Find the degree of homogeneity of function $f (x, y) = a x^{2 / 3} + h x^{1 / 3} y^{1 / 3} + b y^{2 / 3}$ .

2/3

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

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

3/2

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

Open in App

Solution

The correct option is C 2/3
$f (x, y) = a x^{2 / 3} + h x^{1 / 3} y^{1 / 3} + b y^{2 / 3} \Rightarrow f (t x, t y) = a {(t x)}^{2 / 3} + h {(t x)}^{1 / 3} {(t y)}^{1 / 3} + b {(t y)}^{2 / 3} = a t^{2 / 3} x^{2 / 3} + h t^{2 / 3} x^{1 / 3} y^{1 / 3} + b t^{2 / 3} y^{2 / 3} = t^{2 / 3} (a x^{2 / 3} + h x^{1 / 3} y^{1 / 3} + b y^{2 / 3}) = t^{2 / 3} f (x, y)$
Hence degree is $2 / 3$

Suggest Corrections

Similar questions

f (x, y) = x^{3} l n [\sqrt{x + y} / \sqrt{x - y}]

f (x, y) = \frac{1}{x + y}

x = 2^{3} \times 3 \times 5^{2}, y = 2^{2} \times 3^{3},

H C F (x, y)

x = 2^{3} \times 3 \times 5^{2}

y = 2^{2} \times 3^{3}

L C M (x, y)

f (x, y) = \frac{1}{x + y}

Join BYJU'S Learning Program