NCERT Solutions
Class 12 Maths
Application of Derivatives
Ex.6.2 Q.15
Let I be any interval disjoint from (−1, 1). Prove that the function f given by 
is strictly increasing on I.
We have, 
So, 
Now, 
1
x= ±1
The points x = 1 and x = −1 divide the real line in three disjoint intervals i.e. (-∞, -1), (-1, 1)
and (1, ∞).
In interval (−1, 1), it is observed that:
-1 < x < 1
x2 < 1
, x ≠ 0
< 0, x ≠ 0
So, f’(x) = < 0 on (-1, 1) – {0}
Hence, f is strictly decreasing on (-1, 1) – {0}.
In intervals (-∞, -1) and (1, ∞), it is observed that:
x < -1 or 1 < x
x2 > 1
1 >
1 – > 0
So, f’(x) = 1 – > 0 on (-∞, -1) and (1, ∞)
Hence, f is strictly increasing on (-∞, -1) and (1, ∞).
Hence, function f is strictly increasing in interval I disjoint from (−1, 1).
Hence, the given result is proved.