NCERT Solutions for Class 12 Maths Chapter 6 Application of Derivatives

Question:10: A ladder 5 m long is leaning against a wall. The bottom of the ladder is pulled along the ground, away from the wall, at the rate of 2cm/s. How fast is its height on the wall decreasing when the foot of the ladder is 4 m away from the wall?

Answer:

Let h be the height of the ladder and x be the distance of the foot of the ladder from the wall
It is given that $\frac{dx}{dt}=2$ cm/s
We need to find the rate at which the height of the ladder decreases $(\frac{dh}{dt})$
length of ladder(L) = 5m and x = 4m (given)
By Pythagoras' theorem, we can say that
$h^2+x^2=L^2$
$\Rightarrow h^2=L^2-x^2$
$\Rightarrow h=\sqrt{L^2-x^2}$
Differentiate on both sides with respect to t,
$\frac{dh}{dt}=\frac{d(\sqrt{L^2-x^2})}{dx}.\frac{dx}{dt}=\frac{1}{2}\frac{-2x}{\sqrt{5^2-x^2}}.\frac{dx}{dt}=\frac{-x}{\sqrt{25-x^2}}\frac{dx}{dt}$
at x = 4
$\frac{dh}{dt}=\frac{-4}{\sqrt{25-16}}\times 2=\frac{-4}{3}\times 2=-\frac{8}{3}$ cm/s
Hence, the rate at which the height of the ladder decreases is $\frac{8}{3}$ cm/s.

Question:11: A particle moves along the curve $6y=x^3+2$. Find the points on the curve at which the y-coordinate is changing 8 times as fast as the x-coordinate.

Answer:

We need to find the point at which $\frac{dy}{dt}=8\frac{dx}{dt}$
Given the equation of curve = $6y=x^3+2$
Differentiate both sides w.r.t. t
$6\frac{dy}{dt}=\frac{d(x^3)}{dx}.\frac{dx}{dt}+0$
$=3x^2.\frac{dx}{dt}$
$\frac{dy}{dt}=8\frac{dx}{dt}$ (required condition)
$\Rightarrow 6\times 8\frac{dx}{dt}=3x^2.\frac{dx}{dt}$
$\Rightarrow 3x^2.\frac{dx}{dt}=48\frac{dx}{dt}$
$\Rightarrow x^2=\frac{48}{3}=16$
$\Rightarrow x=\pm 4$
when x = 4, $y=\frac{4^3+2}{6}=\frac{64+2}{6}=\frac{66}{6}=11$

and
when x = -4, $y=\frac{(-4{)}^3+2}{6}=\frac{-64+2}{6}=\frac{-62}{6}=-\frac{31}{3}$

So, the coordinates are $(4,11)$ and $(-4,\frac{-31}{3})$.

Question:12: The radius of an air bubble is increasing at the rate of $\frac{1}{2}$ cm/s. At what rate is the volume of the bubble increasing when the radius is 1 cm?

Answer:

It is given that $\frac{dr}{dt}=\frac{1}{2}$ cm/s
We know that the shape of the air bubble is spherical
So, volume(V) = $\frac{4}{3}\pi r^3$
$\frac{dV}{dt}=\frac{dV}{dr}.\frac{dr}{dt}=\frac{d(\frac{4}{3}\pi r^3)}{dr}.\frac{dr}{dt}=4\pi r^2\times \frac{1}{2}=2\pi r^2=2\pi \times (1{)}^2=2\pi$ cm3/s
Hence, the rate of change in volume is $2\pi$ cm3/s.

Question:13 A balloon, which always remains spherical, has a variable diameter $\frac{3}{2}(2x+1)$. Find the rate of change of its volume with respect to x.

