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Limits Continuity and Differentiability question

2019 · 12 Jan · Shift 2 · Q36
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  5. /2019 · 12 Jan · Shift 2 · Q36

Limits Continuity and Differentiability question

2019 · 12 Jan · Shift 2 · Q36

JEE MainMathematicsLimits Continuity and DifferentiabilityMCQ+4 / −1
Let f be a differentiable function such that f(1) = 2 and f '(x) = f(x) for all x ∈\in∈ R R. If h(x) = f(f(x)), then h'(1) is equal to :
  1. A
    4e
  2. B
    2e2
  3. C
    4e2
  4. D
    2e
View written solutionFree

Correct answer: A

  1. We are given:

    \quad \text{for all } x\in \mathbb{R},$$ and $$f(1)=2.$$

  2. Solve the differential equation.

    Since f′(x)=f(x),f'(x)=f(x),f′(x)=f(x), the general solution is f(x)=Cex,f(x)=Ce^x,f(x)=Cex, where CCC is a constant.

  3. Use the condition f(1)=2f(1)=2f(1)=2.

    f(1)=Ce1=Ce=2f(1)=Ce^1=Ce=2f(1)=Ce1=Ce=2 ⇒C=2e.\Rightarrow C=\frac{2}{e}.⇒C=e2​.

    Hence, f(x)=2eex=2ex−1.f(x)=\frac{2}{e}e^x=2e^{x-1}.f(x)=e2​ex=2ex−1.

  4. Define h(x)=f(f(x)).h(x)=f(f(x)).h(x)=f(f(x)).

    We need h′(1)h'(1)h′(1). Using the chain rule: h′(x)=f′(f(x))⋅f′(x).h'(x)=f'(f(x))\cdot f'(x).h′(x)=f′(f(x))⋅f′(x).

  5. Since f′(t)=f(t)f'(t)=f(t)f′(t)=f(t) for every input ttt, f′(f(x))=f(f(x))f'(f(x))=f(f(x))f′(f(x))=f(f(x)) and f′(x)=f(x).f'(x)=f(x).f′(x)=f(x).

    Therefore, h′(x)=f(f(x)) f(x).h'(x)=f(f(x))\,f(x).h′(x)=f(f(x))f(x).

  6. Evaluate at x=1x=1x=1.

    First, f(1)=2.f(1)=2.f(1)=2.

    Next, f(f(1))=f(2).f(f(1))=f(2).f(f(1))=f(2).

    Using f(x)=2ex−1f(x)=2e^{x-1}f(x)=2ex−1, f(2)=2e2−1=2e.f(2)=2e^{2-1}=2e.f(2)=2e2−1=2e.

    So, h′(1)=f(f(1))⋅f(1)=f(2)⋅2=(2e)(2)=4e.h'(1)=f(f(1))\cdot f(1)=f(2)\cdot 2=(2e)(2)=4e.h′(1)=f(f(1))⋅f(1)=f(2)⋅2=(2e)(2)=4e.

  7. Compare with options: 4e\boxed{4e}4e​ which is option A.

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