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UPSEE 2018 Solved Paper

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Question : 124 of 150

Marks: +1, -0

The slope of the tangent at

\left(\frac{\pi}{4}, 0\right)

1+16 x^{2} y=\tan (x-2 y)

is

$\frac{2}{\pi+2}$
$\frac{1}{\pi^{2}+4}$
$\frac{1}{\pi+4}$
$\frac{2}{\pi^{2}+4}$

Solution:

We have,

1+16 x^{2} y=\tan (x-2 y)

On differentiating w.r.t.

x

, we get

32 x y+16 x^{2} \frac{dy}{dx}=\sec ^{2}(x-2 y)

\left(1-2 \frac{dy}{dx}\right)

\Rightarrow \frac{dy}{dx} (16 x^{2}+2 \sec ^{2}(x-2 y)

=\sec ^{2}(x-2 y)-32 x y

\Rightarrow \frac{dy}{dx} = \frac{\sec ^{2}(x-2y)-32xy}{16 x^{2}+2 \sec ^{2}(x-2y)}

Slope of tangent at

\left(\frac{\pi}{4}, 0\right)

i.e.

\left(\frac{dy}{dx}\right)_{\left(\frac{\pi}{4},0\right)}

= \frac{\sec ^{2}\left(\frac{\pi}{4}-0\right)-32\left(\frac{\pi}{4}\right)(0)}{16 \times \left(\frac{\pi}{4}\right)^{2}+2\sec ^{2}\left(\frac{\pi}{4}-0\right)}

\therefore

Slope of tangent

= \frac{2}{\pi^{2}+4}

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