12th NCERT Application of Derivatives Exercise6.2 Questions 19

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Question (1)

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

Solution

F(x) = 3x + 17
So f'(x) = 3 > 0 for all values of x.
So f(x) is strictly increasing.

Question (2)

Show that the functions given by f(x) = e^2x is strictly increasing on R.

Solution

\[f(x) = {e^{2x}}\] \[f'(x) = {e^{2x}}\frac{d}{{dx}}2x\] \[f'(x) = 2{e^{2x}}\] For all values of x, f'(x) >0,
So f(x) is strictly increasing function.

Question (3)

Show that the function given by f(x) = sin x is
(a) strictly increasing in (0. π/2)

Solution

f(x) = sin x
f'(x) = cos x.
For x ∈ ( 0, π/2) , cos x > 0.
So f(x) is strictly increasing in interval (0, π/2)
(b) strictly decreasing in (π/2, π)

Solution

f'(x) = cos x.
For x ∈ ( π/2, π), cos x <0, so f(x) is strictly decreasing in interval ( π/2, π).
(c) neither increasing nor decreasing in ( 0, π)

Solution

Since f(x) is strictly increasing in interval (0, π/2) and is strictly decreasing in interval ( π/2, π).
So in the interval (0, π) it is increasing as well as decreasing. So neither strictly increasing nor decreasing.

Question (4)

Find the intervals in which the function f given by 2x² - 3x is
(a) strictly increasing (b)strictly decreasing.

Solution

\[f(x) = 2{x^2} - 3x\] \[f'(x) = 4x - 3\] At x = 3/4, f'(x) = 0. So intervals are break into two parts (-∞ , 3/4) and (3/4, ∞).

interval	4x - 3	f'(x)
( -∞, 3/4)	- ve	- ve
(3/4, ∞)	+ ve	+ ve

So decreasing in interval ( - ∞, 3/4) and increases in interval ( 3/4, ∞).

Question (5)

Find the intervals in which the function f given by f(x) = 2 x³ - 3x² - 36x + 7 is
(a) strictly increasing (b) strictly decreasing

Solution

\[f(x) = 2{x^3} - 3{x^2} - 36x + 7\] \[f'(x) = 6{x^2} - 6x - 36\] \[f'(x) = 6({x^2} - x - 6)\] \[f'(x) = 6(x - 3)(x + 2)\] At x = -2 and x = 3, f'(x) = 0. So intervals we get are ( -∞, -2), (-2, 3),(3, ∞)

interval	6	x + 2	x - 3	f'(x)
(- ∞, -2)	+ ve	- ve	- ve	+ve;
(-2, 3)	+ve	+ve	- ve	-ve
(3, ∞)	+ve	+ve	+ve	+ve

So (a) f(x) is strictly increasing in intervals ( - ∞, -2) and (3, ∞).
(b) f(x) is strictly decreasing in interval (-2, 3).

Question (6)

Find the intervals in which the following functions are strictly increasing or decreasing.
(a) x ² + 2x - 5

Solution

\[f(x) = {x^2} + 2x - 5\] \[f'(x) = 2x + 2\] At x = -1, f'(x) = 0. So intervals are ( -∞, -1)and ( -1, ∞).
In the interval ( - ∞, -1) f'(x) < 0. so f(x) is strictly decreasing.
In the interval ( -1, ∞) f'(x) > 0. so f(x) is strictly increasing.
(b) 10 - 6x - 2x²

Solution

\[f(x) = 10 - 6x - 2{x^2}\] \[f'(x) = - 6 - 4x\] At x = -3/2, f'(x) = 0. So intervals are ( -∞, -3/2)and ( -3/2, ∞).
In the interval ( - ∞, -3/2) f'(x) > 0. so f(x) is strictly increasing
In the interval ( -3/2, ∞) f'(x) < 0. so f(x) is strictly decreasing.
(c) -2x³ - 9x² - 12x + 1

Solution

\[f(x) = - 2{x^3} - 9{x^2} - 12x + 1\] \[f'(x) = - 6{x^2} - 18x - 12\] \[f'(x) = - 6({x^2} + 3x + 2)\] \[f'(x) = - 6(x + 2)(x + 1)\] So at x = -1, x = -2 f'(x) = 0. So intervals are (- ∞, -2), (-2, -1) and (-1, ∞)

interval	- 6	x + 1	x + 2	f'(x)
( - ∞, -2)	-ve	-ve	-ve ;	-ve
(-2, -1)	-ve	-ve	+ve	+ve
( -1, ∞)	- ve	+ve	+ve	-ve

f(x) is strictly decreasing in the intervals ( -∞, -2) and (-1, ∞) and strictly increasing in interval ( -2, -1).
(d) 6 - 9x - x²

