MHT CET · Maths · Application of Derivatives
The minimum value of \(f(x)=a^{2} \cos ^{2} x+b^{2} \sin ^{2} x\) if \(a^{2}>b^{2}\), is
- A \(a^{2}-b^{2}\)
- B \(b^{2}\)
- C \(a^{2}+b^{2}\)
- D \(a^{2}\)
Answer & Solution
Correct Answer
(D) \(a^{2}\)
Step-by-step Solution
Detailed explanation
Given \(f(x)\)
\(=a^{2} \cos ^{2} x+b^{2} \sin ^{2} x\)
\(=a^{2}\left(\frac{1+\cos 2 x}{2}\right)+b^{2}\left(\frac{1-\cos 2 x}{2}\right)\)
\(=\left(\frac{a^{2}+b^{2}}{2}\right)+\left(\frac{a^{2}-b^{2}}{2}\right) \cos 2 x\)
\(f(x)\) will be maximum when \(\cos 2 x=1\)
\(\mathrm{f}(\mathrm{x})\) will be minimum when \(\cos 2 \mathrm{x}=-1\)
Hence minimum value of \(f(x)\) is
\(
\begin{aligned}
f(x) &=\frac{a^{2}+b^{2}}{2}+\left(\frac{a^{2}-b^{2}}{2}\right)(-1) \\
&=\frac{a^{2}+b^{2}}{2}-\frac{a^{2}-b^{2}}{2}=b^{2}
\end{aligned}
\)
\(=a^{2} \cos ^{2} x+b^{2} \sin ^{2} x\)
\(=a^{2}\left(\frac{1+\cos 2 x}{2}\right)+b^{2}\left(\frac{1-\cos 2 x}{2}\right)\)
\(=\left(\frac{a^{2}+b^{2}}{2}\right)+\left(\frac{a^{2}-b^{2}}{2}\right) \cos 2 x\)
\(f(x)\) will be maximum when \(\cos 2 x=1\)
\(\mathrm{f}(\mathrm{x})\) will be minimum when \(\cos 2 \mathrm{x}=-1\)
Hence minimum value of \(f(x)\) is
\(
\begin{aligned}
f(x) &=\frac{a^{2}+b^{2}}{2}+\left(\frac{a^{2}-b^{2}}{2}\right)(-1) \\
&=\frac{a^{2}+b^{2}}{2}-\frac{a^{2}-b^{2}}{2}=b^{2}
\end{aligned}
\)
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