JEE 2023 — Mathematics PYQ
JEE | Mathematics | 2023Let y(x)=(1+x)(1+x2)(1+x4)(1+x8)(1+x16). Then y′−y′′ at x=−1 is equal to:
Choose the correct answer:
- A.
976
- B.
944
- C.
464
- D.
496
(Correct Answer)
496
Explanation
Solving
1. y(x) ko simplify karna:
Dono taraf (1−x) se multiply aur divide karne par:
y(x)=1−x(1−x)(1+x)(1+x2)(1+x4)(1+x8)(1+x16)
y(x)=1−x1−x32
2. y′(−1) nikaalna:
Quotient rule se ya substitution se:
y′(x)=(1−x)2(1−x)(−32x31)−(1−x32)(−1)
At x=−1:
y′(−1)=4(2)(32)−(0)=464=16
3. y′′(−1) nikaalna:
Differentiate karne par aur x=−1 rakhne par:
y′′(−1)=−480
4. Final Value:
y′(−1)−y′′(−1)=16−(−480)=496
Correct Option: (4)
Explanation
Solving
1. y(x) ko simplify karna:
Dono taraf (1−x) se multiply aur divide karne par:
y(x)=1−x(1−x)(1+x)(1+x2)(1+x4)(1+x8)(1+x16)
y(x)=1−x1−x32
2. y′(−1) nikaalna:
Quotient rule se ya substitution se:
y′(x)=(1−x)2(1−x)(−32x31)−(1−x32)(−1)
At x=−1:
y′(−1)=4(2)(32)−(0)=464=16
3. y′′(−1) nikaalna:
Differentiate karne par aur x=−1 rakhne par:
y′′(−1)=−480
4. Final Value:
y′(−1)−y′′(−1)=16−(−480)=496
Correct Option: (4)