Hello Students! In this post, you will find the complete NCERT Solutions for Maths Class 12 Exercise 6.2.
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NCERT Solutions for Maths Class 12 Exercise 6.1
NCERT Solutions for Maths Class 12 Exercise 6.3
NCERT Solutions for Maths Class 12 Exercise 6.2
Maths Class
12 Ex 6.2 Question 1.
Show that the
function given by f(x) = 3x + 17 is strictly increasing on R.
Solution:
We have, f(x) =
3x + 17
⇒ f’(x) = 3 > 0 ∀ x ∈ R
Therefore, f is strictly increasing on R.
Maths Class
12 Ex 6.2 Question 2.
Show that the
function given by f(x) = e2x is strictly increasing on R.
Solution:
We have, f(x) = e2x
⇒ f’(x) = 2e2x
Case I When x > 0, then f’(x) = 2e2x
Maths Class
12 Ex 6.2 Question 3.
Show that the
function given by f(x) = sin x is
(a) strictly increasing in (0, π/2)
(b) strictly decreasing in (π/2, π)
(c) neither increasing nor decreasing in (0, π)
Solution:
We have, f(x) =
sin x
∴ f’(x) = cos x
(a) Since f’(x) = cos x is +ve in the interval (0, Ï€/2)
⇒ f(x) is strictly increasing in (0,
Ï€/2)
(b) Since f’(x) = cos x is -ve in the interval (Ï€/2, Ï€)
⇒ f(x) is strictly decreasing in (Ï€/2,
Ï€)
(c) Since f’(x) = cos x is +ve in the interval (0, Ï€/2)
while f’(x) is -ve in the interval (Ï€/2, Ï€)
∴ f(x) is neither increasing nor
decreasing in (0, π).
Maths Class
12 Ex 6.2 Question 4.
Find the
intervals in which the function f given by f(x) = 2x² – 3x is
(a) strictly increasing
(b) strictly decreasing
Solution:
We have, f(x) =
2x² – 3x
⇒ f’(x) = 4x – 3
⇒ f’(x) = 0 at x = ¾
The point x
= ¾ divides the real number line in two disjoint intervals.
Maths Class 12 Ex 6.2 Question 5.
Find the
intervals in which the function f given by f(x) = 2x3 – 3x² –
36x + 7 is
(a) strictly increasing
(b) strictly decreasing
Solution:
We have, f(x) =
2x3 – 3x² – 36x + 7
⇒ f’(x) = 6x2 – 6x – 36
= 6(x2 – x – 6)
⇒ f’(x) = 6(x – 3) (x + 2)
⇒ f’(x) = 0 at x = 3 and x = –2
The points x = 3 and x = –2 divide the real number line into three disjoint
intervals, i.e., (–∞, –2), (–2, 3), (3, ∞).
Now, f’(x) is +ve in the intervals (–∞, –2) and (3, ∞). Since in the interval (–∞,
–2), each factor (x – 3) and (x + 2) is -ve.
⇒ f’(x) = +ve.
(a) f is strictly increasing in (–∞, –2) and (3, ∞)
(b) In the interval (–2, 3), (x + 2) is +ve and (x – 3) is -ve.
f’(x) = 6(x – 3)(x + 2) = (+) × (–) = -ve
∴ f is strictly decreasing in the
interval (–2, 3).
Maths Class
12 Ex 6.2 Question 6.
Find the
intervals in which the following functions are strictly increasing or
decreasing:
(a) x² + 2x – 5
(b) 10 – 6x – 2x²
(c) –2x3 – 9x² – 12x
+ 1
(d) 6 – 9x – x²
(e) (x + 1)3 (x – 3)3
Solution:
(a) Given: f(x) = x² + 2x – 5
⇒
f’(x) = 2x
+ 2 = 2(x + 1) …. (i)
Therefore,
⇒ f′(x) = 0 ⇒ x = −1
Now, x = −1 divides the real
number line into intervals (−∞, −1) and (−1, ∞).
In (−∞, −1), f′(x) = 2x + 2 < 0
Thus, f is
strictly decreasing in (−∞, −1).
In (−1, ∞), f′(x) = 2x + 2 > 0
Thus, f is
strictly increasing in (−1, ∞)
(b) Given: f(x) = 10 − 6x − 2x2
Therefore, f′(x) = −6
− 4x
⇒ f′(x) = 0 ⇒ x = −3/2
Now, x = −3/2, divides the real number
line into two intervals (−∞, −3/2) and (−3/2, ∞).
In (−∞, −3/2), f′(x) = −6 −4x < 0
Hence, f is strictly increasing for x < −3/2.
