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

If f'(x) > 0 for given values of x , then f(x) is strictly increasing.and if f'(x) < 0 for given values of x , then f(x) is strictly decreasing .
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) = e2x 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.