of inflection
Solution:
(d) has no points of inflection
Question 20.
The point of inflection of the curve $y=(x-1)^3$ is
(a) $(0,0)$
(b) $(0,1)$
(c) $(1,0)$
(d) $(1,1)$
Solution:
(c) $(1,0)$
Hint: