143

You visited us 143 times! Enjoying our articles? Unlock Full Access!

Product Law

If 2x + y =...

Question

If $2 x + y = 3$ and $x y = 1$ , the value of $(x + y)^{x - y}$ is

100

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

10

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

0

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

1

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

Open in App

Solution

The correct option is C $1$
Given $2 x + y = 3$ . and $x y = 1 \Rightarrow y = 1 / x$

$\Rightarrow 2 x + (1 / x) = 3$

$\Rightarrow 2 x^{2} - 3 x + 1 = 0$

$\Rightarrow 2 x^{2} - 2 x - x + 1 = 0$

$\Rightarrow 2 x (x - 1) - 1 (x - 1) = 0$

$\Rightarrow (2 x - 1) (x - 1) = 0$

$\Rightarrow x = 1 / 2$ or $x = 1$

$\Rightarrow y = 2$ or $y = 1$

If $x = 1 / 2$ and $y = 1$ $\Rightarrow (x + y)^{(x - y)} = (3 / 2)^{- 1 / 2}$

If $x = 1$ and $y = 1$ $\Rightarrow (x + y)^{(x - y)} = 2^{0} = 1$

Suggest Corrections

Similar questions

x y + y x = - 1 (x, y \neq 0)

x^{3} - y^{3}

t a n \frac{1}{2} [s i n^{- 1} \frac{2 x}{1 + x^{2}} + c o s^{- 1} \frac{1 - y^{2}}{1 + y^{2}}]

| x | < 1, y > 0 a n d x y < 1

y = e^{x} + 1

y^{''} - y = 0

y = x^{2} + 2 x + C

y^{'} - 2 x - 2 = 0

y = cos x + C

y^{'} + sin x = 0

y = \sqrt{1 + x^{2}}

y^{'} = \frac{x y}{1 + x^{2}}

y = A x

x y = y (x \neq 0)

y = x sin x

x y = y + x \sqrt{x^{2} y^{2}} (x \neq 0 a n d x > y o r x < y)

x y = log y + C

y^{'} = \frac{y^{2}}{1 - x y} (x y \neq 1)

y - cos y = x

(y sin y + cos y + x) y^{'} = y

x + y = {tan}^{- 1} y

y^{2} y^{'} + y^{2} + 1 = 0

y = \sqrt{a^{2} - x^{2}} x ϵ (- a, a)

x + y \frac{d y}{d x} = 0 (y \neq 0)

x, y \in N,

3^{2 x - y + 1} = 3^{y - 2 x + 1} - 8

{log}_{6} ∣ ∣ 2 x^{2} y - x y^{2} ∣ ∣ = 1 + {log}_{36} (x y),

(x - y)

\frac{x}{y} + \frac{y}{x} = - 1

\frac{1}{2}

Join BYJU'S Learning Program