If y - 2- then find the value ofxy2 - 2x-x-

Given, the expression is xy^2+2x/x. Taking x common from the numerator, we get x(y^2+2)/x=(y^2+2). Now substituting y=2 in the above equation, we get, (y^ ...

You visited us 1 times! Enjoying our articles? Unlock Full Access!

Byju's Answer

Standard XI

Business Studies

Outsourcing and Its Types

If y = 2, the...

Question

$If y = 2, then find the value of \frac{(x y^{2} + 2 x)}{x} .$

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!

Open in App

Solution

The correct option is A 6
$Given, the expression is \frac{x y^{2} + 2 x}{x} .$

$Taking x common from the numerator, we get \frac{x (y^{2} + 2)}{x} = (y^{2} + 2) .$

$Now substituting y = 2 in the above equation, we get, (y^{2} + 2) = 2^{2} + 2 = 4 + 2 = 6$

Suggest Corrections

Similar questions

Q. Find

X

and

Y

, if
(i)

X + Y = [\begin{matrix} 7 & 0 2 & 5 \end{matrix}]

and

X - Y = [\begin{matrix} 3 & 0 0 & 3 \end{matrix}]

(ii)

2 X + 3 Y = [\begin{matrix} 2 & 3 4 & 0 \end{matrix}]

and

3 X + 2 Y = [\begin{matrix} 2 & - 2 - 1 & 5 \end{matrix}]

Find X and Y, if

(i) and

(ii) and

Q. Find x, y, a and b if
(i)

[\begin{matrix} 2 x - 3 y & a - b & 3 \\ 1 & x + 4 y & 3 a + 4 b \end{matrix}] = [\begin{matrix} 1 & - 2 & 3 \\ 1 & 6 & 29 \end{matrix}]

Q. Differentiate

\sin^{- 1} (2 x \sqrt{1 - x^{2}})

with respect to

\sec^{- 1} (\frac{1}{\sqrt{1 - x^{2}}})

, if
(i)

x \in (0, \frac{1}{\sqrt{2}})

(ii)

x \in (\frac{1}{\sqrt{2}}, 1)

f(x) ${\begin{matrix} \frac{k c o s x}{π - 2 x}, i f x \neq \frac{π}{2} 3, i f x = \frac{π}{2} . \end{matrix}$

Join BYJU'S Learning Program

If y=2,then find the value of(xy2+2x)x.

The correct option is A 6Given, the expression is xy2+2xx. Taking x common from the numerator, we get x(y2+2)x=(y2+2). Now substituting y=2 in the aboveequation, we get, (y2+2)=22+2=4+2=6

$If y = 2, then find the value of \frac{(x y^{2} + 2 x)}{x} .$

The correct option is A 6
$Given, the expression is \frac{x y^{2} + 2 x}{x} .$

$Taking x common from the numerator, we get \frac{x (y^{2} + 2)}{x} = (y^{2} + 2) .$

$Now substituting y = 2 in the above equation, we get, (y^{2} + 2) = 2^{2} + 2 = 4 + 2 = 6$