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  • The maximum value of sin x cos x is A.1/4

The maximum value of sin x cos x is
A. \frac{1}{4}
B. \frac{1}{2}
C. \sqrt 2
D. 2\sqrt 2

Answers (1)

Let f(x)= sin x cos x

sin2x=2sin x cos x

\Rightarrow f(x)=\frac{1}{2} \sin 2 x
Apply first derivative and get
\\ \mathrm{f}^{\prime}(\mathrm{x})=\frac{\mathrm{d}\left(\frac{1}{2} \sin 2 \mathrm{x}\right)}{\mathrm{dx}}$\\ $\Rightarrow f^{\prime}(x)=\frac{1}{2} \frac{d(\sin 2 x)}{d x}
Apply derivative,
\Rightarrow f^{\prime}(x)=\frac{1}{2} \cdot \cos 2 x \cdot \frac{d(2 x)}{d x}

\\ \Rightarrow \mathrm{f}^{\prime}(\mathrm{x})=\frac{1}{2} \cdot \cos 2 \mathrm{x} \cdot 2$ $\\\Rightarrow f(x)=\cos 2 x \ldots \ldots(i)
Put f^{\prime}(x)=0 and get
\\\cos 2 x=0 \\ \Rightarrow \cos 2 x=\cos \frac{\pi}{2}
Equate the angles and get
\\\Rightarrow 2 x=\frac{\pi}{2}$ \\$\Rightarrow x=\frac{\pi}{4}
Now we find second derivative by deriving equation (i) and get
\mathrm{f}^{\prime \prime}(\mathrm{x})=\frac{\mathrm{d}(\cos 2 \mathrm{x})}{\mathrm{dx}}

Apply derivative,
\\\Rightarrow \mathrm{f}^{\prime \prime}(\mathrm{x})=-\sin 2 \mathrm{x} \cdot \frac{\mathrm{d}(2 \mathrm{x})}{\mathrm{dx}}$ \\$\Rightarrow f^{\prime}(x)=-\sin 2 x .2$ \\$\Rightarrow f^{\prime}(x)=-2 \sin 2 x
Now we find the value of f^{\prime \prime}(x)$ at $x=\frac{\pi}{4}, we get
\\\Rightarrow\left(f^{\prime}(x)\right)_{x=\frac{\pi}{4}}=-2 \sin 2 x$ \\$\Rightarrow\left(f^{\prime}(x)\right)_{x=\frac{\pi}{4}}=-2 \sin 2\left(\frac{\pi}{4}\right)$ \\$\Rightarrow\left(f^{\prime}(x)\right)_{x=\frac{\pi}{4}}=-2 \sin \left(\frac{\pi}{2}\right)
But \sin \left(\frac{\pi}{2}\right)=1 so above equation becomes
\Rightarrow\left(f^{\prime}(x)\right)_{x=\frac{\pi}{4}}=-2 \times 1=-2<0
Hencee at\mathrm{x}=\frac{\pi}{4}, \mathrm{f}(\mathrm{x}) is maximum and \frac{\pi}{4} is the point of maxima.

Now we will find the maximum value of \sin \mathrm{x} \cos \mathrm{x} by substituting \mathrm{x}=\frac{\pi}{4}, in \mathrm{f}(\mathrm{x}), we get
\\ f(x)=\sin x \cos x$ \\$f\left(\frac{\pi}{4}\right)=\sin \left(\frac{\pi}{4}\right) \cos \left(\frac{\pi}{4}\right)$ \\$\Rightarrow \mathrm{f}\left(\frac{\pi}{4}\right)=\frac{1}{\sqrt{2}} \cdot \frac{1}{\sqrt{2}}$ \\$\Rightarrow \mathrm{f}\left(\frac{\pi}{4}\right)=\frac{1}{2}
So, maximum value of  \sin \mathrm{x} \cos \mathrm{x} is \frac{1}{2}
So, the correct answer is option B

Posted by

infoexpert22

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