If $$\frac{dy}{dx} = \frac{2^x y + 2^y \cdot 2^x}{2^x + 2^x + y\log_e 2}$$, $$y(0) = 0$$, then for $$y = 1$$, the value of $$x$$ lies in the interval:

We are given the differential equation

$$\frac{dy}{dx}= \frac{2^{x}y+2^{y}\,2^{x}}{2^{x}+2^{x}+y\log_e2}$$

and the initial condition $$y(0)=0.$$  Our task is to find the value of $$x$$ for which $$y=1$$ and then decide in which of the four stated intervals that value lies.

First note that every quantity occurring on the right-hand side is positive whenever $$x\ge 0$$ and $$y\ge 0$$. Hence for all points that the solution curve can reach starting from $$(0,0)$$ the derivative $$\dfrac{dy}{dx}$$ is positive, so $$y$$ increases strictly with $$x$$. We may therefore look for the desired $$x$$ by successively estimating the rise of $$y$$ as $$x$$ moves away from the origin.

Write the right-hand side in a slightly shorter form:

$$\frac{dy}{dx} =\frac{2^{x}(y+2^{y})}{2\!\cdot\!2^{x}+y\log 2}. $$

At the point $$(x,y)=(0,0)$$ we have

$$2^{x}=2^{0}=1,\qquad 2^{y}=2^{0}=1.$$

Therefore

$$\left.\frac{dy}{dx}\right|_{(0,0)} =\frac{\,1(0+1)}{2\!\cdot\!1+0\cdot\log2} =\frac12.$$

So at $$x=0$$ the curve starts with slope $$\dfrac12$$. If that slope were to remain constant, an increase of $$y$$ from $$0$$ to $$1$$ would require an increase of $$x$$ by exactly $$2$$. In reality the slope will change, and we must see whether the actual $$x$$ lies below or above $$2$$.

Because an analytic closed‐form integration looks cumbersome, we proceed with a careful numerical bracketing that shows every step of algebra and arithmetic.

Step 1 : from $$x=0$$ to $$x=0.5$$

With slope $$\dfrac12$$ over a small interval $$\Delta x=0.5$$, an Euler estimate gives

$$y(0.5)\approx 0+ \frac12\cdot0.5=0.25.$$

Now evaluate the derivative at the new point $$(x,y)=(0.5,0.25).$$ We need every elementary number:

$$2^{x}=2^{0.5}=\sqrt2\approx1.4142, \qquad 2^{y}=2^{0.25}=2^{\tfrac14}\approx1.1892.$$

The numerator of the derivative is

$$2^{x}y+2^{y}\,2^{x} =1.4142\cdot0.25+1.1892\cdot1.4142 =0.3535+1.6831 \approx2.0366.$$

The denominator is

$$2\!\cdot\!2^{x}+y\log2 =2\!\cdot\!1.4142+0.25\cdot0.6931 =2.8284+0.1733 \approx3.0017.$$

Hence

$$\left.\frac{dy}{dx}\right|_{(0.5,0.25)} =\frac{2.0366}{3.0017}\approx0.6787.$$

Step 2 : from $$x=0.5$$ to $$x=1$$

Using $$\Delta x=0.5$$ again,

$$y(1)\approx0.25+0.6787\cdot0.5 =0.25+0.3394 \approx0.5894.$$

Step 3 : exact value of the slope at $$(1,0.5894)$$

Compute the powers:

$$2^{x}=2^{1}=2,\qquad 2^{y}=2^{0.5894}=e^{0.5894\log2}\;(\log2\approx0.6931) =e^{0.5894\cdot0.6931}\approx e^{0.409} \approx1.505.$$

The numerator is

$$2^{x}y+2^{y}\,2^{x}=2\cdot0.5894+1.505\cdot2 =1.1788+3.0096 \approx4.1884.$$

The denominator is

$$2\!\cdot\!2^{x}+y\log2 =2\!\cdot\!2+0.5894\cdot0.6931 =4+0.4085 \approx4.4085.$$

Thus

$$\left.\frac{dy}{dx}\right|_{(1,0.5894)} =\frac{4.1884}{4.4085}\approx0.9499\ (\text{about }0.95).$$

Step 4 : from $$x=1$$ to $$x=1.2$$

With $$\Delta x=0.2$$,

$$y(1.2)\approx0.5894+0.9499\cdot0.2 =0.5894+0.18998 \approx0.7794.$$

Step 5 : exact slope at $$(1.2,0.7794)$$

Powers of two:

$$2^{1.2}=2^{1}2^{0.2}=2\cdot2^{0.2} =2\cdot e^{0.2\log2} =2\cdot e^{0.1386} \approx2\cdot1.1487 \approx2.2974,$$

$$2^{0.7794}=e^{0.7794\log2}=e^{0.7794\cdot0.6931} =e^{0.540}\approx1.716.$$

Numerator:

$$2^{x}y+2^{y}\,2^{x} =2.2974\cdot0.7794+1.716\cdot2.2974 =1.7917+3.9456 \approx5.7373.$$

Denominator:

$$2\!\cdot\!2^{x}+y\log2 =2\!\cdot\!2.2974+0.7794\cdot0.6931 =4.5948+0.5403 \approx5.1351.$$

Hence

$$\left.\frac{dy}{dx}\right|_{(1.2,0.7794)} =\frac{5.7373}{5.1351}\approx1.1173.$$

Step 6 : from $$x=1.2$$ to $$x=1.4$$

Again with $$\Delta x=0.2$$,

$$y(1.4)\approx0.7794+1.1173\cdot0.2 =0.7794+0.2235 \approx1.0029.$$

Thus at $$x=1.4$$ the computed value of $$y$$ already exceeds $$1$$. Because the function $$y(x)$$ was strictly increasing all along, $$y$$ must cross the level $$y=1$$ at some $$x$$ strictly between $$1.2$$ and $$1.4$$.

Therefore

$$1<x<2.$$

The only interval among the options that contains every number between $$1.2$$ and $$1.4$$ is the interval $$(1,2).$$

Hence, the correct answer is Option A.