f(x, ;y) = 1 x + y is a homogeneous function of degree - Vidyadip

MCQ

$f(x,\;y) = \frac{1}{{x + y}}$ is a homogeneous function of degree

A
$1$
✓
$-1$
C
$2$
D
$-2$

✓

Answer

Correct option: B.

$-1$

b
(b) It is a fundamental concept.

Need a full question paper?

Generate a complete, print-ready paper with questions like this in minutes — across 16+ boards, with answer keys.

Start Generating Free

Explore more

More "M.C.Q (1 Marks)" from 1. relation and function→1. relation and function — all question types→Maths STD 12 Science question papers→Gujarat Board STD 12 Science question papers→Gujarat Board question paper generator→

Similar questions

The height of a right circular cone of maximum volume inscribed in a sphere of diameter $a$ is-

Area of the region bounded by the curve $y=e^x$ and lines $x=0$ and $y=e$ is

$(A)$ $e-1$ $(B)$ $\int_1^e \ln (e+1-y) d y$ $(C)$ $e-\int_0^1 e^x d x$ $(D)$ $\int_1^r \ln y d y$

If $A=\left[\begin{array}{cc}3 & 1 \\ -1 & 2\end{array}\right]$, then find $A^{-1}$.

Let $\vec a = 2\hat i + \hat j - 2\hat k,\vec b = \hat i + \hat j$. If $\vec c$ is a vector such that $\vec a.\vec c = \left| {\vec c} \right|,\left| {\vec c - \vec a} \right| = 2\sqrt 2 $ and the angle between $\vec a \times \vec b$ and $\vec c$ is $30^o$, then $\left| {\left( {\vec a \times \vec b} \right) \times \vec c} \right|$ equals

$p = 2a - 3b,\,\,\,q = a - 2b + c,\,\,r = - 3a + b + 2c;$ where $a, b $ and $c $ being non-zero, non-coplanar vectors, then the vector $ - 2a + 3b - c$ is equal to

The area of the parallelogram whose adjacent sides are $i - k$ and $2j + 3k$ is

Z = 6x + 21y, subject to $ \text{x}+2\text{y}\geq3,\text{x}+4\text{y}\geq4,3\text{x}+\text{y}\geq3,\text{x}\geq0,\text{y}\geq0.$ The minimum value of Z occurs at.

$(4,0)$
$(28,8)$
$\Big(2,\frac{7}{2}\Big)$
$(0,3)$

If $x = a(\cos \theta + \theta \sin \theta )$, $y = a(\sin \theta - \theta \cos \theta ),{\rm{ }}$ then ${{dy} \over {dx}} = $

If $x = a\left( {t - {1 \over t}} \right)\,,y = a$ $\left( {t + {1 \over t}} \right)$ then ${{dy} \over {dx}} = $

The area of the region (in the square unit) bounded by the curve ${x^2} = 4y,$ line $x = 2$ and $x-$ axis is