Answer:

Volume of sphere(V) = $\frac{4}{3}\pi r^3$
Diameter = $\frac{3}{2}(2x+1)$
So, radius(r) = $\frac{3}{4}(2x+1)$
$\frac{dV}{dx}=\frac{d(\frac{4}{3}\pi r^3)}{dx}=\frac{d(\frac{4}{3}\pi (\frac{3}{4}(2x+1){)}^3)}{dx}=\frac{4}{3}\pi \times 3\times \frac{27}{64}(2x+1{)}^2\times 2$
$=\frac{27}{8}\pi (2x+1{)}^2$

Question:14: Sand is pouring from a pipe at the rate of 12 cm3/s. The falling sand forms a cone on the ground in such a way that the height of the cone is always one-sixth of the radius of the base. How fast is the height of the sand cone increasing when the height is 4 cm?

Answer:

Given = $\frac{dV}{dt}=12$ cm3/s and $h=\frac{1}{6}r$
To find = $\frac{dh}{dt}$ at h = 4 cm
Volume of cone(V) = $\frac{1}{3}\pi r^{2}h$
$\frac{dV}{dt}=\frac{dV}{dh}.\frac{dh}{dt}=\frac{d(\frac{1}{3}\pi (6h{)}^{2}h)}{dh}.\frac{dh}{dt}=\frac{1}{3}\pi \times 36\times 3h^2.\frac{dh}{dt}=36\pi \times (4{)}^2.\frac{dh}{dt}$
$\Rightarrow \frac{dV}{dt}=576\pi .\frac{dh}{dt}$

Question:15 The total cost C(x) in Rupees associated with the production of x units of an item is given by $C(x)=0.007x^3-0.003x^2+15x+4000$. Find the marginal cost when 17 units are produced.

Answer:

Marginal cost (MC) = $\frac{dC}{dx}$
$C(x)=0.007x^3-0.003x^2+15x+4000$
$\frac{dC}{dx}=\frac{d(0.007x^3-0.003x^2+15x+400)}{dx}=3\times 0.007x^2-2\times 0.003x+15$
$=.021x^2-0.006x+15$
Now, at x = 17
MC $=0.021(17{)}^2-0.006(17)+15$
$=6.069-0.102+15$
$=20.967$
Hence, the marginal cost when 17 units are produced is 20.967.

Question:16 The total revenue in Rupees received from the sale of x units of a product is given by $R(x)=13x^2+26x+15$. Find the marginal revenue when x = 7

Answer:

Marginal revenue = $\frac{dR}{dx}$
$R(x)=13x^2+26x+15$
$\frac{dR}{dx}=\frac{d(13x^2+26x+15)}{dx}=13\times 2x+26=26(x+1)$
at x = 7
$\frac{dR}{dx}=26(7+1)=26\times 8=208$
Hence, marginal revenue when x = 7 is 208.

Question:17 The rate of change of the area of a circle with respect to its radius r at r = 6 cm is (A) 10π (B) 12π (C) 8π (D) 11π

Answer:

Area of circle(A) = $\pi r^2$
$\frac{dA}{dr}=\frac{d(\pi r^2)}{dr}=2\pi r$
Now, at r = 6 cm
$\frac{dA}{dr}=2\pi \times 6=12\pi$ cm2/s
Hence, the rate of change of the area of a circle with respect to its radius r at r = 6 cm is $12\pi$ cm2/s
Hence, the correct answer is B.

Question:18 The total revenue in Rupees received from the sale of x units of a product is given by $R(x)=3x^2+36x+5$. The marginal revenue, when x = 15, is (A) 116 (B) 96 (C) 90 (D) 126

Answer:

Marginal revenue = $\frac{dR}{dx}$
$R(x)=3x^2+36x+5$
$\frac{dR}{dx}=\frac{d(3x^2+36x+5)}{dx}=3\times 2x+36=6(x+6)$
at x = 15
$\frac{dR}{dx}=6(15+6)=6\times 21=126$
Hence, marginal revenue when x = 15 is 126.
Hence, the correct answer is D.

Question:1. Show that the function given by f (x) = 3x + 17 is increasing on R.

Answer:
Let $x_1$ and $x_2$ be two numbers in R.
$x_1<x_2$
$\Rightarrow 3x_1<3x_2$
$\Rightarrow 3x_1+17<3x_2+17$
$\Rightarrow f(x_1)<f(x_2)$
Hence, f is strictly increasing on R.

Question:2. Show that the function given by $f(x)=e^{2x}$ is increasing on R.

