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Ncert Solutions Maths class 11th Maths Class 11th Inverse Trigonometric Functions

If $\frac{d y}{d x} = \frac{2^{x} y + 2^{y} . 2^{x}}{2^{x} + 2^{x + y} l o g_{e} 2}, y (0) = 0,$ then for y = 1, the value of x lies in the interval:

Option 1 -

(1, 2)

Option 2 -

$(\frac{1}{2}, 1]$

Option 3 -

$(0, \frac{1}{2}]$

Option 4 -

(2, 3)

2 Views | Posted a month ago

Asked by Shiksha User

1 Answer

Answered by

Vishal Baghel | Contributor-Level 10

a month ago

Correct Option - 1

Detailed Solution:

$\frac{d y}{d x} = \frac{2^{x} y + 2^{y} . 2^{x}}{2^{x} + 2^{x + y} l o g_{e} 2}$

$\int \frac{1 + 2^{y} l o g_{e} 2}{y + 2^{y}} d y = \int d x$

=> ln|y + 2^y| = x + c

y (0) = 0

ln |y + 2^y| = x

y = 1

x = ln 3

$x \in (1, 2)$

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If $\frac{d y}{d x} = \frac{2^{x} y + 2^{y} . 2^{x}}{2^{x} + 2^{x + y} l o g_{e} 2}, y (0) = 0,$ then for y = 1, the value of x lies in the interval:

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If d y d x = 2 x y + 2 y . 2 x 2 x + 2 x + y l o g e 2 , y ( 0 ) = 0 , then for y = 1, the value of x lies in the interval:

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If $\frac{d y}{d x} = \frac{2^{x} y + 2^{y} . 2^{x}}{2^{x} + 2^{x + y} l o g_{e} 2}, y (0) = 0,$ then for y = 1, the value of x lies in the interval: