JEE Main Previous Year Questions (2023): Indefinite Integral | Chapter-wise Tests for JEE Main & Advanced PDF Download

[JEE Main MCQs]

Q1: For and C is constant of integration, then α + 2β + 3γ - 4δ is equal to
(a) -8
(b) -4
(c) 1
(d) 4
Ans: (d)
We have,

Let,

∴ α + 2β + 3γ - 4δ = 2 + 2 × 2 + 3 × 2 − 4 × 2 = 4

Q2: If I(x) =  (cos x sin 2x - sin x)dx and I (0) = 1, then I(π/3) is equal to:
(a)
(b)
(c)
(d)
Ans: (d)




Q3: The integral  is equal to:
(a)
(b)
(c)
(d) None
Ans: (a)



Q4: Let  then I (1) is equal to:
(a)
(b)
(c)
(d)
Ans: (c)

Comparing coefficients of t², t and constant terms, we get
A + B = 0 , C − B = 0 , − C = 1
On solving above equations, we get
C = -1, = B, A = 1


⇒ 0 + 0 + C = 0 ⇒ C = 0


Q5: Let  If I (0) = 0, then I (π/4) is equal to:
(a)
(b)
(c)
(d)
Ans: (a)
We have,

Now, let

On putting x sin ⁡ x + cos ⁡ x = t
⇒ (x cos ⁡ x + sin x − sin x) dx = dt
⇒ x cos ⁡ x dx = dt

 = 2 log ⁡ (x sin⁡ x + cosx) + c
 
When, x = 0 , then
I ( 0 ) = 0 + 2 log ⁡ ( 1 ) + c = 0
⇒ c = 0


Q6: Let  is equal to
(a)
(b)
(c)
(d)
Ans: (d)

[JEE Main Numericals]

Q7: Let . If f(0) = 0
and  is equal to ____________.
Ans: 28




Q8: Let  then α⁴ is equal to _________.
Ans: 4
Given integral:  
Let's make the substitution x = t². Then, dx = 2t dt.
Substituting these values, the integral becomes:

Now, let's evaluate this integral:
 
Substituting back t = √x, we have:
 
Simplifying further:
 
We are given that I (9) = 12 + 7 ln ⁡ 7 .
Let's substitute x = 9 and solve for the constant C:
 
From this equation, we can see that C = 0.
Now, we need to calculate I (1):
 
Therefore, α = 8.
Finally, to find α⁴:
α⁴ = ( 8 ) 4
⇒ α⁴ = 8 2
⇒ α⁴ = 64
Hence, α⁴ is equal to 64.

Q9: If  is equal to ____________.
Ans: 1