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Test: Single Correct MCQs: Definite Integrals and Applications of Integrals | JEE Advanced


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41 Questions MCQ Test Mathematics For JEE | Test: Single Correct MCQs: Definite Integrals and Applications of Integrals | JEE Advanced

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Test: Single Correct MCQs: Definite Integrals and Applications of Integrals | JEE Advanced - Question 1

The value of the definite integra  equal to a.

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


Test: Single Correct MCQs: Definite Integrals and Applications of Integrals | JEE Advanced - 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 Test: Single Correct MCQs: Definite Integrals and Applications of Integrals | JEE Advanced - 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).

Test: Single Correct MCQs: Definite Integrals and Applications of Integrals | JEE Advanced - 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 Test: Single Correct MCQs: Definite Integrals and Applications of Integrals | JEE Advanced - 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)

Test: Single Correct MCQs: Definite Integrals and Applications of Integrals | JEE Advanced - Question 4

The value of the integral 

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


Test: Single Correct MCQs: Definite Integrals and Applications of Integrals | JEE Advanced - Question 5

For any integer n the integral –– 

has the value

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


Test: Single Correct MCQs: Definite Integrals and Applications of Integrals | JEE Advanced - Question 6

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

Detailed Solution for Test: Single Correct MCQs: Definite Integrals and Applications of Integrals | JEE Advanced - 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

Test: Single Correct MCQs: Definite Integrals and Applications of Integrals | JEE Advanced - Question 7

The value of 

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


Test: Single Correct MCQs: Definite Integrals and Applications of Integrals | JEE Advanced - Question 8

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

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


Test: Single Correct MCQs: Definite Integrals and Applications of Integrals | JEE Advanced - Question 9

The  value of  represents the greatest integer function is

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



Test: Single Correct MCQs: Definite Integrals and Applications of Integrals | JEE Advanced - Question 10

equals

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




Test: Single Correct MCQs: Definite Integrals and Applications of Integrals | JEE Advanced - Question 11

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

We have




Adding (1) and (2), we get

Test: Single Correct MCQs: Definite Integrals and Applications of Integrals | JEE Advanced - 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 Test: Single Correct MCQs: Definite Integrals and Applications of Integrals | JEE Advanced - Question 12

In the range   we have to find the value of

Test: Single Correct MCQs: Definite Integrals and Applications of Integrals | JEE Advanced - Question 13

 where f is such that 

Then g(2) satisfies the inequality

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


(applying line integral on inequality)

     … (1)

(applying line integral on inequality)

    … (2)

From (1) and (2), we get

Test: Single Correct MCQs: Definite Integrals and Applications of Integrals | JEE Advanced - Question 14

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



[∵ ecos x sin x is an odd function.]

Test: Single Correct MCQs: Definite Integrals and Applications of Integrals | JEE Advanced - Question 15

The value of the integral 

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


We know that for    and hence 

and for 

Test: Single Correct MCQs: Definite Integrals and Applications of Integrals | JEE Advanced - Question 16

The value of 

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

               .....(1)
 
Adding (1) and (2),



 (even function)
             ...(3)

          ....(4) 
Adding (3) and (4) 
∴ I = π /2

Test: Single Correct MCQs: Definite Integrals and Applications of Integrals | JEE Advanced - Question 17

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

Detailed Solution for Test: Single Correct MCQs: Definite Integrals and Applications of Integrals | JEE Advanced - 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.

Test: Single Correct MCQs: Definite Integrals and Applications of Integrals | JEE Advanced - Question 18

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

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

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

Test: Single Correct MCQs: Definite Integrals and Applications of Integrals | JEE Advanced - Question 19

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

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

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


⇒ f is a periodic function of period T





Test: Single Correct MCQs: Definite Integrals and Applications of Integrals | JEE Advanced - Question 20

The integral 

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



Test: Single Correct MCQs: Definite Integrals and Applications of Integrals | JEE Advanced - Question 21

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

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


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

Test: Single Correct MCQs: Definite Integrals and Applications of Integrals | JEE Advanced - Question 22

increases in

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





∴ f(x) increases when  x < 0

Test: Single Correct MCQs: Definite Integrals and Applications of Integrals | JEE Advanced - Question 23

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

Detailed Solution for Test: Single Correct MCQs: Definite Integrals and Applications of Integrals | JEE Advanced - 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

Test: Single Correct MCQs: Definite Integrals and Applications of Integrals | JEE Advanced - Question 24

If f (x) is differentiable and  equals

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

Differentiating both sides w.r.t. t  

[Using Leibnitz theorem]

Test: Single Correct MCQs: Definite Integrals and Applications of Integrals | JEE Advanced - Question 25

The value of the integral 

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


Test: Single Correct MCQs: Definite Integrals and Applications of Integrals | JEE Advanced - 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 Test: Single Correct MCQs: Definite Integrals and Applications of Integrals | JEE Advanced - Question 26

y = ax2 and x = ay2

Points of intersection are O (0, 0) and 


Test: Single Correct MCQs: Definite Integrals and Applications of Integrals | JEE Advanced - Question 27

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

Test: Single Correct MCQs: Definite Integrals and Applications of Integrals | JEE Advanced - 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 Test: Single Correct MCQs: Definite Integrals and Applications of Integrals | JEE Advanced - 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)

Test: Single Correct MCQs: Definite Integrals and Applications of Integrals | JEE Advanced - Question 29

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

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

The given curves are 



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





Also when  

Test: Single Correct MCQs: Definite Integrals and Applications of Integrals | JEE Advanced - Question 30

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

and f (0) = 0, then 

Detailed Solution for Test: Single Correct MCQs: Definite Integrals and Applications of Integrals | JEE Advanced - 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.

Test: Single Correct MCQs: Definite Integrals and Applications of Integrals | JEE Advanced - Question 31

The value of 

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

Applying L’ Hospital’s rule, we get

Test: Single Correct MCQs: Definite Integrals and Applications of Integrals | JEE Advanced - 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 Test: Single Correct MCQs: Definite Integrals and Applications of Integrals | JEE Advanced - Question 32

Now on differentiating, we get


Test: Single Correct MCQs: Definite Integrals and Applications of Integrals | JEE Advanced - Question 33

The value of 

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

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

Test: Single Correct MCQs: Definite Integrals and Applications of Integrals | JEE Advanced - 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 Test: Single Correct MCQs: Definite Integrals and Applications of Integrals | JEE Advanced - Question 34



Test: Single Correct MCQs: Definite Integrals and Applications of Integrals | JEE Advanced - 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 Test: Single Correct MCQs: Definite Integrals and Applications of Integrals | JEE Advanced - Question 35

We have


Test: Single Correct MCQs: Definite Integrals and Applications of Integrals | JEE Advanced - Question 36

The value of the integral 

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


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

Test: Single Correct MCQs: Definite Integrals and Applications of Integrals | JEE Advanced - Question 37

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

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

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



Test: Single Correct MCQs: Definite Integrals and Applications of Integrals | JEE Advanced - 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 Test: Single Correct MCQs: Definite Integrals and Applications of Integrals | JEE Advanced - Question 38


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

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

Test: Single Correct MCQs: Definite Integrals and Applications of Integrals | JEE Advanced - Question 39

The following integral 

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




Test: Single Correct MCQs: Definite Integrals and Applications of Integrals | JEE Advanced - Question 40

The value of  is equal to

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



Test: Single Correct MCQs: Definite Integrals and Applications of Integrals | JEE Advanced - Question 41

Area of the region   is equal to

Detailed Solution for Test: Single Correct MCQs: Definite Integrals and Applications of Integrals | JEE Advanced - 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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