Q55 / 100 · BITSAT 2024 (Memory Based Paper 1) · Mathematics

If the function $f(x)$, defined below, is continuous on the interval $[0,8]$, then $f(x)=\left\{\begin{array}{cc}x^2+a x+b & , 0 \leq x \lt 2 \\ 3 x+2, & 2 \leq x \leq 4 \\ 2 a x+5 b, & 4 \lt x \leq 8\end{array}\right.$

  • Correct answer
    A
    $a=3, b=-2$
  • B
    $a=-3, b=2$
  • C
    $a=-3, b=-2$
  • D
    $a=3, b=2$

Solution

Here, $f(x)$ is continuous on the interval $[0,8]$. So it will also be continuous on 2 and 4. At $x=2$, $\lim _{x \rightarrow 2^{-}} f(x)=\lim _{x \rightarrow 2^{+}} f(x)$ $\lim _{x \rightarrow 2^{-}}\left(x^2+\mathrm{ax}+\mathrm{b}\right)=\lim _{x \rightarrow 2^{+}}(3 x+2)$ $4+2 a+b=3(2)+2$ $\therefore 2 a+b=4$ ...(i) At $x=4$ $\lim _{x \rightarrow 4^{-}} f(x)=\lim _{x \rightarrow 4^{+}} f(x)$ $\lim _{x \rightarrow 4^{-}}(3 x+2)=\lim _{x \rightarrow 4^{+}} 2 a x+5 b$ $3(4)+2=2 a(4)+5 b$ $\therefore 8 a+5 b=14$ .......(ii) On solving Eqs. (i) and (ii), we get $a=3$ and $b=-2$.

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

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