NCERT Solutions
Class 12 Maths
Application of Derivatives
Ex.6.2 Q.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
(a) We have,
f(x) = x2 + 2x – 5
f’(x) = 2x + 2
Now, f’(x) = 0
⟹
2x + 2 = 0
⟹ x = -1
Point x = −1 divides the real line into two disjoint intervals i.e. (-∞, -1) and (-1, ∞)
In interval (-∞, -1), f’(x) = 2x + 2 < 0
So, f is strictly decreasing in interval (-∞, -1).
Thus, f is strictly decreasing for x < −1
In interval (-1, ∞), f’(x) = 2x + 2 > 0
So, f is strictly increasing in interval (-1, ∞).
Thus, f is strictly increasing for x > −1
(b) We have,
f(x) = 10 − 6x − 2x2
f’(x) = -6 – 4x
Now, f’(x) > 0
⟹ -6 – 4x = 0
⟹ x = -32
The point divides the real line into two disjoint intervals (-∞, -32 ) and (-32 , ∞)
In interval (-∞, -32 ), f’(x) = -6 – 4x < 0
So, f is strictly decreasing in interval (-∞, -32 ).
Thus, f is strictly decreasing for x < -32
In interval (-32 , ∞), f’(x) = - 6 – 4x < 0
So, f is strictly decreasing in interval (-32 , ∞).
Thus, f is strictly increasing for x > −3/2
(c) We have,
f(x) = −2x3 − 9x2 − 12x + 1
f’(x) = −6x2 − 18x – 12
Now, f’(x) = 0
⟹ −6x2 − 18x – 12 = 0
⟹ x2 + 3x + 2 = 0
⟹ (x + 1)(x + 2) = 0
⟹ x = -1, -2
Points x = −1 and x = −2 divide the real line into three disjoint intervals
i.e. (-∞, -2), (-2, -1) and (-1, ∞)
In intervals (-∞, -2) and (-1, ∞) i.e., when x < −2 and x > −1
f’(x) = −6x2 − 18x – 12 < 0
So, f is strictly decreasing for x < −2 and x > −1
f’(x) = −6x2 − 18x – 12 < 0
Now, in interval (−2, −1) i.e., when −2 < x < −1,
f’(x) = −6x2 − 18x – 12 > 0
So, f is strictly increasing for −2 < x < −1
(d) We have,
f(x) = 6 − 9x − x2
f’(x) = -9 – 2x
Now, f’(x) = 0
⟹ -9 – 2x = 0
⟹ x = -92
The point divides the real line into two disjoint intervals i.e. (-∞, -92 ) and (-92 , ∞).
In interval (-∞, -92 )i.e. for x < -92
f’(x) = -9 – 2x > 0
So, f is strictly increasing for x < -92
In interval (-9/2, ∞) i.e., for x > -92
f’(x) = -9 – 2x < 0
So, f is strictly decreasing for x > -92
(e) We have,
f(x) = (x + 1)3(x − 3)3
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)
Now, f’(x) = 0
⟹ 6(x + 1)2(x − 3)2(x - 1) = 0
⟹ x = -1, 3, 1
The points x = −1, x = 1, and x = 3 divide the real line into four disjoint intervals i.e. (-∞, -1),
(−1, 1), (1, 3), and (3, ∞).
In intervals (-∞, -1), (−1, 1)
f’(x) = 6(x + 1)2(x − 3)2(x - 1) < 0
So, f is strictly decreasing in intervals (-∞, -1) and (−1, 1).
In intervals (1, 3) and (3, ∞)
f’(x) = 6(x + 1)2(x − 3)2(x - 1) > 0
So, f is strictly increasing in intervals (1, 3) and (3, ∞).