$f = a_{0} x^{n} + a_{1} x^{n - 1} y + . . . + a_{n - 1} x y^{n - 1} + a_{n} y^{n}$ where $a_{i} (i = 0 t o n) a r e c o n s t a n t t h e n x \frac{\partial f}{\partial x} + y \frac{\partial f}{\partial y} =$

\frac{f}{n}

No worries! We‘ve got your back. Try BYJU‘S free classes today!

\frac{n}{f}

No worries! We‘ve got your back. Try BYJU‘S free classes today!

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

n \sqrt{f}

No worries! We‘ve got your back. Try BYJU‘S free classes today!

Open in App

Solution

The correct option is C nf
Given : $f = a_{0} x^{n} + a_{1} x^{n - 1} y + . . . + a_{n - 1} x y^{n - 1} + a_{n} y^{n}$

$∵ f (\partial x, \partial y) = \partial^{n} f (x, y)$

$\Rightarrow$ f is a homogeneous function in x and y of degree 'n'

$∴$ By Euler's theorem for homogeneous function, we have

$x \frac{\partial f}{\partial x} + y \frac{\partial f}{\partial y} = n f$

Suggest Corrections

Similar questions

f (x) = a_{0} x^{n} + a_{1} x^{n - 1} + . . . + a_{n - 1} x + a_{n}, where a_{0}, a_{1}, a_{2}, . . ., a_{n}

a_{0} = a_{1} = a_{2} = . . . = a_{n - 1} = 0 and a_{n} \neq 0,

f : A \to B

f (x) = \frac{x - 1}{x - 2}

A = R - {2}

B = R - {1}

f

f (x) = a^{x} (a > 0)

f (x) = f_{1} (x) + f_{2} (x)

f_{1} (x)

f_{2} (x)

f_{1} (x + y) + f_{1} (x - y)

| x | + a_{0} x^{n} + a_{1} x^{n - 1} + a_{2} x^{n - 2} + . . . . + a_{n - 1} x + a_{n}

Join BYJU'S Learning Program