In (−3/2, ∞), f′(x) = −6 − 4x > 0
Hence, f is strictly increasing for x > −3/2.
(c) Given: f(x) = –2x3 – 9x2 –
12x + 1
∴ f’(x) = –6x2 – 18x –
12 = –6(x2 + 3x + 2)
f'(x) = –6(x + 1)(x + 2)
Now, f’(x) = 0
gives x = –1 or x = –2
The points x = – 2 and x = – 1 divide the real number line into three disjoint intervals
namely ( –∞, –2) ( –2, –1) and( –1, ∞).
In the interval (–∞, –2), i.e., –∞
< x < -2, (x + 1) and (x + 2) are –ve.
∴ f’(x) = (–) (–) (–) = –ve.
⇒ f(x) is decreasing in (–∞,
–2)
In the interval (–2, –1), i.e., – 2 < x < –1, (x + 1) is –ve and (x + 2)
is +ve.
∴ f'(x) = (–)(–) (+) = +ve.
⇒ f(x) is increasing in (–2, –1)
In the interval (–1, ∞), i.e., –1 < x < ∞,
(x + 1) and (x + 2) are both positive.
∴ f’(x) = (–) (+) (+) = –ve.
⇒ f(x) is decreasing in (–1, ∞)
Hence, f(x) is increasing for –2 < x < –1 and decreasing for x < –2
and x > –1.
(d) Given: f(x) = 6 − 9x − x2
Hence, f′(x) = −9 − 2x
⇒ f′(x) = 0 ⇒ x = −9/2
In (−9/2, ∞), f′(x) < 0
Hence, f is strictly decreasing for x > −9/2.
In (−∞, −9/2), f′(x) > 0
Hence, f is strictly decreasing in x > −9/2.
(e) Given: f(x) = (x + 1)3 (x − 3)3
Hence, f′(x) = 3(x + 1)2 (x
− 3)3 + 3(x − 3)2 (x + 1)3
= 3(x + 1)2 (x
− 3)2 [x – 3 + x + 1]
= 3(x + 1)2
(x − 3)2 (2x − 2)
= 6(x + 1)2 (x
− 3)2 (x − 1)
Therefore, f′(x) = 0
⇒ x = −1, 3, 1
Now, x = −1, 3, 1 divides the real number
line into four intervals (−∞, −1), (−1, 1), (1, 3) and (3, ∞).
In (−∞, −1) and (−1, 1), f′(x) = 6(x + 1)2 (x
− 3)2 (x − 1) < 0
Hence, f is strictly decreasing in (−∞, −1) and (−1, 1).
In (1, 3) and (3, ∞), f′(x) = 6(x + 1)2 (x
− 3)2 (x − 1) > 0
Hence, f is strictly increasing in (1, 3) and (3, ∞).
Maths Class
12 Ex 6.2 Question 7.
Show that 
,
is an increasing function of x throughout its domain.
Solution:
When, −1 < x
< 0
Then, x < 0 ⇒ x2 > 0
x > −1 ⇒ (2 + x) > 0
⇒ (2 + x)2 > 0
Thus, f is increasing
throughout the domain.
Maths Class
12 Ex 6.2 Question 8.
Find the values
of x for which y = [x (x – 2)]² is an increasing function.
Solution:
We have, y = x4 –
4x3 + 4x2
∴
dy/dx = 4x3 –
12x2 + 8x
For the function to be increasing dy/dx = 0
4x3 – 12x2 + 8x = 0
⇒ 4x(x – 1) (x – 2) = 0
⇒
x = 0, x = 1, x = 2
Now, x = 0, x = 1 and x = 2 divide the
real number line in the intervals (−∞, 0), (0, 1), (1, 2) and (2,∞).
In (−∞, 0) and (1, 2), dy/dx < 0
Hence, y is strictly decreasing in intervals (−∞, 0) and (1, 2)
In (0, 1) and (2, ∞), dy/dx > 0
Hence, y is strictly increasing in intervals (0, 1) and (2, ∞).
Maths Class
12 Ex 6.2 Question 9.
Prove that 
is
an increasing function of θ in [0, π/2].
Solution:
Ex 6.2 Class 12 Maths Question 10.
Prove that the
logarithmic function is strictly increasing on (0, ∞).
Solution:
Let f(x) = log x
Then, f’(x) = 1/x
For x >
0, 1/x > 0,
Therefore, f’(x) > 0
Hence, f(x) is an increasing function in the interval (0, ∞).
