MHT CET · Maths · Trigonometric Ratios & Identities
The number of solutions, of \(2^{1+|\cos x|+|\cos x|^2+}\) __________ \(=4\) in \((-\pi, \pi)\), is
- A 2
- B 3
- C 4
- D 6
Answer & Solution
Correct Answer
(C) 4
Step-by-step Solution
Detailed explanation
\(\begin{aligned} & 2^{1+|\cos x|+|\cos x|^2+\ldots}=4 \\ & \Rightarrow 2^{\frac{1}{1-|\cos x|}}=2^2\end{aligned}\)
\(\begin{aligned}
& \Rightarrow \frac{1}{1-|\cos x|}=2 \\
& \Rightarrow 1-|\cos x|=\frac{1}{2} \\
& \Rightarrow|\cos x|=\frac{1}{2} \\
& \Rightarrow \cos x= \pm \frac{1}{2} \\
& \Rightarrow x=\frac{-2 \pi}{3}, \frac{-\pi}{3}, \frac{\pi}{3}, \frac{2 \pi}{3} \quad \ldots[\because x \in(-\pi, \pi)]
\end{aligned}\)
\(\therefore \quad\) Number of solutions \(=4\)
\(\begin{aligned}
& \Rightarrow \frac{1}{1-|\cos x|}=2 \\
& \Rightarrow 1-|\cos x|=\frac{1}{2} \\
& \Rightarrow|\cos x|=\frac{1}{2} \\
& \Rightarrow \cos x= \pm \frac{1}{2} \\
& \Rightarrow x=\frac{-2 \pi}{3}, \frac{-\pi}{3}, \frac{\pi}{3}, \frac{2 \pi}{3} \quad \ldots[\because x \in(-\pi, \pi)]
\end{aligned}\)
\(\therefore \quad\) Number of solutions \(=4\)
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