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Definite Integrals MCQ - 1


10 Questions MCQ Test Topic-wise Tests & Solved Examples for IIT JAM Mathematics | Definite Integrals MCQ - 1


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This mock test of Definite Integrals MCQ - 1 for Mathematics helps you for every Mathematics entrance exam. This contains 10 Multiple Choice Questions for Mathematics Definite Integrals MCQ - 1 (mcq) to study with solutions a complete question bank. The solved questions answers in this Definite Integrals MCQ - 1 quiz give you a good mix of easy questions and tough questions. Mathematics students definitely take this Definite Integrals MCQ - 1 exercise for a better result in the exam. You can find other Definite Integrals MCQ - 1 extra questions, long questions & short questions for Mathematics on EduRev as well by searching above.
QUESTION: 1

Let f:  be defined as f (t)  =  f(t )dt

Solution:
QUESTION: 2

A minimum value of  is

Solution:

Correct Answer :- d

Explanation : f(x) = ∫(0 to x) te-t^2 dt

f'(x) = xe-x^2 = 0

Therefore, x = 0

f"(x) = e-x^2(1 - 2x2)

f"(0) = 1 > 0

Therefore, the minimum value is f(0) = 0

QUESTION: 3

Let f(x) = ∫ex(x−1)(x−2)dx then f decrease in the interval 

Solution:

f(x) = ∫ex(x−1)(x−2)dx

For decreasing function ,f′(x)<0,f′(x)<0

⇒ex(x−1)(x−2) < 0

⇒(x−1)(x−2) < 0

⇒ ex(x−1)(x−2) < 0

⇒ (x−1)(x−2) < 0
⇒1 < x < 2,

∵ ex > 0 ∀ x ∈ R

QUESTION: 4

The value of x1 and x2 with x1 < x2 such that  has the largest value are 
Solution:
QUESTION: 5

Let f(x) =  Then the real roots of tlie equation x2 — f'(x) = 0 are
Solution:
QUESTION: 6

 is equal to
Solution:

ANSWER :- A

Solution :- limx→0 1/x[∫y→a e^sin2tdt− ∫x+y→a e^sin2tdt]

 limx→0  [∫y→a e^sin2tdt− ∫x+y→a e^sin2tdt]/x

Hence it is 0/0 form, Apply L-hospital rule.

lim x-->0  0 - {e^(sin^2a).0 - e^(sin^2(x+y).1}/1

lim x-->0 e^(sin^2(x+y))

= e^(sin^2y)

QUESTION: 7

If f(x) is differentiable and   equals
Solution:
QUESTION: 8

If 
Solution:
QUESTION: 9

The area of the region between the curves y =  and bounded by the lines x = 0 and  
Solution:
QUESTION: 10

Let f(x) = where g is a real valued continuous function on. Then f′ (x) is equal to 
Solution:

Here f(x) = where g is a real valued continuous function on R.
Apply Lebnitz's rule, we obtain

f′ (x) = 0

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