JEE Advanced (Single Correct MCQs): Definite Integrals and Applications of Integrals - JEE MCQ

# JEE Advanced (Single Correct MCQs): Definite Integrals and Applications of Integrals - JEE MCQ

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## 41 Questions MCQ Test Chapter-wise Tests for JEE Main & Advanced - JEE Advanced (Single Correct MCQs): Definite Integrals and Applications of Integrals

JEE Advanced (Single Correct MCQs): Definite Integrals and Applications of Integrals for JEE 2024 is part of Chapter-wise Tests for JEE Main & Advanced preparation. The JEE Advanced (Single Correct MCQs): Definite Integrals and Applications of Integrals questions and answers have been prepared according to the JEE exam syllabus.The JEE Advanced (Single Correct MCQs): Definite Integrals and Applications of Integrals MCQs are made for JEE 2024 Exam. Find important definitions, questions, notes, meanings, examples, exercises, MCQs and online tests for JEE Advanced (Single Correct MCQs): Definite Integrals and Applications of Integrals below.
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JEE Advanced (Single Correct MCQs): Definite Integrals and Applications of Integrals - Question 1

### The value of the definite integra  equal to a.

Detailed Solution for JEE Advanced (Single Correct MCQs): Definite Integrals and Applications of Integrals - Question 1

JEE Advanced (Single Correct MCQs): Definite Integrals and Applications of Integrals - Question 2

### Let a, b, c be non-zero real numbers such that Then the quadratic equation ax2 + bx +c= 0 has

Detailed Solution for JEE Advanced (Single Correct MCQs): Definite Integrals and Applications of Integrals - Question 2

Now we know that if  then it means that
f (x) is + ve on some part of (α, β) and – ve on other part of (α, β).
But here 1 + cos8 x is always + ve,
∴ ax2 + bx + c is + ve on some part of [1 , 2] and – ve on other part [1, 2]
∴ ax2 + bx + c= 0 has at least one root in (1, 2).
⇒ ax2 + bx + c = 0 has at least one root in (0,2).

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JEE Advanced (Single Correct MCQs): Definite Integrals and Applications of Integrals - Question 3

### The area  bounded by the curves y = f(x), the x-axis and the ordinates x = 1 and x = b is (b – 1) sin (3b + 4). Then f(x) is

Detailed Solution for JEE Advanced (Single Correct MCQs): Definite Integrals and Applications of Integrals - Question 3

Differentiating both sides w.r.t b, we get
f(b) = 3(b – 1) cos(3b + 4) + sin(3b + 4)
⇒ f(x) = 3(x – 1) cos(3x + 4) + sin(3x + 4)

JEE Advanced (Single Correct MCQs): Definite Integrals and Applications of Integrals - Question 4

The value of the integral

Detailed Solution for JEE Advanced (Single Correct MCQs): Definite Integrals and Applications of Integrals - Question 4

JEE Advanced (Single Correct MCQs): Definite Integrals and Applications of Integrals - Question 5

For any integer n the integral ––

has the value

Detailed Solution for JEE Advanced (Single Correct MCQs): Definite Integrals and Applications of Integrals - Question 5

JEE Advanced (Single Correct MCQs): Definite Integrals and Applications of Integrals - Question 6

Let f : R → R and g : R → R be continous functions. Then the value of the integral

Detailed Solution for JEE Advanced (Single Correct MCQs): Definite Integrals and Applications of Integrals - Question 6

We have,

Let F (x) = (f (x) + f (- x)) (g (x) - g (-x))
then F (-x) = (f (- x) + f (x))(g (- x)- g (x)) = - [f (x) + f (- x)][ g(x) - g (-x)]
=-F (x)
∴ F(x) is an odd function, ∴ we get I = 0

JEE Advanced (Single Correct MCQs): Definite Integrals and Applications of Integrals - Question 7

The value of

Detailed Solution for JEE Advanced (Single Correct MCQs): Definite Integrals and Applications of Integrals - Question 7

JEE Advanced (Single Correct MCQs): Definite Integrals and Applications of Integrals - Question 8

