MHT CET · Maths · Quadratic Equation
The Number of values of \(C\) that satisfy the conclusion of Rolle's theorem in case of following function \(\mathrm{f}(x)=\sin 2 \pi x, x \in[-1,1]\) is
- A 02
- B 04
- C 03
- D zero
Answer & Solution
Correct Answer
(B) 04
Step-by-step Solution
Detailed explanation
\(\mathrm{f}(x)=\sin 2 \pi x \)
\( \therefore \mathrm{f}^{\prime}(x)=2 \pi \cos 2 \pi x \)
\( \text {Now, } \mathrm{f}^{\prime}(\mathrm{C})=0 \)
\( \Rightarrow 2 \pi \cos 2 \pi \mathrm{C}=0\)
\(\Rightarrow \cos 2 \pi \mathrm{C}=0 \)
\(\Rightarrow 2 \pi \mathrm{C}=\frac{-3 \pi}{2}, \frac{-\pi}{2}, \frac{\pi}{2}, \frac{3 \pi}{2} \) \( \ldots[\because x \in(-1,1) \Rightarrow 2 \pi x \in(-2 \pi, 2 \pi)] \)
\(\Rightarrow \mathrm{C}=\frac{-3}{4},-\frac{1}{4}, \frac{1}{4}, \frac{3}{4}\)
Number of values of \(\mathrm{C}=4\)
\( \therefore \mathrm{f}^{\prime}(x)=2 \pi \cos 2 \pi x \)
\( \text {Now, } \mathrm{f}^{\prime}(\mathrm{C})=0 \)
\( \Rightarrow 2 \pi \cos 2 \pi \mathrm{C}=0\)
\(\Rightarrow \cos 2 \pi \mathrm{C}=0 \)
\(\Rightarrow 2 \pi \mathrm{C}=\frac{-3 \pi}{2}, \frac{-\pi}{2}, \frac{\pi}{2}, \frac{3 \pi}{2} \) \( \ldots[\because x \in(-1,1) \Rightarrow 2 \pi x \in(-2 \pi, 2 \pi)] \)
\(\Rightarrow \mathrm{C}=\frac{-3}{4},-\frac{1}{4}, \frac{1}{4}, \frac{3}{4}\)
Number of values of \(\mathrm{C}=4\)
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