Maths Class
12 Ex 6.2 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:
Given: f(x) = x²
– x + 1
Then, f’(x) = 2x – 1
f(x) is strictly
increasing when f’(x) > 0.
⇒
2x – 1 > 0
⇒
x > ½
Thus, f(x) is
increasing on (1/2, 1).
f(x) is strictly
decreasing when f’(x) < 0.
⇒
2x – 1 < 0
⇒
x < ½
Thus, f(x) is
decreasing on (–1, 1/2).
Hence, f(x) is neither strictly increasing
nor decreasing on (–1, 1).
Maths Class
12 Ex 6.2 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) We have, f(x) = cos x
∴ f’(x) = – sin x < 0 in [0, Ï€/2]
∴ f’(x) is a decreasing function.
Maths Class
12 Ex 6.2 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:
(D) We have, f(x) = x100 + sin x
– 1
∴ f’(x)= 100x99+ cos x
(a) For (0, 1) i.e. 0 < x < 1,
x99 and cos x are both +ve ∴ f’ (x) > 0
⇒ f(x) is increasing on (0, 1).
Therefore, the
option (D) is correct.
Maths Class
12 Ex 6.2 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:
We have, f(x) =
x² + ax + 1
∴ f’(x) = 2x + a.
Since f(x) is an increasing function on (1, 2)
f’(x) > 0 for all 1 < x < 2
Now, f”(x) = 2
for all x ∈ (1, 2) ⇒ f”(x) > 0 for all x ∈ (1, 2)
⇒ f’(x) is an increasing function on
(1, 2)
⇒ f’(x) is the least value of f’(x) on
(1, 2)
But f’(x) > 0 ∀ x ∈ (1, 2)
∴ f’(1) > 0 ⇒ 2 + a > 0
⇒ a > –2
Thus, the least
value of a is –2.
Maths Class
12 Ex 6.2 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:
Given f(x) = x + 1/x
Maths Class
12 Ex 6.2 Question 16.
Prove that the
function f given by f(x) = log sin x is strictly increasing on (0, π/2) and
strictly decreasing on (π/2, π).
Solution:
We have, f(x) =
log sin x
f’(x) = 
When 0 < x < Ï€/2, f’(x) is +ve; i.e., increasing
When Ï€/2 < x < Ï€, f’(x) is -ve; i.e., decreasing,
∴ f(x) is decreasing.
Hence, f is
increasing on (0, π/2) and strictly decreasing on (π/2, π).
Prove that the
function f given by f(x) = log (cos x) is strictly decreasing on (0, π/2) and
strictly increasing on (π/2, π).
Solution:
We have, f(x) =
log (cos x)
f’(x) = 
In the interval (0, Ï€/2), f’(x) = -ve
∴ f is strictly decreasing.
In the interval (Ï€/2, Ï€), f’(x) is + ve.
∴ f is strictly increasing in the
interval.
Maths Class
12 Ex 6.2 Question 18.
Prove that the
function given by f(x) = x3 – 3x2 + 3x – 100 is
increasing in R.
Solution:
We have, f(x) = x3 –
3x2 + 3x – 100
f’(x) = 3x2 –
6x + 3
= 3(x2 – 2x + 1)
= 3(x – 1)2
Now x ∈ R, f'(x) = (x – 1)2 ≥ 0
i.e. f'(x) ≥ 0 ∀ x ∈ R
Hence, f(x) is
increasing on R.
Maths Class
12 Ex 6.2 Question 19.
The interval in
which y = x2 e-x is increasing in
(A) (–∞, ∞) (B) (–2, 0) (C) (2, ∞) (D) (0, 2)
Solution:
(D) We have, y
= x2 e-x or
f(x) = x2 e-x
f’(x) = 2xe-x + x2(–e-x)
= xe-x(2-x) = e-xx(2 – x)
Now e-x is positive for all x ∈
R; f’(x) = 0 at x = 0, 2
x = 0, x = 2 divide the number line into three
disjoint intervals, viz. (-∞, 0), (0, 2) and (2, ∞).
(a) In interval (-∞, 0), x is +ve and (2 – x) is
+ve
∴ f’(x) = e-xx(2 – x) = (+) (–) (+) = -ve
⇒ f is decreasing in (-∞, 0).
(b) In interval (0, 2), f’(x) = e-x x(2
– x)
= (+) (+) (+) = +ve
⇒ f is increasing in (0, 2).
(c) In interval (2, ∞), f’(x) = e-x x(2
– x) = (+) (+) (–) = -ve
⇒ f is decreasing in the interval (2, ∞)
Hence, the option (D) is correct.
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