If f (x) and  then constants A and B are

Detailed Solution for JEE Advanced (Single Correct MCQs): Definite Integrals and Applications of Integrals - Question 8

JEE Advanced (Single Correct MCQs): Definite Integrals and Applications of Integrals - Question 9

The  value of  represents the greatest integer function is

Detailed Solution for JEE Advanced (Single Correct MCQs): Definite Integrals and Applications of Integrals - Question 9

JEE Advanced (Single Correct MCQs): Definite Integrals and Applications of Integrals - Question 10

equals

Detailed Solution for JEE Advanced (Single Correct MCQs): Definite Integrals and Applications of Integrals - Question 10

JEE Advanced (Single Correct MCQs): Definite Integrals and Applications of Integrals - Question 11

Detailed Solution for JEE Advanced (Single Correct MCQs): Definite Integrals and Applications of Integrals - Question 11

We have

Adding (1) and (2), we get

JEE Advanced (Single Correct MCQs): Definite Integrals and Applications of Integrals - Question 12

If for a real number y, [y] is the greatest integer less than or equal to y, then the value of the integral

Detailed Solution for JEE Advanced (Single Correct MCQs): Definite Integrals and Applications of Integrals - Question 12

In the range   we have to find the value of

JEE Advanced (Single Correct MCQs): Definite Integrals and Applications of Integrals - Question 13

where f is such that

Then g(2) satisfies the inequality

Detailed Solution for JEE Advanced (Single Correct MCQs): Definite Integrals and Applications of Integrals - Question 13

(applying line integral on inequality)

… (1)

(applying line integral on inequality)

… (2)

From (1) and (2), we get

JEE Advanced (Single Correct MCQs): Definite Integrals and Applications of Integrals - Question 14

Detailed Solution for JEE Advanced (Single Correct MCQs): Definite Integrals and Applications of Integrals - Question 14

[∵ ecos x sin x is an odd function.]

JEE Advanced (Single Correct MCQs): Definite Integrals and Applications of Integrals - Question 15

The value of the integral

Detailed Solution for JEE Advanced (Single Correct MCQs): Definite Integrals and Applications of Integrals - Question 15

We know that for    and hence

and for

JEE Advanced (Single Correct MCQs): Definite Integrals and Applications of Integrals - Question 16

The value of

Detailed Solution for JEE Advanced (Single Correct MCQs): Definite Integrals and Applications of Integrals - Question 16

.....(1)

(even function)
...(3)

....(4)
∴ I = π /2

JEE Advanced (Single Correct MCQs): Definite Integrals and Applications of Integrals - Question 17

The area bounded by the curves y = |x| –1 and y = –|x| + 1 is

Detailed Solution for JEE Advanced (Single Correct MCQs): Definite Integrals and Applications of Integrals - Question 17

The given lines are
y = x – 1; y = – x – 1;
y = x  + 1 and  y =  – x + 1
which are two pairs of parallel lines and distance between the lines of each pair is √2 Also non parallel lines are perpendicular. Thus lines represents a square of side √2 Hence, area = (√2)2 = 2 sq. units.

JEE Advanced (Single Correct MCQs): Definite Integrals and Applications of Integrals - Question 18

Then the real roots of the equation x2 – f '(x) = 0 are

Detailed Solution for JEE Advanced (Single Correct MCQs): Definite Integrals and Applications of Integrals - Question 18

Now the given equation x2 - f ' (x)=0 becomes

JEE Advanced (Single Correct MCQs): Definite Integrals and Applications of Integrals - Question 19

Let T > 0 be a fixed real number. Suppose f is a continuous function such that for all

Detailed Solution for JEE Advanced (Single Correct MCQs): Definite Integrals and Applications of Integrals - Question 19

Given that T > 0 is a fixed real number. f is continuous

⇒ f is a periodic function of period T

JEE Advanced (Single Correct MCQs): Definite Integrals and Applications of Integrals - Question 20

The integral

Detailed Solution for JEE Advanced (Single Correct MCQs): Definite Integrals and Applications of Integrals - Question 20

