JEE 2023 — Mathematics PYQ
JEE | Mathematics | 2023If ∫−0.150.15∣100x2−1∣dx=3000k, then k is equal to _____.
Choose the correct answer:
- A.
575
(Correct Answer) - B.
576
- C.
577
- D.
578
575
Explanation
Solution:
Since it is an even function:
I=2∫00.15∣100x2−1∣dx=3000k
I=2[∫00.1(1−100x2)dx+∫0.10.15(100x2−1)dx]
I=2[(x−3100x3)00.1+(3100x3−x)0.10.15]
Solving the definite integrals:
I=2[(101−301)+(3100(0.003375)−0.15)−(301−101)]
3000k=3000575
k=575
Explanation
Solution:
Since it is an even function:
I=2∫00.15∣100x2−1∣dx=3000k
I=2[∫00.1(1−100x2)dx+∫0.10.15(100x2−1)dx]
I=2[(x−3100x3)00.1+(3100x3−x)0.10.15]
Solving the definite integrals:
I=2[(101−301)+(3100(0.003375)−0.15)−(301−101)]
3000k=3000575
k=575