Solution

\[f(x) = 6 - 9x - {x^2}\] \[f'(x) = - 9 - 2x\] At x = -9/2, f'(x) = 0. So intervals are ( -∞, -9/2)and ( -9/2, ∞).
In the interval ( - ∞, -9/2) f'(x) > 0. so f(x) is strictly increasing
In the interval ( -9/2, ∞) f'(x) < 0. so f(x) is strictly decreasing.
(e) (x + 1)³(x - 3)³

Solution

\[f(x) = {(x + 1)^3}{(x - 3)^3}\] \[f'(x) = {(x + 1)^3}\frac{d}{{dx}}{(x - 3)^3} + {(x - 3)^3}\frac{d}{{dx}}{(x + 1)^3}\] \[f'(x) = {(x + 1)^3}3{(x - 3)^2} + {(x - 3)^3}3{(x + 1)^2}\] \[f'(x) = 3{(x + 1)^2}{(x - 3)^2}[x + 1 + x - 3]\] \[f'(x) = 3{(x + 1)^2}{(x - 3)^2}[2x - 2]\] \[f'(x) = 6{(x + 1)^2}{(x - 3)^2}[x - 1]\] \[for\;any\;value\;of\;x,\;6{(x + 1)^2}{(x - 3)^2} > 0\] So increasing or decreasing only depends on ( x - 1). At x = 1, f'(x) = 0.
At x = 1, f'(x) = 0. So intervals are ( -∞, 1)and ( 1, ∞).
In the interval ( - ∞, 1) f'(x) < 0. so f(x) is strictly decreasing.
In the interval ( 1, ∞) f'(x) > 0. so f(x) is strictly increasing.

Question (7)

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

Solution

\[y = \log (1 + x) - \frac{{2x}}{{2 + x}}\] \[Diff.w.r.t.x,\] \[\frac{{dy}}{{dx}} = \frac{1}{{1 + x}} - \frac{{(2 + x)\frac{d}{{dx}}2x - 2x\frac{d}{{dx}}(2 + x)}}{{{{(2 + x)}^2}}}\] \[\frac{{dy}}{{dx}} = \frac{1}{{1 + x}} - \frac{{(2 + x)2 - 2x}}{{{{(2 + x)}^2}}}\] \[\frac{{dy}}{{dx}} = \frac{1}{{1 + x}} - \frac{4}{{{{(2 + x)}^2}}}\] \[\frac{{dy}}{{dx}} = \frac{{{{(2 + x)}^2} - 4(1 + x)}}{{(1 + x){{(2 + x)}^2}}}\] \[\frac{{dy}}{{dx}} = \frac{{4 + 4x + {x^2} - 4 - 4x}}{{(1 + x){{(2 + x)}^2}}}\] \[\frac{{dy}}{{dx}} = \frac{{{x^2}}}{{(1 + x){{(2 + x)}^2}}}\] \[For\;x > - 1,x + 1 > 0,for\;all\;values\;of\;x,{x^2} > 0,{(2 + x)^2} > 0\] \[\frac{{dy}}{{dx}} > 0\] So f(x) is an increasing function of x throughout its domain.

Question (8)

Find the values of x for which y = [x(x - 2)]² is an increasing function.

Solution

\[y = {[x(x - 2)]^2}\] \[Diff.w.r.t.x,\] \[\frac{{dy}}{{dx}} = 2[x(x - 2)]\frac{d}{{dx}}[x(x - 2)]\] \[\frac{{dy}}{{dx}} = 2[x(x - 2)](2x - 2)\] \[\frac{{dy}}{{dx}} = 4x(x - 2)(x - 1)\] At x = 0, x = 1 and x = 2 f'(x) = 0.
So intervals are break into ( -∞, 0), (0 , 1),(1, 2) and (2, ∞).

interval	4x	x - 1	x - 2	f'(x)
(∞,0)	- ve	-ve	-ve	-ve
(0, 1)	+ve	-ve	-ve	+ve
(1, 2)	+ve	+ve	-ve	-ve
(2, ∞)	+ve	+ve	+ve	+ve

So f(x) is increasing in the interval (0 , 1) and (2, ∞).so for the values of x 0< x < 1 and x > 2.