JEE Advanced (Single Correct MCQs): Definite Integrals and Applications of Integrals - Question 21

then the expression for l(m, n) in terms of l(m + 1, n – 1) is

Detailed Solution for JEE Advanced (Single Correct MCQs): Definite Integrals and Applications of Integrals - Question 21

Intergrating by parts considering (1 + t)n as first function, we get

JEE Advanced (Single Correct MCQs): Definite Integrals and Applications of Integrals - Question 22

increases in

Detailed Solution for JEE Advanced (Single Correct MCQs): Definite Integrals and Applications of Integrals - Question 22

∴ f(x) increases when  x < 0

JEE Advanced (Single Correct MCQs): Definite Integrals and Applications of Integrals - Question 23

The area bounded by the curves  and x-axis in the 1st quadrant is

Detailed Solution for JEE Advanced (Single Correct MCQs): Definite Integrals and Applications of Integrals - Question 23

The curves given are

and x-axis y = 0 ...(3)
Eqn. (1), [y2 = x] represents right handed parabola but with +ve values of y i.e., part of curve lying above x-axis.
Solving (1) and (2) we get,

Also (2) meets x-axis at (3, 0)

Shaded area is the required area given by

JEE Advanced (Single Correct MCQs): Definite Integrals and Applications of Integrals - Question 24

If f (x) is differentiable and  equals

Detailed Solution for JEE Advanced (Single Correct MCQs): Definite Integrals and Applications of Integrals - Question 24

Differentiating both sides w.r.t. t

[Using Leibnitz theorem]

JEE Advanced (Single Correct MCQs): Definite Integrals and Applications of Integrals - Question 25

The value of the integral

Detailed Solution for JEE Advanced (Single Correct MCQs): Definite Integrals and Applications of Integrals - Question 25

JEE Advanced (Single Correct MCQs): Definite Integrals and Applications of Integrals - Question 26

The area enclosed between the curves y = ax2 and x = ay2 (a > 0) is 1 sq. unit, then the value of a is

Detailed Solution for JEE Advanced (Single Correct MCQs): Definite Integrals and Applications of Integrals - Question 26

y = ax2 and x = ay2

Points of intersection are O (0, 0) and

JEE Advanced (Single Correct MCQs): Definite Integrals and Applications of Integrals - Question 27

Detailed Solution for JEE Advanced (Single Correct MCQs): Definite Integrals and Applications of Integrals - Question 27

JEE Advanced (Single Correct MCQs): Definite Integrals and Applications of Integrals - Question 28

Th e area bounded by the par abolas y = (x + 1)2 and y = (x – 1)2 and the line y = 1/4 is

Detailed Solution for JEE Advanced (Single Correct MCQs): Definite Integrals and Applications of Integrals - Question 28

The given curves are

y = (x+ 1)2     ...(1)

upward parabola with vertex at (–1 ,0) meeting  y –axis at (0, 1)
y = (x -1)2                 ...(2)

a line parallel to x–axis meeting (1) at

and meeting (2) at

The graph is as shown

The required area is the shaded portion given by ar (BPCQB)  = 2 Ar(PQCP) (by symmetry)

JEE Advanced (Single Correct MCQs): Definite Integrals and Applications of Integrals - Question 29

The area of the region between the curves   and   bounded by the lines x = 0 and   is

Detailed Solution for JEE Advanced (Single Correct MCQs): Definite Integrals and Applications of Integrals - Question 29

The given curves are

∴ The area bounded by the above curves, by the lines

Also when

JEE Advanced (Single Correct MCQs): Definite Integrals and Applications of Integrals - Question 30

Let f be  a non-negative function defined on the interval

and f (0) = 0, then

Detailed Solution for JEE Advanced (Single Correct MCQs): Definite Integrals and Applications of Integrals - Question 30

Given that f is a non negative function defined on

Differentiating both sides with respect to x, we get

Integrating both sides with respect to x, we get

∵ Given that f (0) = 0   ⇒ C = 0

Hence f (x) = ± sin x

But as f (x) is a non negative function on  [0, 1]
∴ f (x) = sinx.