Answer:
Let $x_1$ and $x_2$ be two numbers in R.
$x_1<x_2$
$\Rightarrow 2x_1<2x_2$
$\Rightarrow e^{2x_1}<e^{2x_2}$
$\Rightarrow f(x_1)<f(x_2)$
Hence, the function $f(x)=e^{2x}$ is strictly increasing in R.

Question:3(a) Show that the function given by f (x) = $\sin x$ is increasing in $(0,\frac{\pi }{2})$

Answer:
Given $f(x)=\sin x$
$f^{{}^{\prime }}(x)=\cos x$
Since, $\cos x>0$ for each x $\in (0,\frac{\pi }{2})$
$f^{{}^{\prime }}(x)>0$
Hence, $f(x)=\sin x$ is strictly increasing in $(0,\frac{\pi }{2})$.

Question:3(b) Show that the function given by f (x) = $\sin x$ is decreasing in $(\frac{\pi }{2},\pi )$

Answer:
$f(x)=\sin x$
$f^{{}^{\prime }}(x)=\cos x$
Since, $\cos x<0$ for each $xϵ(\frac{\pi }{2},\pi )$
So, we have $f^{{}^{\prime }}(x)<0$
Hence, $f(x)=\sin x$ is strictly decreasing in $(\frac{\pi }{2},\pi )$

Question:3(c) Show that the function given by f (x) = $\sin x$ is neither increasing nor decreasing in $(0,\pi )$

Answer:
We know that $\sin x$ is strictly increasing in $(0,\frac{\pi }{2})$ and strictly decreasing in $(\frac{\pi }{2},\pi )$
So, by this, we can say that $f(x)=\sin x$ is neither increasing nor decreasing in the range $(0,\pi )$.

Question:4(a). Find the intervals in which the function f given by $f(x)=2x^2-3x$ is increasing

Answer:
$f(x)=2x^2-3x$
$f^{{}^{\prime }}(x)=4x-3$
Now,
$f^{{}^{\prime }}(x)=0$
$\Rightarrow 4x-3=0$
$\Rightarrow x=\frac{3}{4}$

So, the range is $(-\mathrm{\infty },\frac{3}{4})$ and $(\frac{3}{4},\mathrm{\infty })$
So,
$f(x)<0$, when $x\in (-\mathrm{\infty },\frac{3}{4})$ Hence, f(x) is strictly decreasing in this range
and
$f(x)>0$, when $x\in (\frac{3}{4},\mathrm{\infty })$
Hence, f(x) is strictly increasing in this range.
Hence, $f(x)=2x^2-3x$ is strictly increasing in $x\in (\frac{3}{4},\mathrm{\infty })$

Question:4(b) Find the intervals in which the function f given by $f(x)=2x^2-3x$ is decreasing

Answer:
$f(x)=2x^2-3x$
$f^{{}^{\prime }}(x)=4x-3$
Now,
$f^{{}^{\prime }}(x)=0$
$\Rightarrow 4x-3=0$
$\Rightarrow x=\frac{3}{4}$

So, the range is $(-\mathrm{\infty },\frac{3}{4})$ and $(\frac{3}{4},\mathrm{\infty })$.
So,
$f(x)<0$, when $x\in (-\mathrm{\infty },\frac{3}{4})$ Hence, f(x) is strictly decreasing in this range
and
$f(x)>0$, when $x\in (\frac{3}{4},\mathrm{\infty })$ Hence, f(x) is strictly increasing in this range.
Hence, $f(x)=2x^2-3x$ is strictly decreasing in $x\in (-\mathrm{\infty },\frac{3}{4})$.