Question (9)

Prove that \[y = \frac{{4\sin \theta }}{{\left( {2 + \cos \theta } \right)}} - \theta \] is an increasing function of θ in [0, π/2]

Solution

\[y = \frac{{4\sin \theta }}{{2 + \cos \theta }} - \theta \] \[Diff.w.r.t.\theta ,\] \[\frac{{dy}}{{d\theta }} = \frac{{(2 + \cos \theta )\frac{d}{{d\theta }}4\sin \theta - 4\sin \theta \frac{d}{{d\theta }}(2 + \cos \theta) }}{{{{(2 + \cos \theta )}^2}}} - 1\] \[\frac{{dy}}{{d\theta }} = \frac{{(2 + \cos \theta )4\cos \theta - 4\sin \theta ( - \sin \theta )}}{{{{(2 + \cos \theta )}^2}}} - 1\] \[\frac{{dy}}{{d\theta }} = \frac{{8\cos \theta + 4{{\cos }^2}\theta + 4{{\sin }^2}\theta }}{{{{(2 + \cos \theta )}^2}}} - 1\] \[\frac{{dy}}{{d\theta }} = \frac{{8\cos \theta + 4 - {{(2 + \cos \theta )}^2}}}{{{{(2 + \cos \theta )}^2}}}\] \[\frac{{dy}}{{d\theta }} = \frac{{8\cos \theta + 4 - 4 - 4\cos \theta - {{\cos }^2}\theta }}{{{{(2 + \cos \theta )}^2}}}\] \[\frac{{dy}}{{d\theta }} = \frac{{4\cos \theta - {{\cos }^2}\theta }}{{{{(2 + \cos \theta )}^2}}}\] \[\frac{{dy}}{{d\theta }} = \frac{{\cos \theta (4 - \cos \theta )}}{{{{(2 + \cos \theta )}^2}}}\] \[for\;\theta \in \left( {0,\frac{\pi }{2}} \right),cos\theta > 0,\left( {4 - \cos \theta } \right) > 0,{(2 + \cos \theta )^2} > 0\] \[\frac{{dy}}{{d\theta }} > 0\] So y is increasing function of θ in [0, π/2]

Question (10)

Prove that the logarithmic function is strictly increasing on (0, ∞)

Solution

f(x) = log x, x ∈( 0, ∞)
f'(x) = 1/ x.
since x ∈( 0, ∞), 1/x ∈( 0, ∞)
So f'(x) > 0, so f(x) is strictly increasing function.

Question (11)

Prove that the function f given by f(x) = x² - x + 1 is neither strictly increasing nor strictly decreasing on (-1, 1)

Solution

\[f(x) = {x^2} - x + 1\] \[f'(x) = 2x - 1\] At x = 1/2 f'(x) = 0. So interval (-1, 1) is divided in two intervals (-1, 1/2) and (1/2, 1).
When x ∈ (-1, 1/2), f'(x) < 0, so f(x) is strictly decreasing. and when x ∈ (1/2, 1) f(x) > 0, so it is increasing.
So in interval (-1, 1) it is neither strictly increasing nor strictly decreasing.

Question (12)

Which of the following functions are strictly decreasing on (0, π/2)
(a) cos x (b) cos 2x (c) cos 3x (D) tan x

Solution

(A) IF \[f(x) = \cos x\] \[f'(x) = - \sin x\] \[As\;x \in (0,\frac{\pi }{2}), - \sin x < 0\] So f'(x) < 0, so f is strictly decreasing function on (0, π/2).
(B) If \[f(x) = \cos 2x\] \[f'(x) = - 2\sin 2x\] \[x \in (0,\frac{\pi }{2}),2x \in (0,\pi ),\sin 2x > 0\] So f'(x) < 0, so f is strictly decreasing function on (0, π/2).
(C) If \[f(x) = \cos 3x\] \[f'(x) = - 3\sin 3x\] \[As\;x \in (0,\frac{\pi }{2}),3x \in (0,\frac{{3\pi }}{2}),\] \[If\;x \in (0,\pi ),\sin 3x > 0,so\;f'(x) < 0\] Since f'(x) < 0, so f is strictly decreasing function,
\[x \in (\pi ,\frac{{3\pi }}{2}),\sin 3x < 0,so\;f'(x) > 0\] Since f'(x) > 0, so f is strictly increasing function,
so f(x) = cos3x is neither strictly increasing nor strictly decreasing on ( 0, π/2).
(D) \[f(x) = \tan x\] \[f'(x) = {\sec ^2}x\] \[for\;x \in (0,\frac{\pi }{2}),Se{c^2}x > 0\] Since f'(x) > 0, f is strictly increasing on (0, π/2).
So (A) and (B) are strictly decreasing on (0, π/2).

Question (13)

On which of the following intervals is the function f given by f(x) = x¹⁰⁰ + sin x - 1 strictly decreasing?
(A) (0, 1) (B) (π/2, π) (C) (0, π/2) (D) None of these.

