Q104 / 160 · AP EAMCET 2019 (23 Apr Shift 1) · Mathematics

If a function \(f\) is defined by : \(\begin{aligned} f(x) & =0, \text { when } x=1, \\ & =x^3-1, \text { when } 1 < x < \infty, \end{aligned}\) \(=x-1\), when \(-\infty < x < 1\), then at \(x=1, f\) is

  • A
    continuous and differentiable
  • Correct answer
    B
    continuous but not differentiable
  • C
    discontinuous and differentiable
  • D
    discontinuous and not differentiable

Solution

We have, \(f(x)=\left\{\begin{array}{rc} x-1, & -\infty < x < 1 \\ 0, & x=1 \\ x^3-1, & 1 < x < \infty \end{array}\right.\) Now, (LHL at \(x=1\) ) \(\begin{aligned} & =\lim _{x \rightarrow 1}(x-1) \\ & =\mathbf{l}-\mathbf{l}=\mathbf{0} \end{aligned}\) (RHL at \(x=1\) ) \(=\lim _{x \rightarrow 1} x^3-1\) \(=(\mathrm{l})^3-\mathbf{1}=0 \text { and } f(\mathrm{l})=0\) \(\therefore \quad \text { LHL }=\text { RHL }=f(\mathrm{l})\) So, \(f(x)\) is continuous at \(x=1\) Now, (LHD at \(x=1\) ) \(=\lim _{x \rightarrow 1} \frac{(x-1)-0}{x-1}=1\) and (RHD at \(x=1\) ) \(\begin{aligned} & =\lim _{x \rightarrow 1} \frac{\left(x^3-1\right)-0}{x-1} \\ & =\lim _{x \rightarrow 1}\left(x^2+x+1\right)=3 \end{aligned}\) \(\therefore\) LHD \(\neq\) RHD So, \(f(x)\) is not differentiable at \(x=1\)

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

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