Question:5(a) Find the intervals in which the function f given by $f(x)=2x^3-3x^2-36x+7$ is increasing

Answer:
It is given that
$f(x)=2x^3-3x^2-36x+7$
So,
$f^{{}^{\prime }}(x)=6x^2-6x-36$
Also, $f^{{}^{\prime }}(x)=0$
$\Rightarrow 6x^2-6x-36=0$
$\Rightarrow 6(x^2-x-6)$
$\Rightarrow x^2-x-6=0$
$\Rightarrow x^2-3x+2x-6=0$
$\Rightarrow x(x-3)+2(x-3)=0$
$\Rightarrow (x+2)(x-3)=0$
$\therefore x=-2,x=3$
So, three ranges are there $(-\mathrm{\infty },-2),(-2,3)$ and $(3,\mathrm{\infty })$
Function $f^{{}^{\prime }}(x)=6x^2-6x-36$ is positive in interval $(-\mathrm{\infty },-2),(3,\mathrm{\infty })$ and negative in the interval $(-2,3)$.
Hence, $f(x)=2x^3-3x^2-36x+7$ is strictly increasing in $(-\mathrm{\infty },-2)\cup (3,\mathrm{\infty })$ and strictly decreasing in the interval (-2, 3).

Question:5(b) Find the intervals in which the function f given by $f(x)=2x^3-3x^2-36x+7$ is decreasing

Answer:
We have $f(x)=2x^3-3x^2-36x+7$

Differentiating the function with respect to x, we get :

$f^{\prime }(x)=6x^2-6x-36=6(x-3)(x+2)$

When $f^{\prime }(x)=0$ , we have :

$\Rightarrow 0=6(x-3)(x+2)$
$\Rightarrow (x-3)(x+2)=0$
So, three ranges are there $(-\mathrm{\infty },-2),(-2,3)$ and $(3,\mathrm{\infty })$
Function $f^{{}^{\prime }}(x)=6x^2-6x-36$ is positive in the interval $(-\mathrm{\infty },-2),(3,\mathrm{\infty })$ and negative in the interval (-2,3)

So, f(x) is decreasing in (-2, 3).

Question:6(a) Find the intervals in which the following functions are strictly increasing or
decreasing: $x^2+2x-5$

Answer:

f(x) = $x^2+2x-5$
$f^{{}^{\prime }}(x)=2x+2=2(x+1)$
Now,
$$\begin{gathered} f^{{}^{\prime }}(x)=0 \\ 2(x+1)=0 \\ x=-1 \end{gathered}$$
The range is from $(-\mathrm{\infty },-1)and(-1,\mathrm{\infty })$.
In interval $(-\mathrm{\infty },-1)$, $f^{{}^{\prime }}(x)=2(x+1)$ is -ve
Hence, function f(x) = $x^2+2x-5$ is strictly decreasing in interval $(-\mathrm{\infty },-1)$.
In interval $(-1,\mathrm{\infty })$, $f^{{}^{\prime }}(x)=2(x+1)$ is +ve.
Hence, function f(x) = $x^2+2x-5$ is strictly increasing in interval $(-1,\mathrm{\infty })$.

Question:6(b) Find the intervals in which the following functions are strictly increasing or
Decreasing: $10-6x-2x^2$

Answer:
Given function is,
$f(x)=10-6x-2x^2$
$f^{{}^{\prime }}(x)=-6-4x$
Now,
$f^{{}^{\prime }}(x)=0$
$\Rightarrow 6+4x=0$
$\Rightarrow x=-\frac{3}{2}$
So, the range is $(-\mathrm{\infty },-\frac{3}{2})$ and $(-\frac{3}{2},\mathrm{\infty })$.
In interval $(-\mathrm{\infty },-\frac{3}{2})$ , $f^{{}^{\prime }}(x)=-6-4x$ is +ve.
Hence, $f(x)=10-6x-2x^2$ is strictly increasing in the interval $(-\mathrm{\infty },-\frac{3}{2})$.
In interval $(-\frac{3}{2},\mathrm{\infty })$ , $f^{{}^{\prime }}(x)=-6-4x$ is -ve.
Hence, $f(x)=10-6x-2x^2$ is strictly decreasing in interval $(-\frac{3}{2},\mathrm{\infty })$.

