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Standard XII

Mathematics

Stationary Point

fx=sin-1x2+co...

Question

$f (x) = ({sin}^{- 1} x)^{2} + ({cos}^{- 1} x)^{2}$ is stationary at

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

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x = \frac{π}{4}

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x = 1

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x = 0

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Solution

The correct option is A $x = \frac{1}{\sqrt{2}}$
$f^{'} (x) = 2 s i n^{- 1} (x) \frac{1}{\sqrt{1 - x^{2}}} + 2 c o s^{- 1} (x) (\frac{- 1}{\sqrt{1 - x^{2}}})$
$= 2 \frac{1}{\sqrt{1 - x^{2}}} (s i n^{- 1} (x) - c o s^{- 1} (x))$
$f^{'} (x) = 0$ when $x = \frac{1}{\sqrt{2}}$

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f(x)=(sin−1x)2+(cos−1x)2 is stationary at

The correct option is A x=1√2f′(x)=2sin−1(x)1√1−x2+2cos−1(x)(−1√1−x2)=21√1−x2(sin−1(x)−cos−1(x))f′(x)=0 when x=1√2

$f (x) = ({sin}^{- 1} x)^{2} + ({cos}^{- 1} x)^{2}$ is stationary at

The correct option is A $x = \frac{1}{\sqrt{2}}$
$f^{'} (x) = 2 s i n^{- 1} (x) \frac{1}{\sqrt{1 - x^{2}}} + 2 c o s^{- 1} (x) (\frac{- 1}{\sqrt{1 - x^{2}}})$
$= 2 \frac{1}{\sqrt{1 - x^{2}}} (s i n^{- 1} (x) - c o s^{- 1} (x))$
$f^{'} (x) = 0$ when $x = \frac{1}{\sqrt{2}}$