COMEDK · Maths · 25. Continuity and Differentiability
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 \(b=1\)
- C \(a=-1\) and \(b=\frac{3}{2}\)
- D \(a=-1\) and \(b=\frac{-3}{2}\)
Answer & Solution
Correct Answer
(C) \(a=-1\) and \(b=\frac{3}{2}\)
Step-by-step Solution
Detailed explanation
If \(f(x)\) is continuous, then \(\begin{array}{lll} \therefore f\left(0^{-}\right) =f\left(0^{+}\right) \\ 0+1 =0-a \\ -1 =a \\ f\left(1^{-}\right) =f\left(1^{+}\right) \\ 1-a =2 b-1 \\ =2 b-1 \\ \frac{3}{2} =b \end{array}\)
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