JEE Main 2021 (31 August Shift 2)

LEVELJEE Main

Animated Solution for Mathematics - Differential Equations: If $\frac{d y}{d x} = \frac{2 ^{x} y + 2 ^{y} \cdot 2 ^{x}}{2 ^{x} + 2 ^{x + y} l o g _{e} 2}, y (0) = 0$ , then for $y = 1$ , the value of $x$ lies in the interval:

Q: Why can we just cancel $2^x$ from the numerator and denominator? Is it possible for $2^x$ to be zero?

That is a fantastic question! In the realm of real numbers, the exponential function $2^x$ is strictly positive for all $x$. It never touches the x-axis, meaning it is never zero. Because it is never zero, we are mathematically safe to divide both sides by it without losing any solutions. Always check for domain restrictions, but here, we are safe!

Q: I got stuck at the integration step. How do I know when to use substitution?

Pattern recognition is the heart of calculus! Whenever you see a fraction where the numerator looks like the derivative of the denominator (or a constant multiple of it), that is a huge signal to use substitution. Here, the derivative of $y + 2^y$ is exactly $1 + 2^y \ln 2$. Always check the derivative of your denominator first!

Q: Why is the interval $(1, 2)$? How do I estimate $\ln 3$ without a calculator?

In JEE Advanced, you won't have a calculator, so you must rely on your knowledge of constants. We know $e \approx 2.718$. Since $2.718 < 3 < 7.389$ (which is $e^2$), it follows that $\ln(e) < \ln(3) < \ln(e^2)$. This simplifies to $1 < \ln 3 < 2$. It is all about bounding the value between known powers of $e$.

Select Answer:

Visualized Solution

The Given Differential Equation

Given: $\frac{d y}{d x} = \frac{2 ^{x} y + 2 ^{y} \cdot 2 ^{x}}{2 ^{x} + 2 ^{x + y} l n 2}$
Initial condition: $y (0) = 0$
Goal: Find the interval for $x$ when $y = 1$ .

Factoring the Numerator

Look at the numerator: $2^{x} y + 2^{y} \cdot 2^{x}$
Both terms share a common factor of $2^{x}$ .
Factoring it out: $2^{x} (y + 2^{y})$

Factoring the Denominator

Look at the denominator: $2^{x} + 2^{x + y} ln 2$
Recall exponent rules: $2^{x + y} = 2^{x} \cdot 2^{y}$
The denominator becomes: $2^{x} + 2^{x} \cdot 2^{y} ln 2$
Factoring out $2^{x}$ : $2^{x} (1 + 2^{y} ln 2)$

Simplifying the Equation

Substitute the factored forms back into the equation.
$\frac{d y}{d x} = \frac{2 ^{x} ( y + 2 ^{y} )}{2 ^{x} ( 1 + 2 ^{y} l n 2 )}$
Cancel out $2^{x}$ (since $2^{x} \neq = 0$ for any real $x$ ).
Simplified equation: $\frac{d y}{d x} = \frac{y + 2 ^{y}}{1 + 2 ^{y} l n 2}$

Separating Variables

We need to group all $y$ terms with $d y$ and $x$ terms with $d x$ .
Cross-multiply to separate variables:
$\frac{1 + 2 ^{y} l n 2}{y + 2 ^{y}} d y = d x$
The equation is now ready for integration.

Integrating Both Sides

Integrate both sides of the equation:
$\int \frac{1 + 2 ^{y} l n 2}{y + 2 ^{y}} d y = \int 1 d x$
The right side is simply $x + c$ .
The left side requires a closer look.

Substitution Method

Let the denominator be $u = y + 2^{y}$ .
Differentiate $u$ with respect to $y$ :
$\frac{d u}{d y} = 1 + 2^{y} ln 2$
This perfectly matches our numerator!
So, $d u = (1 + 2^{y} ln 2) d y$ .

