Q213 / 270 · NDA 2023 (Phase 2) · Mathematics

Let $f(x)=\left\{\begin{array}{cc}a x(x+1)+b, & x \lt 1 \\ x-1 & 1 \leq x \leq 2\end{array}\right.$ If the function $f(x)$ is differentiable at $x=1$, then what is the value of $(a+b)$ ?

  • Correct answer
    A
    $-\frac{1}{3}$
  • B
    -1
  • C
    0
  • D
    1

Solution

Since, $\mathrm{f}(\mathrm{x})=\left\{\begin{array}{cc} \mathrm{ax}(\mathrm{x}+1)+\mathrm{b} ; & \mathrm{x} \lt 1 \\ \mathrm{x}-1 ; & 1 \leq \mathrm{x} \leq 2 \end{array}\right.$ Since, $f(x)$ is differential at $x=1$ $\begin{aligned} & \mathrm{f}^{\prime}(\mathrm{x})=\left\{\begin{array}{cc} \mathrm{a}(2 \mathrm{x}+1) ; & \mathrm{x} \lt 1 \\ 1 ; & 1 \leq \mathrm{x} \leq 2 \end{array}\right. \\ & \Rightarrow \mathrm{a}(3)=1 \Rightarrow \mathrm{a}=\frac{1}{3} \end{aligned}$ since $f(x)$ is continuous at $x=1$ $\Rightarrow 2 \mathrm{a}+\mathrm{b}=0 \Rightarrow \mathrm{~b}=\frac{-2}{3}$ so, $a+b=\frac{-1}{x}$

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

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