Question:6(c) Find the intervals in which the following functions are strictly increasing or
decreasing: $-2x^3-9x^2-12x+1$

Answer:
Given function is,
$f(x)=-2x^3-9x^2-12x+1^{}$
$f^{{}^{\prime }}(x)=-6x^2-18x-12$
Now,
$$\begin{gathered} f^{{}^{\prime }}(x)=0 \\ -6x^2-18x-12=0 \end{gathered}$$
$\Rightarrow -6(x^2+3x+2)=0$
$\Rightarrow x^2+3x+2=0$
$\Rightarrow x^2+x+2x+2=0$
$\Rightarrow x(x+1)+2(x+1)=0$
$\Rightarrow (x+2)(x+1)=0$
$\therefore x=-2$ and $x=-1$
So, the range is $(-\mathrm{\infty },-2),(-2,-1)$ and $(-1,\mathrm{\infty })$.
In interval $(-\mathrm{\infty },-2)\cup (-1,\mathrm{\infty })$ , $f^{{}^{\prime }}(x)=-6x^2-18x-12$ is -ve.
Hence, $f(x)=-2x^3-9x^2-12x+1^{}$ is strictly decreasing in interval $(-\mathrm{\infty },-2)\cup (-1,\mathrm{\infty })$.
In interval (-2,-1) , $f^{{}^{\prime }}(x)=-6x^2-18x-12$ is +ve.
Hence, $f(x)=-2x^3-9x^2-12x+1^{}$ is strictly increasing in the interval (-2,-1).

Question:6(d) Find the intervals in which the following functions are strictly increasing or
decreasing: $6-9x-x^2$

Answer:
Given function is,
$f(x)=6-9x-x^2$
$f^{{}^{\prime }}(x)=-9-2x$
Now,
$f^{{}^{\prime }}(x)=0$
$\Rightarrow -9-2x=0$
$\Rightarrow 2x=-9$
$\Rightarrow x=-\frac{9}{2}$
So, the range is $(-\mathrm{\infty },-\frac{9}{2})$ and $(-\frac{9}{2},\mathrm{\infty })$
In interval $(-\mathrm{\infty },-\frac{9}{2})$ , $f^{{}^{\prime }}(x)=-9-2x$ is +ve.
Hence, $f(x)=6-9x-x^2$ is strictly increasing in interval $(-\mathrm{\infty },-\frac{9}{2})$.
In interval $(-\frac{9}{2},\mathrm{\infty })$ , $f^{{}^{\prime }}(x)=-9-2x$ is -ve.
Hence, $f(x)=6-9x-x^2$ is strictly decreasing in interval $(-\frac{9}{2},\mathrm{\infty })$.

Question:6(e) Find the intervals in which the following functions are strictly increasing or
decreasing: $(x+1{)}^3(x-3{)}^3$

Answer:
Given function is,
$f(x)=(x+1{)}^3(x-3{)}^3$
$f^{{}^{\prime }}(x)=3(x+1{)}^2(x-3{)}^3+3(x-3{)}^2(x+1{)}^3$
Now,
$f^{{}^{\prime }}(x)=0$
$\Rightarrow 3(x+1{)}^2(x-3{)}^3+3(x-3{)}^2(x+1{)}^3=0$
$\Rightarrow 3(x+1{)}^2(x-3{)}^2((x-3)+(x+1))=0$
$\Rightarrow (x+1)(x-3)=0$ or $(2x-2)=0$
So, $x=-1$ and $x=3$ or, $x=1$
So, the intervals are $(-\mathrm{\infty },-1),(-1,1),(1,3)$ and $(3,\mathrm{\infty })$.
Our function $f^{{}^{\prime }}(x)=3(x+1{)}^2(x-3{)}^3+3(x-3{)}^2(x+1{)}^3$ is +ve in the interval $(1,3)$ and $(3,\mathrm{\infty })$.
Hence, $f(x)=(x+1{)}^3(x-3{)}^3$ is strictly increasing in the interval $(1,3)$ and $(3,\mathrm{\infty })$.
Our function $f^{{}^{\prime }}(x)=3(x+1{)}^2(x-3{)}^3+3(x-3{)}^2(x+1{)}^3$ is -ve in the interval $(-\mathrm{\infty },-1)$ and $(-1,1)$
Hence, $f(x)=(x+1{)}^3(x-3{)}^3$ is strictly decreasing in interval $(-\mathrm{\infty },-1)$ and $(-1,1)$.

