Find the best tutors and institutes for Class 12 Tuition
Search in
Prove that the function f given by f(x) = x2 − x + 1 is neither strictly increasing nor strictly decreasing on (−1, 1).
For strictly increasing function f'(x) >0 for given interval
For strictly decreasing function f'(x) <0 for given interval
Here
f'(x) =2x-1 interval (-1, 1)
at x= - 1/2 , f'(x) = -2
at x=3/4 , f'(x) = 1/2
Hence prove function neither strictly increasing nor strictly decreasing in given interval (−1, 1).
Which of the following functions are strictly decreasing on
?
(A) cos x (B) cos 2x (C) cos 3x (D) tan x
(A) Let
In interval
is strictly decreasing in interval
.
(B) Let
is strictly decreasing in interval.
(C) Let
The point 
divides the intervalinto two disjoint intervals
i.e., 0
∴ f3 is strictly increasing in interval
.
Hence, f3 is neither increasing nor decreasing in interval.
(D) Let
In interval
∴ f4 is strictly increasing in interval
Therefore, functions cos x and cos 2x are strictly decreasing in
Hence, the correct answers are A and B.
Prove that the function given by 
is increasing in R.
Show that the function given by f(x) = 3x + 17 is strictly increasing on R.
Let
be any two numbers in R.
Then, we have:
Hence, f is strictly increasing on R.
Show that the function given by f(x) = e2x is strictly increasing on R.
Let
be any two numbers in R.
Then, we have:
Hence, f is strictly increasing on R.
Show that the function given by f(x) = sin x is
(a) strictly increasing in 
(b) strictly decreasing in 
(c) neither increasing nor decreasing in (0, π)
The given function is f(x) = sin x.
(a) Since for each
we have
.
Hence, f is strictly increasing in.
(b) Since for each
, we have
.
Hence, f is strictly decreasing in.
(c) From the results obtained in (a) and (b), it is clear that f is neither increasing nor decreasing in (0, π).
Find the intervals in which the function f given by f(x) = 2x2 − 3x is
(a) strictly increasing (b) strictly decreasing
The given function is f(x) = 2x2 − 3x.
Now, the point
divides the real line into two disjoint intervals i.e., 
and
In interval
Hence, the given function (f) is strictly decreasing in interval.
In interval
Hence, the given function (f) is strictly increasing in interval
.
Find the intervals in which the function f given by f(x) = 2x3 − 3x2 − 36x + 7 is
(a) strictly increasing (b) strictly decreasing
The given function is f(x) = 2x3 − 3x2 − 36x + 7.
∴
x = − 2, 3
The points x = −2 and x = 3 divide the real line into three disjoint intervals i.e.,
In intervals
is positive while in interval
(−2, 3), 
is negative.
Hence, the given function (f) is strictly increasing in intervals
, while function (f) is strictly decreasing in interval
(−2, 3).
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,
Now,
x = −1
Point x = −1 divides the real line into two disjoint intervals i.e., 
In interval
∴f is strictly decreasing in interval
Thus, f is strictly decreasing for x < −1.
In interval
∴ f is strictly increasing in interval
(c) We have,
f(x) = −2x3 − 9x2 − 12x + 1
Points x = −1 and x = −2 divide the real line into three disjoint intervals i.e.,
In intervals
i.e., when x < −2 and x > −1,
.
∴ f is strictly decreasing for x < −2 and x > −1.
(e) We have,
f(x) = (x + 1)3 (x − 3)3
The points x = −1, x = 1, and x = 3 divide the real line into four disjoint intervals i.e.,
, (−1, 1), (1, 3), and
.
In intervalsand (−1, 1), 
.
∴ f is strictly decreasing in intervalsand (−1, 1).
In intervals (1, 3) and, 
.
∴ f is strictly increasing in intervals (1, 3) and.
Now, in interval (−2, −1) i.e., when −2 < x < −1, 
.
∴ f is strictly increasing for 
.
Thus, f is strictly increasing for x > −1.