JEE Advanced (Single Correct MCQs): Definite Integrals and Applications of Integrals - Question 31

The value of

Detailed Solution for JEE Advanced (Single Correct MCQs): Definite Integrals and Applications of Integrals - Question 31

Applying L’ Hospital’s rule, we get

JEE Advanced (Single Correct MCQs): Definite Integrals and Applications of Integrals - Question 32

Let f  be a real-valued function defined on the interval (–1, 1) such that   and let  f –1  be the inverse function of f. Then (f –1)' (2) is equal to

Detailed Solution for JEE Advanced (Single Correct MCQs): Definite Integrals and Applications of Integrals - Question 32

Now on differentiating, we get

JEE Advanced (Single Correct MCQs): Definite Integrals and Applications of Integrals - Question 33

The value of

Detailed Solution for JEE Advanced (Single Correct MCQs): Definite Integrals and Applications of Integrals - Question 33

Adding values of I in equation (1) and (2)

JEE Advanced (Single Correct MCQs): Definite Integrals and Applications of Integrals - Question 34

Let the straight lin e x = b divide the area enclosed by y = (1 – x)2 , y = 0, and x = 0 into two parts R1 (0 < x < b) and R2 (b < x < 1) such that R1 - R2 = 1/4. Then b equals

Detailed Solution for JEE Advanced (Single Correct MCQs): Definite Integrals and Applications of Integrals - Question 34

JEE Advanced (Single Correct MCQs): Definite Integrals and Applications of Integrals - Question 35

Let f : [– 1, 2] → [0, ∞) be a continuous function such that f (x) = f (1 – x) for all x ∈ [–1,  2]

and R2 be the area of the region bounded by y = f (x), x = –1, x = 2, and the x-axis.
Then

Detailed Solution for JEE Advanced (Single Correct MCQs): Definite Integrals and Applications of Integrals - Question 35

We have

JEE Advanced (Single Correct MCQs): Definite Integrals and Applications of Integrals - Question 36

The value of the integral

Detailed Solution for JEE Advanced (Single Correct MCQs): Definite Integrals and Applications of Integrals - Question 36

[as x2 cos x is an even  cos x is an odd function]

JEE Advanced (Single Correct MCQs): Definite Integrals and Applications of Integrals - Question 37

The area enclosed by the curves y = sin x + cos x and y = cosx - sinx over the interval

Detailed Solution for JEE Advanced (Single Correct MCQs): Definite Integrals and Applications of Integrals - Question 37

The rough graph of y = sin x + cos x and y = |cos x – sin x| suggest the required area is

JEE Advanced (Single Correct MCQs): Definite Integrals and Applications of Integrals - Question 38

(the set of all real number) be a positive, non-constant and differentiable function such that  Then the value of    in the interval

Detailed Solution for JEE Advanced (Single Correct MCQs): Definite Integrals and Applications of Integrals - Question 38

⇒  e-2 x f (x) is strictly decreasing function on

Also given that f(x) is positive function so f(x) > 0

JEE Advanced (Single Correct MCQs): Definite Integrals and Applications of Integrals - Question 39

The following integral

Detailed Solution for JEE Advanced (Single Correct MCQs): Definite Integrals and Applications of Integrals - Question 39

JEE Advanced (Single Correct MCQs): Definite Integrals and Applications of Integrals - Question 40

The value of  is equal to

Detailed Solution for JEE Advanced (Single Correct MCQs): Definite Integrals and Applications of Integrals - Question 40

JEE Advanced (Single Correct MCQs): Definite Integrals and Applications of Integrals - Question 41

Area of the region   is equal to

Detailed Solution for JEE Advanced (Single Correct MCQs): Definite Integrals and Applications of Integrals - Question 41

Solving (i) and (ii), we get intersection points as  (1, 2), (6, 3), (– 4, 1), (–39, –6)
The graph of given region is as follows-

Required area = Area (trap PQRS) – Area (PST + TQR)

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