Question:7 Show that $y=\log (1+x)-\frac{2x}{2+x},x>-1$ is an increasing function of x throughout its domain.

Answer:
Given function is,
$f(x)\Rightarrow y=\log (1+x)-\frac{2x}{2+x}$
$f^{{}^{\prime }}(x)\Rightarrow \frac{dy}{dx}=\frac{1}{1+x}-\frac{2(2+x)-(1)(2x)}{(2+x{)}^2}=\frac{1}{1+x}-\frac{4+2x-2x}{(2+x{)}^2}$
$=\frac{1}{1+x}-\frac{4}{(2+x{)}^2}$
$=\frac{(2+x{)}^2-4(x+1)}{(x+1)(2+x{)}^2}$
$=\frac{4+x^2+4x-4x-4}{(x+1)(2+x{)}^2}$
$=\frac{x^2}{(x+1)(2+x{)}^2}$
So, $f^{{}^{\prime }}(x)=\frac{x^2}{(x+1)(x+2{)}^2}$
Now, for $x>-1$, it is clear that $f^{{}^{\prime }}(x)=\frac{x^2}{(x+1)(x+2{)}^2}>0$
Hence, $f(x)\Rightarrow y=\log (1+x)-\frac{2x}{2+x}$ strictly increasing when $x>-1$

Question:8 Find the values of x for which $y=[x(x-2){]}^2$ is an increasing function.

Answer:
Given function is,
$f(x)\Rightarrow y=[x(x-2){]}^2$
$f^{{}^{\prime }}(x)\Rightarrow \frac{dy}{dx}=2[x(x-2)][(x-2)+x]$
$=2(x^2-2x)(2x-2)$
$=4x(x-2)(x-1)$
Now,
$f^{{}^{\prime }}(x)=0$
$\Rightarrow 4x(x-2)(x-1)=0$
So, $x=0,x=2$ and $x=1$
So, the intervals are $(-\mathrm{\infty },0),(0,1),(1,2)$ and $(2,\mathrm{\infty })$
In interval $(0,1)$ and $(2,\mathrm{\infty })$, $f^{{}^{\prime }}(x)>0$
Hence, $f(x)\Rightarrow y=[x(x-2){]}^2$ is an increasing function in the interval $(0,1)\cup (2,\mathrm{\infty })$.

Question:9 Prove that $y=\frac{4\sin \theta }{(2+\cos \theta )}-\theta$ is an increasing function of $\theta \text{in}[0,\frac{\pi }{2}]$

Answer:
Given function is,
$f(x)=y=\frac{4\sin \theta }{(2+\cos \theta )}-\theta$

$f^{{}^{\prime }}(x)=\frac{dy}{d\theta }=\frac{4\cos \theta (2+\cos \theta )-(-\sin \theta )4\sin \theta )}{(2+\cos \theta {)}^2}-1$
$=\frac{8\cos \theta +4{\cos }^2\theta +4{\sin }^2\theta -(2+\cos \theta {)}^2}{(2+\cos \theta {)}^2}$
$=\frac{8\cos \theta +4({\cos }^2\theta +{\sin }^2\theta )-4-{\cos }^2\theta -4\cos \theta }{(2+\cos \theta {)}^2}$
$=\frac{8\cos \theta +4-4-{\cos }^2\theta -4\cos \theta }{(2+\cos \theta {)}^2}$
$=\frac{4\cos \theta -{\cos }^2\theta }{(2+\cos \theta {)}^2}$
Now, for $\theta \in [0,\frac{\pi }{2}]$
$4\cos \theta \ge {\cos }^2\theta$
$\Rightarrow 4\cos \theta -{\cos }^2\ge 0$ and $(2+\cos \theta {)}^2>0$
So, $f^{{}^{\prime }}(x)>0$ for $\theta$ in $[0,\frac{\pi }{2}]$
Hence, $f(x)=y=\frac{4\sin \theta }{(2+\cos \theta )}-\theta$ is increasing function in $\theta \in [0,\frac{\pi }{2}]$.