JEE Mains · Maths · STD 12 - 5. continuity and differentiation
Let \(f(x) = \begin{cases} \dfrac{1}{3}, & x \leq \pi/2 \\ \dfrac{b(1-\sin x)}{(\pi-2x)^2}, & x > \pi/2 \end{cases}\). If \(f\) is continuous at \(x=\pi/2\), then the value of \(\displaystyle\int_{0}^{3b-6} |x^2+2x-3|\,dx\) is:
- A \(5\)
- B \(2\)
- C \(3\)
- D \(4\)
Answer & Solution
Correct Answer
(D) \(4\)
Step-by-step Solution
Detailed explanation
Since \(f(x)\) is continuous at \(x = \pi/2\), the right-hand limit must equal the value of the function at \(x = \pi/2\). \(\lim_{x \to \pi/2^+} f(x) = f(\pi/2)\) \(\lim_{x \to \pi/2^+} \dfrac{b(1-\sin x)}{(\pi-2x)^2} = \dfrac{1}{3}\) Let \(x = \pi/2 + h\). As…
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