Q68 / 180 · COMEDK 2024 (Memory Based) · Mathematics
If $f(x)=\left\{\begin{array}{lc}x^2+1 & x \lt 0 \\ x^3-a & 0 \leq x \lt 1 \text { and it is continuous } \\ 2 b-x & x \geq 1\end{array}\right.$ $\forall x \in R$, then
- A$a=1$ and $b=-1$
- B$a=1$ and $b=1$
- Correct answerC$a=-1$ and $b=\frac{3}{2}$
- D$a=-1$ and $b=\frac{-3}{2}$
Solution
If $f(x)$ is continuous, then
$\begin{array}{lll}<br/>\therefore f\left(0^{-}\right) =f\left(0^{+}\right) \\<br/>0+1 =0-a \\<br/>-1 =a \\<br/>f\left(1^{-}\right) =f\left(1^{+}\right) \\<br/>1-a =2 b-1 \\<br/>=2 b-1 \\<br/>\frac{3}{2} =b<br/>\end{array}$
$\begin{array}{lll}<br/>\therefore f\left(0^{-}\right) =f\left(0^{+}\right) \\<br/>0+1 =0-a \\<br/>-1 =a \\<br/>f\left(1^{-}\right) =f\left(1^{+}\right) \\<br/>1-a =2 b-1 \\<br/>=2 b-1 \\<br/>\frac{3}{2} =b<br/>\end{array}$
Hence, the correct answer is option: 'C'.