JEE MAIN - Mathematics (2007 - No. 10)
The solution for $$x$$ of the equation $$\int\limits_{\sqrt 2 }^x {{{dt} \over {t\sqrt {{t^2} - 1} }} = {\pi \over 2}} $$ is
$${{\sqrt 3 } \over 2}$$
$$2\sqrt 2 $$
$$2$$
None
Explanation
$$\int_{\sqrt 2 }^x {{{dt} \over {t\sqrt {{t^2} - 1} }}} = {\pi \over 2}$$
$$\therefore$$ $$\left[ {{{\sec }^{ - 1}}t} \right]_{\sqrt 2 }^x = {\pi \over 2}$$
$$\left[ {} \right.$$ As $$\int {{{dx} \over {x\sqrt {{x^2} - 1} }}} = {\sec ^{ - 1}}x$$ $$\left. {} \right]$$
$$ \Rightarrow {\sec ^{ - 1}}x - {\sec ^{ - 1}}\sqrt 2 = {\pi \over 2}$$
$$ \Rightarrow {\sec ^{ - 1}}x - {\pi \over 4} = {\pi \over 2}$$
$$ \Rightarrow {\sec ^{ - 1}}x = {\pi \over 2} + {\pi \over 4}$$
$$ \Rightarrow {\sec ^{ - 1}}x = {{3\pi } \over 4}$$
$$ \Rightarrow x = \sec {{3\pi } \over 4}$$
$$ \Rightarrow x = - \sqrt 2 $$
$$\therefore$$ $$\left[ {{{\sec }^{ - 1}}t} \right]_{\sqrt 2 }^x = {\pi \over 2}$$
$$\left[ {} \right.$$ As $$\int {{{dx} \over {x\sqrt {{x^2} - 1} }}} = {\sec ^{ - 1}}x$$ $$\left. {} \right]$$
$$ \Rightarrow {\sec ^{ - 1}}x - {\sec ^{ - 1}}\sqrt 2 = {\pi \over 2}$$
$$ \Rightarrow {\sec ^{ - 1}}x - {\pi \over 4} = {\pi \over 2}$$
$$ \Rightarrow {\sec ^{ - 1}}x = {\pi \over 2} + {\pi \over 4}$$
$$ \Rightarrow {\sec ^{ - 1}}x = {{3\pi } \over 4}$$
$$ \Rightarrow x = \sec {{3\pi } \over 4}$$
$$ \Rightarrow x = - \sqrt 2 $$
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