

NCERT Solutions for Class 12 Maths Chapter 6 Miscellaneous Exercise
Question 1. Show that the function given by f(x) = e 2x is strictly increasing on R. Solution : Given: f'(x) = e 2x Now, x ∈ R Since the value of e 2x is always positive for any real value of x, e 2x > 0. ⇒2e 2x > 0 ⇒f'(x) > 0 So f(x) is incerasing on R.NCERT Solutions for Class 12 Maths Chapter 6 Exercise 6.1
Question 2. Show that the function given by f ( x ) = 3 x + 17 is strictly increasing on R. Solution :
Question
3.
Show that the function given by
f
(
x
) = sin
x
is
(a) strictly increasing (0,π/2) (b) strictly decreasing in (π/2,π) (c) neither increasing nor decreasing in (0,π)
Solution :
Given:
The given function is
f
(
x
) = sin
x
.
NCERT Solutions for Class 12 Maths Chapter 6 Exercise 6.3
Question 4. Find the intervals in which the function F given by 2x 2 - 3x is (a) strictly increasing, (b) strictly decreasing. Solution : Given:
NCERT Solutions for Class 12 Maths Chapter 6 Exercise 6.4
Question 5. Find the intervals in which the function F given by f ( x ) = 2 x 3 − 3 x 2 − 36 x + 7 is (a) strictly increasing, (b) strictly decreasing. Solution : (a)Given:
NCERT Solutions for Class 12 Maths Chapter 6 Exercise 6.5
Question 6. Find the intervals in which the following functions are strictly increasing or decreasing:(a) x 2 + 2 x − 5 (b) 10 − 6 x − 2 x 2
(c) −2 x 3 − 9 x 2 − 12 x + 1 (d) 6 − 9 x − x 2
(e) ( x + 1) 3 ( x − 3) 3
Solution : (a) Given:
Question
7. Show that
is an increasing function of x throughout its domain.
Solution :
Given:
∴dydx=11+x-(2+x)(2)-2x(1)(2+x)2=11+x-4(2+x)2=x2(1+x)(2+x)2
Now, dydx=0
⇒x2(1+x)(2+x)2=0⇒x2=0 [(2+x)≠0 as x>-1]⇒x=0
Since x > -1 , point = 0 divides the domain (−1, ∞) in two disjoint intervals i.e., −1 < x < 0 and x > 0
When −1 < x < 0, we have:
x<0⇒x2>0x>-1⇒(2+x)>0⇒(2+x2)>0
∴ y'=x2(1+x)(2+x)2>0
Also, when x > 0
x>0⇒x2>0, (2+x)2>0
∴ y'=x2(1+x)(2+x)2>0
Hence, function f is increasing throughout this domain
Question
8. Find the value of x for which
is an increasing function.
Solution :
Question
9. Prove that
is an increasing function of θ in [0, π/2]
Solution :
Question
10. Prove that the logarithmic function is strictly increasing on (0,∞)
Solution :
Given: The given function is f(x) = logx
∴ f'(x) = 1/x
It is clear that for x > 0, f'(x) = 1/x > 0.
Hence, f(x) = log x is strictly increasing in interval (0, ∞).
Question
11. Prove that the function f given by f (x) = x
2
- x + 1 is neither strictly increasing nor strictly decreasing on (-1,1)
Solution :
Given: The given function f (x) = x
2
- x + 1
hence,f is neither strictly increasing nor decreasing on the interval (-1,1)
Question
12. Which of the following functions are strictly decreasing on (0,π/2)
(A) cos x
(B) cos 2 x
(C) cos 3 x
(D) tan x
Solution :
Question
13. On which of the following intervals is the function f given by f(x) = x
100
+ sinx - 1 is strictly decreasing:
(A) (0, 1)
(B) (π/2,π)
(C) (0,π/2)
(D) None of these
Solution :
Given:
Question
14. Find the least value of a such that the function f given by f(x) = x
2
+ ax + 1 strictly increasing on (1, 2).
Solution :
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 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.
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 :
Given:
Question
17. Prove that the function f given by f(x) = log cos x is strictly decreasing on (0,π/2) and strictly decreasing on (π/2,π)
Solution :
Given:
On the
Question
18. Prove that the function given by f(x) = x
3
- 3x
2
+ 3x - 100 is increasing in R.
Solution :
Given:
Question
19. The interval in which y = x
2
e
-x
is increasing in:
(A) (-∞, ∞)
(B)(-2,0)
(C) (2, ∞)
(D) (0, 2)
Solution :
Given:
