JEE 2024 — Mathematics PYQ
JEE | Mathematics | 2024Let y=loge(1+x21−x2), -1<x<1. Then at x=21, the value of 225(y′−y′′) is equal to
Choose the correct answer:
- A.
732
- B.
746
- C.
742
- D.
736
(Correct Answer)
736
Explanation
y = \log_{e}\left(\frac{1-x^{2}}{1+x^{2}}\right), \quad -1 < x < 1
y′=(1+x21−x2)1⋅(1+x2)2(1+x2)(−2x)−(1−x2)(2x)
=(1−x2)(1+x2)2x(−1−x2−1+x2)=1−x4−4x
y′′=(1−x4)2(1−x4)(−4)−(−4x)(−4x3)=(1−x4)24[−1+x4−4x4]
=4(1−x4)2(−3x4−1)=−4[(1−x4)23x4+1]
At x=21
⇒y′=1−161−4×21
=15−2×16
=−1532
y′′=−4(1−101)2(103+1)
=16−4×19×15×1516×16=225−64×19
Now 225(y′−y′′)
=225(15−32+22564×19)
=32(−15+38)
=32×23
=736
Explanation
y = \log_{e}\left(\frac{1-x^{2}}{1+x^{2}}\right), \quad -1 < x < 1
y′=(1+x21−x2)1⋅(1+x2)2(1+x2)(−2x)−(1−x2)(2x)
=(1−x2)(1+x2)2x(−1−x2−1+x2)=1−x4−4x
y′′=(1−x4)2(1−x4)(−4)−(−4x)(−4x3)=(1−x4)24[−1+x4−4x4]
=4(1−x4)2(−3x4−1)=−4[(1−x4)23x4+1]
At x=21
⇒y′=1−161−4×21
=15−2×16
=−1532
y′′=−4(1−101)2(103+1)
=16−4×19×15×1516×16=225−64×19
Now 225(y′−y′′)
=225(15−32+22564×19)
=32(−15+38)
=32×23
=736