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Application of Derivatives

Multiple Choice Questions

781.

If the tangent to the curve y = x + siny at a point (a, b) is parallel to the line joining $(0, \frac{3}{2}) and (\frac{1}{2}, 2)$ then

$|b - a| = 1$
$b = \frac{π}{2} + a$
$|a + b| = 1$
b = a

782.
$The derivative of \tan^{- 1} (\frac{\sqrt{1 + x^{2}} - 1}{x}) with respect to \tan^{- 1} (\frac{2 x \sqrt{1 - x^{2}}}{1 - 2 x^{2}}) at x = \frac{1}{2} is :$

$\frac{\sqrt{3}}{10}$

$\frac{\sqrt{3}}{12}$

$\frac{2 \sqrt{3}}{5}$

$\frac{2 \sqrt{3}}{3}$

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$\frac{\sqrt{3}}{10}$

$Let x = \tan (θ) y_{1} = \tan^{- 1} (\frac{\sec (θ) - 1}{\tan (θ)}) = \tan^{- 1} (\tan (\frac{θ}{2})) = \frac{θ}{2} = \frac{1}{2} \tan^{- 1} (x) x = \sin (φ), y_{2} = \tan^{- 1} (\frac{2 \sin (ϕ) \cos (ϕ)}{\cos (2 ϕ)}) = \tan^{- 1} (\tan (2 ϕ)) = 2 ϕ = 2 \sin^{- 1} (x) \frac{{dy}_{1}}{{dy}_{2}} = \frac{\frac{{dy}_{1}}{dx}}{\frac{{dy}_{2}}{dx}} = \frac{(\frac{1}{1 + x^{2}}) . \frac{1}{2}}{2 . \frac{1}{\sqrt{1 - x^{2}}}} = \frac{\sqrt{1 - x^{2}}}{4 (1 + x^{2})} = \frac{\sqrt{1 - \frac{1}{4}}}{4 . (1 + \frac{1}{4})} = \frac{\sqrt{3}}{10}$

Previous Year Papers

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Test Yourself

Multiple Choice Questions

782. The derivative of tan-11 + x2 - 1x with respect to tan-12x1 - x21 - 2x2 at x = 12 is : 310 312 235 233

782.
$The derivative of \tan^{- 1} (\frac{\sqrt{1 + x^{2}} - 1}{x}) with respect to \tan^{- 1} (\frac{2 x \sqrt{1 - x^{2}}}{1 - 2 x^{2}}) at x = \frac{1}{2} is :$

$\frac{\sqrt{3}}{10}$

$\frac{\sqrt{3}}{12}$

$\frac{2 \sqrt{3}}{5}$

$\frac{2 \sqrt{3}}{3}$