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Differentiation question

2022 · 25 Jun · Shift 1 · Q27
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  5. /2022 · 25 Jun · Shift 1 · Q27

Differentiation question

2022 · 25 Jun · Shift 1 · Q27

JEE MainMathematicsDifferentiationMCQ+4 / −1
Let f : R →\to→ R be defined as f(x)=x3+x−5f(x) = {x^3} + x - 5f(x)=x3+x−5. If g(x) is a function such that f(g(x))=x,∀′x′∈Rf(g(x)) = x,\forall 'x' \in Rf(g(x))=x,∀′x′∈R, then g'(63) is equal to ‾\underline{\hspace{2cm}}​.
  1. A
    149{1 \over {49}}491​
  2. B
    349{3 \over {49}}493​
  3. C
    4349{43 \over {49}}4943​
  4. D
    9149{91 \over {49}}4991​
View written solutionFree

Correct answer: A

  1. We are given f(x)=x3+x−5f(x)=x^3+x-5f(x)=x3+x−5 and a function ggg such that f(g(x))=x ∀x∈R.f(g(x))=x \,\forall x\in \mathbb R.f(g(x))=x∀x∈R.

    This means ggg is the inverse function of fff, i.e. g=f−1.g=f^{-1}.g=f−1.

  2. To find g′(63)g'(63)g′(63), use the derivative formula for an inverse function: g′(y)=1f′(g(y)).g'(y)=\frac{1}{f'(g(y))}.g′(y)=f′(g(y))1​.

    So, g′(63)=1f′(g(63)).g'(63)=\frac{1}{f'(g(63))}.g′(63)=f′(g(63))1​.

  3. First find g(63)g(63)g(63). Since f(g(63))=63f(g(63))=63f(g(63))=63, let g(63)=a.g(63)=a.g(63)=a. Then f(a)=63f(a)=63f(a)=63 a3+a−5=63a^3+a-5=63a3+a−5=63 a3+a−68=0.a^3+a-68=0.a3+a−68=0.

    Try integer roots. For a=4a=4a=4, 43+4−5=64+4−5=63.4^3+4-5=64+4-5=63.43+4−5=64+4−5=63. Hence, g(63)=4.g(63)=4.g(63)=4.

  4. Now compute f′(x)f'(x)f′(x): f′(x)=3x2+1.f'(x)=3x^2+1.f′(x)=3x2+1.

    Therefore, f′(4)=3(4)2+1=3⋅16+1=48+1=49.f'(4)=3(4)^2+1=3\cdot 16+1=48+1=49.f′(4)=3(4)2+1=3⋅16+1=48+1=49.

  5. So, g′(63)=1f′(4)=149.g'(63)=\frac{1}{f'(4)}=\frac{1}{49}.g′(63)=f′(4)1​=491​.

  6. Comparing with the options, the correct option is: A 149\boxed{\text{A } \frac{1}{49}}A 491​​

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