PYQSidePowered byPrepblink ExamsMy practice

Integrals

JEE Main 2024 (Online) 27th January Evening ShiftMCQMedium📈 Topic in 21% papers+4−1
Ques. 1JEE Main 2024, 27 January (Evening Shift)

The integral (x8-x2 )dx (x12+3x6+1 ) -1 (x3+ 1x3 ) is equal to :

  1. e ( | -1 (x3+ 1x3 ) | )1/3+C
  2. e ( | -1 (x3+ 1x3 ) | )+C
  3. e ( | -1 (x3+ 1x3 ) | )1/2+C
  4. e ( | -1 (x3+ 1x3 ) | )3+C

Answer

(1) e ( | -1 (x3+ 1x3 ) | )1/3+C

Solution

Put t= -1 (x3+ 1x3 )

dt= 3 (x2- 1x4 )1+ (x3+ 1x3 )2 dx

= 3 (x6-1 )/x4 (x12+3x6+1 )/x6 dx

= 3 (x8-x2 )x12+3x6+1 dx

1+ (x3+ 1x3 )2= x12+3x6+1x6

I= 13 dtt= 13 |t|+C

= e | -1 (x3+ 1x3 ) |1/3+C

Answer: (1)

Keep practising — Integrals

Ques. 2JEE Main 2024, 27 January (Evening Shift)

For 0<a<1, the value of the integral 0 dx1-2a x+a2 is :

  1. 2 +a2
  2. 2 -a2
  3. 1-a2
  4. 1+a2
Ques. 3JEE Main 2024, 27 January (Evening Shift)

Let f(x)= 0x g(t) e ( 1-t1+t )dt, where g is a continuous odd function. If - /2 /2 (f(x)+ x2 x1+ex )dx= ( )2- , then is equal to .

All Integrals PYQs →

PDFDownload Integrals PYQs as PDF Free, with answer key and a QR code to every solution — print it or study offline⬇
All JEE Main Mathematics chapters