What positive value(s) of x, less than $360^{o}$ , will give a minimum value for 4 - 2 sin x cos x?

\frac{π}{4}

only

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\frac{5 π}{4}

only

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\frac{π}{2}

and

\frac{5 π}{2}

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\frac{3 π}{2}

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\frac{π}{4}

and

\frac{5 π}{4}

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Solution

The correct option is E $\frac{π}{4}$ and $\frac{5 π}{4}$
$4 - 2 s i n x c o s x = 4 - s i n 2 x$ , It will be minimum if $s i n 2 x$ is maximum
The maximum value of $s i n 2 x$ is $1$ and occurs at $2 x = 90, 450$
Which implies $x = 45, 225$
Therefore the correct option is $E$

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Similar questions

If y = - sin2x. Find the values of x at which the tangents drawn to the graph of this function is parallel to the x- axis.

{sin}^{- 1} \frac{1}{2} + {cos}^{- 1} \frac{1}{2} + {tan}^{- 1} \frac{1}{\sqrt{3}}

(a) \frac{2 π}{3} (b) \frac{π}{2}

(c) \frac{3 π}{4} (d) \frac{5 π}{6}

{x \in R : | cos x | \geq sin x} \cap [0, \frac{3 π}{2}] =

x

cos x = \frac{- 1}{\sqrt{2}}

tan x = 1

y = \sqrt{5 - x^{2}}

y = | x - 1 |

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