JEE MAIN - Mathematics (2020 - 4th September Morning Slot - No. 6)
If $$\left( {a + \sqrt 2 b\cos x} \right)\left( {a - \sqrt 2 b\cos y} \right) = {a^2} - {b^2}$$
where a > b > 0, then $${{dx} \over {dy}}\,\,at\left( {{\pi \over 4},{\pi \over 4}} \right)$$ is :
where a > b > 0, then $${{dx} \over {dy}}\,\,at\left( {{\pi \over 4},{\pi \over 4}} \right)$$ is :
$${{a - 2b} \over {a + 2b}}$$
$${{a - b} \over {a + b}}$$
$${{a + b} \over {a - b}}$$
$${{2a + b} \over {2a - b}}$$
Explanation
$$(a + \sqrt 2 b\cos x)(a - \sqrt 2 b\cos y) = {a^2} - {b^2}$$
$$ \Rightarrow {a^2} - \sqrt 2 ab\cos y + \sqrt 2 ab\cos x - 2{b^2}\cos x\cos y = {a^2} - {b^2}$$
Differentiating both sides :
$$0 - \sqrt 2 ab\left( { - \sin y{{dy} \over {dx}}} \right) + \sqrt 2 ab( - \sin x)$$
$$ - 2{b^2}\left[ {\cos x\left( { - \sin y{{dy} \over {dx}}} \right) + \cos y( - \sin x)} \right] = 0$$
At $$\left( {{\pi \over 4},{\pi \over 4}} \right)$$ :
$$ab{{dy} \over {dx}} - ab - 2{b^2}\left( { - {1 \over 2}{{dy} \over {dx}} - {1 \over 2}} \right) = 0$$
$$ \Rightarrow {{dx} \over {dy}} = {{ab + {b^2}} \over {ab - {b^2}}} = {{a + b} \over {a - b}}$$; a, b > 0
$$ \Rightarrow {a^2} - \sqrt 2 ab\cos y + \sqrt 2 ab\cos x - 2{b^2}\cos x\cos y = {a^2} - {b^2}$$
Differentiating both sides :
$$0 - \sqrt 2 ab\left( { - \sin y{{dy} \over {dx}}} \right) + \sqrt 2 ab( - \sin x)$$
$$ - 2{b^2}\left[ {\cos x\left( { - \sin y{{dy} \over {dx}}} \right) + \cos y( - \sin x)} \right] = 0$$
At $$\left( {{\pi \over 4},{\pi \over 4}} \right)$$ :
$$ab{{dy} \over {dx}} - ab - 2{b^2}\left( { - {1 \over 2}{{dy} \over {dx}} - {1 \over 2}} \right) = 0$$
$$ \Rightarrow {{dx} \over {dy}} = {{ab + {b^2}} \over {ab - {b^2}}} = {{a + b} \over {a - b}}$$; a, b > 0
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