For the curve $2 y=x^y+y^x$ The value of $\frac{\frac{d^2 y}{d x^2}}{\frac{d^2 x}{d y^2}}$ at $\mathrm{x}=1$ is $\alpha$ then $|\alpha|=$

JEE MATHEMATICS MCQ: DIFFERENTIATION

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Analyzing +4 | -1

For the curve

2 y=x^y+y^x

The value of

\frac{\frac{d^2 y}{d x^2}}{\frac{d^2 x}{d y^2}}

\mathrm{x}=1

\alpha

then

|\alpha|=

View answer and explanation

Correct option: A

-1

Option C (12) is correct because the least value of the sum of squares is found using Cauchy-Schwarz inequality.

Practice All Chapter MCQs →

f(x)=|x|^{|\tan x|}

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x

y

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[x]

x

x=k, k

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y = {\sec ^{ - 1}}\left[ {\mathrm{cosec}x} \right] + \mathrm{cose}{\mathrm{c}^{ - 1}}\left[ {\sec x} \right] + {\sin ^{ - 1}}[\cos x] + {\cos ^{ - 1}}[\sin x],

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\frac{d}{d x} \sin ^{-1}\left(3 x-4 x^{3}\right)

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v = \left( {\frac{x}{y}} \right),

v\frac{{\partial z}}{{\partial x}} + \frac{{\partial z}}{{\partial y}}