Q42 / 155 · WBJEE 2023 · Mathematics

Suppose \(f: \mathbb{R} \rightarrow \mathbb{R}\) be given by \(f(x)=\left\{\begin{array}{cc}1, \quad \text { if } x=1 \\ e^{\left(x^{10}-1\right)}+(x-1)^2 \sin \frac{1}{x-1}\end{array}\right., \text { if } x \neq 1 \text { then }\)

  • Correct answer
    A
    \(f^{\prime}(1)\) does not exist
  • B
    \(f^{\prime}(1)\) exists and is zero
  • C
    \(f^{\prime}(1)\) exist and is 9
  • D
    \(f^{\prime}(1)\) exists and is 10

Solution

Hint: \(f: R \rightarrow R f(x)=\left\{\begin{array}{cc} 1 & \text { if } x=1 & \\ e^{\left(x^{10}-1\right)} &+(x-1)^2 &\cdot \sin \left(\frac{1}{x-1}\right) \text { if } x \neq 1\end{array}\right.\) \(\begin{aligned} & f\left(1^{-}\right)=1+0=1 \\ & f\left(1^{+}\right)=1+0=1 \\ & f(1)=1 \\ & \therefore \quad f(x) \text { is continuous at } x=1 \\ & f^{\prime}(x)=\left\{e^{\left(x^{10}-1\right)} \cdot\left(10 x^5\right)+2(x-1) \cdot \sin \left(\frac{1}{x-1}\right)-(x-1)^2 \cdot \cos \left(\frac{1}{x-1}\right) \cdot\left(\frac{1}{x-1}\right)^2\right\} \end{aligned}\) which does not exist.

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

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