Monotonicity JEE Advanced previous year questions with solutions are given on this page. These are prepared by our experts in a step-by-step manner. One of the best methods to prepare for the JEE Advanced exam is to go through the important questions and thus, students can strengthen the basic concept and can improve their problem-solving skills. Students are advised to solve these questions and then cross-check their answers with the provided solutions. This will help you to identify the topics in which you need more preparation.
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Question 1: The number of points in (-∞, ∞) for which x2 – x sin x – cos x = 0 is
(a) 6
(b) 4
(c) 2
(d) 0
Answer: c
Solution:
Let f(x) = x2 – x sin x – cos x = 0
f’(x) = 2x – sin x – x cos x + sin x
=> 2x – x cos x = 0
When x> 0, f(x) is increasing.
When x< 0, f(x) is decreasing.
f(0) = -1
f(∞) = ∞
f(-∞) = ∞
lim x→ +∞f(x) = ∞
lim x→ -∞f(x) = ∞
Using the intermediate value theorem, it will cut the x-axis at 2 points.
Thus 2 solutions.
Hence option (c) is the answer.
Question 2: The function f(x) = (3x – 7)x2/3, x ∈ R, is increasing for all x lying in:
(a) (-∞, -14/15) U (0,∞)
(b) (-∞, 14/15)
(c) (-∞,0) U (14/15, ∞)
(d) (-∞,0) U (3/7, ∞)
Answer: c
Solution:
f(x) = (3x – 7)x2/3
f’(x) = (3x – 7)(â…”)x-1/3 + 3x2/3
= (15x – 14)/3x1/3
For increasing function, f’(x) > 0, then
x ∈ (-∞, 0) U (14/15, ∞)
Hence option (c) is the answer.
Question 3: If f: R→ R is a differentiable function such that f’(x) > 2f(x) for all x belongs to R, and f(0) = 1, then
(a) f(x) is increasing in (0, ∞)
(b) f(x) is decreasing in (0, ∞)
(c) f(x) > e2x in (0, ∞)
(d) f’(x) < e2x in (0, ∞)
Answer: (a, c)
Solution:
f’(x) – 2f(x) > 0
=> e-2x (df(x)/dx) – 2e-2x f(x) > 0
=> (d/dx) (e-2x f(x)) > 0
=> e-2x f(x) is an increasing function.
For x>0, f(x) > f(0)
=> f’(x) > 2e2x > 0
So f is an increasing function in (0, ∞)
Hence option a and c are the answers.
Question 4: Let h(x) = f(x) – (f(x))2 + (f(x))3 for every real number x. Then
(a) h is increasing whenever f is increasing
(b) h is increasing whenever f is decreasing
(c) h is decreasing whenever f is decreasing
(d) none of these
Answer: a, c
Solution:
h(x) = f(x) – (f(x))2 + (f(x))3
h’(x) = f’(x) (1 – 2f(x) + 3 (f(x))2]
= 3f’(x) [ (f(x))2 – â…” f(x) + â…“ ]
= 3f’(x) [ {f(x) – â…“ }2 + 2/9 ]
Here h’(x) < 0 when f’(x) < 0
h’(x) > 0, when f’(x) > 0
h(x) increases when f(x) increases.
h(x) decreases when f(x) decreases.
Hence option a and c are correct.
Question 5: If f(x) = [(p2-1)/(p2+1)] [x3 – 3x + log 2] is a strictly decreasing function for x belongs to R, then the set of possible values of P is
(a) (-1, 1)
(b) (1, ∞)
(c) ( -∞, 1)
(d) R
Answer: a
Solution:
Given f(x) = [(p2-1)/(p2+1)] [x3 – 3x + log 2]
f’(x) = [(p2-1)/(p2+1)](3x2 – 3)
For decreasing function, f’(x) < 0
So [(p2-1)/(p2+1)](3x2 – 3) < 0
=> [(p2-1)/(p2+1)] x2 < 1
=> (p2-1)x2 < p2+1
=> (p2-1)x2 – (p2+1) < 0
=> p2-1 < 0
=> p2-1 is negative
So p ∈ (-1, 1)
Hence option a is the answer.
Question 6: The real number k for which the equation, 2x3 + 3x + k = 0 has two distinct real roots in [0, 1]
(a) lies between 1 and 2
(b) lies between 2 and 3
(c) lies between 1 and 0
(d) does not exist
Answer: d
Solution:
f(x) = 2x3 + 3x + k
f’(x) = 6x2 + 3 > 0
So f(x) is strictly increasing.
f(x) = 0 has only one real root.
So two roots are not possible.
Hence option d is the answer.
Question 7: If f(x) = xex(1-x), then f(x) is
(a) increasing on [-½, 1]
(b) decreasing on R
(c) increasing on R
(d) decreasing on [-½, 1]
Answer: a
Solution:
Given f(x) = xex(1-x)
f’(x) = ex(1-x) + (1 – 2x) x ex(1-x)
= -ex(1-x) (2x+1)(x-1)
Critical points are x = -½ and 1.
Hence f(x) is increasing in [-½, 1]
Hence option a is the answer.
Question 8: If the function f: R – {1, -1} → A defined by f(x) = x2/(1+x2) is surjective, then A is equal to
Answer: [0, 1)
Solution:
f(x) = x2/(1+x2)
For surjective function codomain = Range
f(x) = (x2 +1 – 1)/(1+x2)
= 1 – 1/(1+x2)
f(x)min = 1 – 1/(1+0) = 1-1 = 0
f(x)max ≈ 1
Range of f(x) = [0, 1)
So A = [0, 1)
Question 9: The function f defined by f(x) = x3 – 3x2 + 5x + 7 is
(a) increasing in R
(b) decreasing in R
(c) decreasing in (0, ∞) and increasing in (-∞, 0)
(d) increasing in (0, ∞) and decreasing in (-∞, 0)
Answer: a
Solution:
Given f(x) = x3 – 3x2 + 5x + 7
f’(x) = 3x2 – 6x + 5
Discriminant of above equation = 36 – 60 = -24 < 0
For x> 0, f’(x) > 0
For x< 0, f’(x) > 0
So the function is increasing in R.
Hence option a is the answer.
Question 10: If the normal to the curve y = f(x) at the point (3, 4) makes an angle 3π/4 with the positive x-axis, then f’(3) =
(a) -1
(b) -3/4
(c) 4/3
(d) 1
Answer: d
Solution:
Slope of tangent = dy/dx
= f’(x)(3,4)
Slope of normal = -1/f’(x)(3,4)
=> -1/f’(3) = tan 3π/4
= tan (π/2 + π/4 )
= -1
So f’(3) = 1
Hence option d is the answer.
Also Read
JEE Main Maths Limits, Continuity and Differentiability Previous Year Questions With Solutions
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