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प्रश्न
The maximum value of sin x . cos x is ______.
विकल्प
`1/4`
`1/2`
`sqrt(2)`
`2sqrt(2)`
उत्तर
The maximum value of sin x . cos x is `1/2`.
Explanation:
We have f(x) = sin x cos x
⇒ f(x) = `1/2 * 2 sin x cos x`
= `1/2 sin 2x`
f'(x) = `1/2 * 2 cos 2x`
⇒ f'(x) = cos 2x
Now for local maxima and local minima f'(x) = 0
∴ cos 2x = 0
2x = `("n" + 1) pi/2`, n ∈ I
⇒ x = `(2"n" + 1) pi/4`
∴ x = `pi/4, (3pi)/4` .....
f"(x) = – 2 sin 2x
`"f''"(x)_(x = pi/4)` = `-2 sin 2 * pi/4`
= `- 2 sin pi/2`
= – 2 < 0 maxima
`"f''"(x)_(x = (3pi)/4) = - 2 sin 2 * (3pi)/4`
= `-2 sin (3pi)/4`
= 2 > 0 minima
So f(x) is maximum at x = `pi/4`
∴ Maximum value of f(x) = `sin pi/4 * cos pi/4`
= `1/sqrt(2) * 1/sqrt(2)`
= `1/sqrt(2)`.
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