Solution

\[f(x) = {x^{100}} + \sin x - 1\] \[f'(x) = 100{x^{99}} + \cos x\] \[(A)\;x \in (0,1),f'(x) > 0\] So f(x) is strictly increasing function.
\[(B)\;x \in \left( {\frac{\pi }{2},\pi } \right)\cos x < 0,but\;f'(x) > 0\] So f(x) is strictly increasing function.
\[(C)\;x \in (0,\frac{\pi }{2}),f'(x) > 0\] So f(x) is strictly increasing function.
So f(x) is not decreasing in any of the given intervals.
So (D) None of these is the correct option.

Question (14)

Find the least value of a such that the function f given by f(x) = x² + ax + 1 is strictly increasing on (1, 2)

Solution

\[f(x) = {x^2} + ax + 1\] \[f'(x) = 2x + a\] \[x \in (1,2),\min .x = 1\] \[f'(1) = 2(1) + a = 2 + a\] \[As\;f\;\text{increasing}\;f'(x) > 0\] \[2 + a > 0\] \[a > - 2\] So least value of a for which f is strictly increasing function is -2.

Question (15)

Let I be any interval disjoint from (-1,1). Prove that the function f given by f(x) = x + (1/x) is strictly increasing on I.

Solution

\[As\;x \in I,\text{which is disjoint}{\mathop{\rm int}} \;of\;( - 1,1)\] \[So\;x \ge 1,\;or\;x \le - 1\] \[f(x) = x + \frac{1}{x}\] \[f'(x) = 1 - \frac{1}{{{x^2}}}\] \[f'(x) = \frac{{{x^2} - 1}}{{{x^2}}}\] \[As\;x \ge 1,\;or\;x \le - 1 \Rightarrow {x^2} \ge 1\] \[so\;{x^2} - 1 \ge 0\] \[f'(x) = \frac{{{x^2} - 1}}{{{x^2}}} > 0\] So f is strictly increasing on I , where I is any interval disjoint from (-1,1).

Question (16)

Prove that the function f given by f(x) = logsin x is strictly increasing on (0, π/2) and strictly decreasing on (π/2, π)

Solution

\[f(x) = \log \sin x\] \[f'(x) = \frac{1}{{\sin x}}\frac{d}{{dx}}\sin x\] \[f'(x) = \frac{1}{{\sin x}}\cos x = \cot x\] \[x \in (0,\frac{\pi }{2}),\cot x > 0\] Since f'(x) > 0, f is strictly increasing on (0, π/2) \[x \in (\frac{\pi }{2},\pi ),\cot x < 0\] Since f'(x) < 0, f is strictly decreasing on (π/2, π)

Question (17)

Prove that the function given by f(x) = log cos x is strictly decreasing on (0. π/2) and strictly increasing on (π/2, π)

Solution

\[f(x) = \log \cos x\] \[f'(x) = \frac{1}{{\cos x}}\frac{d}{{dx}}\cos x\] \[f'(x) = \frac{1}{{\cos x}}( - \sin x) = - \tan x\] \[x \in (0,\frac{\pi }{2}),\tan x > 0\] So f'(x) < 0, So f is strictly decreasing on (0, π/2). \[x \in (\frac{\pi }{2},\pi ),\tan x < 0\] So f'(x) > 0, So f is strictly increasing on (π/2, π).

Question (18)

Prove that the function given by f(x) = x³ - 3x² + 3x - 100 is increasing in R.

Solution

\[f(x) = {x^3} - 3{x^2} + 3x - 100\] \[f'(x) = 3{x^2} - 6x + 3\] \[f'(x) = 3({x^2} - 2x + 1)\] \[f'(x) = 3{(x - 1)^2}\] \[for\;x \in R,{(x - 1)^2} > 0\] So f'(x) > 0, So f(x) is increasing function in R.

Question (19)

The interval in which y = x²e^-x is increasing is
(A) (∞, ∞) (B) (-2, 0) (C) (2, ∞) (D) (0, 2)

Solution

\[f(x) = {x^2}{e^{ - x}}\] \[f'(x) = {x^2}{e^{ - x}}( - 1) + {e^{ - x}}2x\] \[f'(x) = x(2 - x){e^{ - x}}\] \[for\;x \in R,{e^{ - x}} > 0\] At x = 0 and x = 2, f'(x) = 0
So intervals are break into (-∞ , 0), (0, 2) and (2, ∞)

interval	x	2 - x	f'(x)
(-∞, 0)	- ve	+ve;	-ve
( 0, 2)	+ve	+ve	+ve
(2, ∞)	+ve	-ve	-ve

If x ∈ (-∞, 0) , f'(x) < 0, so decreasing. If x ∈ (0,2) f'(x) > 0, so it is increasing. while when x ∈ (2, ∞) f'(x) <0, so decreasing.
So increasing in the interval (0,2) . So D is the correct option.