Show that
, is an increasing function of x throughout its domain.
We have,
∴dydx=11+x−(2+x)(2)−2x(1)(2+x)2=11+x−4(2+x)2=x2(1+x)(2+x)2dydx=11+x-(2+x)(2)-2x(1)(2+x)2=11+x-4(2+x)2=x2(1+x)(2+x)2
Now, dydx=0dydx=0
⇒x2(1+x)(2+x)2=0⇒x2=0 [(2+x)≠0 as x>−1]⇒x=0⇒x2(1+x)(2+x)2=0⇒x2=0 [(2+x)≠0 as x>-1]⇒x=0
Since x > −1, point x = 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)>0x<0⇒x2>0x>-1⇒(2+x)>0⇒(2+x2)>0
∴ y′=x2(1+x)(2+x)2>0y'=x2(1+x)(2+x)2>0
Also, when x > 0:
x>0⇒x2>0, (2+x)2>0x>0⇒x2>0, (2+x)2>0
∴ y′=x2(1+x)(2+x)2>0y'=x2(1+x)(2+x)2>0
Hence, function f is increasing throughout this domain.
Find the values of x for which
is an increasing function.
We have,
The points x = 0, x = 1, and x = 2 divide the real line into four disjoint intervals i.e., 
In intervals
, 
.
∴ y is strictly decreasing in intervals .
However, in intervals (0, 1) and (2, ∞), 
∴ y is strictly increasing in intervals (0, 1) and (2, ∞).
y is strictly increasing for 0 < x < 1 and x > 2.
Prove that 
is an increasing function of θ in
.
We have,
Since cos θ ≠ 4, cos θ = 0.
Now,
In interval
, we have cos θ > 0. Also, 4 > cos θ ⇒ 4 − cos θ > 0. 
Therefore, y is strictly increasing in interval.
Also, the given function is continuous at 
Hence, y is increasing in interval
.
Prove that the logarithmic function is strictly increasing on (0, ∞).
It is clear that for x > 0, 
Hence, f(x) = log x is strictly increasing in interval (0, ∞).
On which of the following intervals is the function f given by 
strictly decreasing?
(A) 
(B) 
(C) 
(D) None of these
We have,
In interval
Thus, function f is strictly increasing in interval (0, 1).
In interval 
Thus, function f is strictly increasing in interval.
∴ f is strictly increasing in interval
.
Hence, function f is strictly decreasing in none of the intervals.
The correct answer is D.
Let I be any interval disjoint from (−1, 1). Prove that the function f given by
is strictly increasing on I.
We have,
The points x = 1 and x = −1 divide the real line in three disjoint intervals i.e., 
.
In interval (−1, 1), it is observed that:
∴ f is strictly decreasing on 
.
In intervals
, it is observed that:
∴ f is strictly increasing on
.
Hence, function f is strictly increasing in interval I disjoint from (−1, 1).
Hence, the given result is proved.
Prove that the function f given by f(x) = log sin x is strictly increasing on 
and strictly decreasing on
We have,
In interval
∴ f is strictly increasing in
.
In interval
∴f is strictly decreasing in
Prove that the function f given by f(x) = log cos x is strictly decreasing on 
and strictly increasing on
We have,
In interval
∴f is strictly decreasing on.
In interval
∴f is strictly increasing on
.
The interval in which 
is increasing is
(A) 
(B) (−2, 0) (C) 
(D) (0, 2)
We have,
The points x = 0 and x = 2 divide the real line into three disjoint intervals i.e.,
In intervals
is always positive.
∴f is decreasing on
In interval (0, 2),
∴ f is strictly increasing on (0, 2).
Hence, f is strictly increasing in interval (0, 2).
The correct answer is D.
How helpful was it?
How can we Improve it?
Please tell us how it changed your life *
Please enter your feedback
UrbanPro.com helps you to connect with the best Class 12 Tuition in India. Post Your Requirement today and get connected.
Find best tutors for Class 12 Tuition Classes by posting a requirement.
Get started now, by booking a Free Demo Class
