Q242 / 270 · NDA 2024 (Phase 1) · Mathematics

A differentiable function $f(x)$ has a local maximum at $x=0$. Let $y=2 f(x)+a x-b$. The function $y$ has a relative maxima at $x=0$ for

  • A
    $a \gt 0, b=0$
  • Correct answer
    B
    for all $b$ and $a=0$
  • C
    for all $b \gt 0$ only
  • D
    for all $a$ and $b=0$

Solution

Given, $\quad y=2 f(x)+a x-b$ $\qquad ....(i)$ Differentiating both side w.r. to $x$ $\begin{aligned} \Rightarrow \quad \frac{d y}{d x} & =2 f^{\prime}(x)+a(1) \qquad ....(ii) \\ \frac{d^2 y}{d x^2} & =2 f^{\prime \prime}(x) \end{aligned} \qquad ....(iii)$ $\because y$ has a relative maxima at $x=0$ $\begin{aligned} & \therefore & \left(\frac{d y}{d x}\right)_{x=0} & =0 \\ & & & \\ \Rightarrow & & 2 f^{\prime}(0)+a & =0 \\ \Rightarrow & & 2(0)+a & =0 \\ & & a & =0 \end{aligned}$ Also, $\quad\left(\frac{d^2 y}{d x^2}\right)_{x=0} \lt 0$ $\begin{aligned} \Rightarrow & 2 f^{\prime \prime}(0) & \lt 0 \\ \Rightarrow & f^{\prime \prime}(0) & =0\end{aligned}$ Clearly, $y$ has relative maxima for $a=0$ and all values of $b$.

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

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