KCET · Maths · Differentiation
If \(f(x)=\cos ^{-1}\left[\frac{1}{\sqrt{13}}(2 \cos x-3 \sin x)\right]\). Then, \(f^{\prime}(0.5)\) is equal to
- A \(0.5\)
- B 1
- C 0
- D \(-1\)
Answer & Solution
Correct Answer
(B) 1
Step-by-step Solution
Detailed explanation
Given, \(f(x)=\cos ^{-1}\left\{\frac{1}{\sqrt{13}}(2 \cos x-3 \sin x)\right\}\)
\(\Rightarrow f(x)=\cos ^{-1}\left\{\frac{2}{\sqrt{13}} \cos x-\frac{3}{\sqrt{13}} \sin x\right\}\)
\(\Rightarrow \quad f(x)=\cos ^{-1}\{\cos \alpha \cdot \cos x-\sin \alpha \cdot \sin x\}\)
\(\begin{aligned}
&=\cos ^{-1}\{\cos (x+\alpha)\} \\
&=x+\alpha
\end{aligned}\)
On differentiating w.r.t. \(x\), we get
\(\begin{aligned}
f^{\prime}(x) &=1=\text { constant value } \\
\therefore \quad f^{\prime}(0.5) &=1
\end{aligned}\)
\(\Rightarrow f(x)=\cos ^{-1}\left\{\frac{2}{\sqrt{13}} \cos x-\frac{3}{\sqrt{13}} \sin x\right\}\)
\(\Rightarrow \quad f(x)=\cos ^{-1}\{\cos \alpha \cdot \cos x-\sin \alpha \cdot \sin x\}\)
\(\begin{aligned}
&=\cos ^{-1}\{\cos (x+\alpha)\} \\
&=x+\alpha
\end{aligned}\)
On differentiating w.r.t. \(x\), we get
\(\begin{aligned}
f^{\prime}(x) &=1=\text { constant value } \\
\therefore \quad f^{\prime}(0.5) &=1
\end{aligned}\)
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