Q44 / 155 · WBJEE 2023 · Mathematics

\(f(x)\) is a differentiable function and given \(f^{\prime}(2)=6\) and \(f^{\prime}(1)=4\), then \(L=\lim _{h \rightarrow 0} \frac{f\left(2+2 h+h^2\right)-f(2)}{f\left(1+h-h^2\right)-f(1)}\)

  • A
    does not exist
  • B
    equal to -3
  • Correct answer
    C
    equal to 3
  • D
    equal to \(3 / 2\)

Solution

Hint: \(\operatorname{It}_{h \rightarrow 0} \frac{f\left(2+2 h+h^2\right)-f(2)}{f\left(1+h-h^2\right)-f(1)} \quad[\div(\) form \()]\) by L. Hospital rule. \(\begin{aligned} \operatorname{It}_{h \rightarrow 0} \frac{f^{\prime}\left(2+2 h+h^2\right)(2 h+2)}{f^{\prime}\left(1+h-h^2\right)(1-2 h)} & =\frac{f^{\prime}(2)}{f^{\prime}(1)} \times\left(\frac{2}{1}\right) \\ & =\frac{6}{4} \times 2=3 \end{aligned}\)

Hence, the correct answer is option: 'C'.

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