Executing the Integration

Substitute $u$ and $d u$ into the integral:
$\int \frac{1}{u} d u = \int 1 d x$
$ln ∣ u ∣ = x + c$
Substitute back $u = y + 2^{y}$ :
$ln ∣ y + 2^{y} ∣ = x + c$

Applying Initial Condition

We are given $y (0) = 0$ . This means when $x = 0$ , $y = 0$ .
Substitute these values into our general solution:
$ln ∣0 + 2^{0} ∣ = 0 + c$
$ln ∣0 + 1∣ = c$

Finding the Constant $c$

Simplify the equation:
$ln (1) = c$
Since $ln (1) = 0$ , we get $c = 0$ .
The particular solution is: $x = ln (y + 2^{y})$

Finding $x$ for $y = 1$

The question asks for the value of $x$ when $y = 1$ .
Substitute $y = 1$ into our particular solution:
$x = ln (1 + 2^{1})$
$x = ln (1 + 2)$
$x = ln 3$

Estimating the Interval

We need to find which interval contains $ln 3$ .
Recall the value of Euler's number: $e \approx 2.718$ .
Compare $3$ with powers of $e$ :
$e^{1} < 3 < e^{2}$ (since $2.718 < 3 < 7.389$ )

Final Conclusion

Take the natural logarithm ( $ln$ ) on all parts of the inequality:
$ln (e^{1}) < ln (3) < ln (e^{2})$
$1 < ln 3 < 2$
Therefore, $x \in (1, 2)$ .
The correct option is $(1, 2)$ .

00:00 / 00:00

The Sigma Insight: Variable Separable Method

Analyzing the Setup

Welcome, future engineers! Today, we are going to dismantle a problem that, at first glance, looks like a chaotic mess of exponents and logarithms.

When you see an expression like

\frac{d y}{d x} = \frac{2 ^{x} y + 2 ^{y} \cdot 2 ^{x}}{2 ^{x} + 2 ^{x + y} ln 2}

it is natural to feel a moment of hesitation. But remember, in the JEE Advanced arena, complexity is often just a mask for elegance. Our job is to peel back the layers.

The Power of Factoring

Before we rush into integration, we must simplify. Look at the numerator:

2^{x} y + 2^{y} \cdot 2^{x}

Both terms are dancing around the factor

2^{x}

. When we pull that out, we get

2^{x} (y + 2^{y})

Now, turn your gaze to the denominator:

2^{x} + 2^{x + y} ln 2

. Recall your exponent laws:

2^{x + y} = 2^{x} \cdot 2^{y}

Suddenly, the denominator reveals its true form:

2^{x} + 2^{x} \cdot 2^{y} ln 2

. Factoring out

2^{x}

here gives us

2^{x} (1 + 2^{y} ln 2)

When we place these back into our differential equation, the

2^{x}

terms cancel out beautifully. We are left with a much cleaner, more manageable equation:

\frac{d y}{d x} = \frac{y + 2 ^{y}}{1 + 2 ^{y} ln 2}

The Detective Work of Integration

Now that we have separated our variables, we arrive at:

\int \frac{1 + 2 ^{y} ln 2}{y + 2 ^{y}} d y = \int d x

This is where the magic happens. If you look closely at the left-hand integral, you might notice something profound.

Let

u = y + 2^{y}

. If we differentiate

u

with respect to

y

, we get:

\frac{d u}{d y} = 1 + 2^{y} ln 2

This is not a coincidence! The numerator is exactly the derivative of the denominator. Our integral transforms into the standard form

\int \frac{1}{u} d u

, which is simply

ln ∣ u ∣

Thus, our solution becomes:

ln ∣ y + 2^{y} ∣ = x + c

The Final Estimation

We are given the initial condition

y (0) = 0

. Plugging these values in, we find

ln ∣0 + 2^{0} ∣ = 0 + c

, which simplifies to

ln (1) = c

, meaning

c = 0

Our particular solution is:

x = ln (y + 2^{y})

Finally, we need to find

x

when

y = 1

. Substituting this, we get:

x = ln (1 + 2^{1}) = ln 3

To place this in an interval, we use the fact that

e \approx 2.718

. Since

e^{1} < 3 < e^{2}

, it follows that

1 < ln 3 < 2

Therefore, $x$ lies in the interval $(1, 2)$ .

Common Traps & FAQs

Why can we just cancel

2^{x}

from the numerator and denominator? Is it possible for

2^{x}

to be zero?

That is a fantastic question! In the realm of real numbers, the exponential function

2^{x}

is strictly positive for all

x

. It never touches the x-axis, meaning it is never zero. Because it is never zero, we are mathematically safe to divide both sides by it without losing any solutions. Always check for domain restrictions, but here, we are safe!

I got stuck at the integration step. How do I know when to use substitution?

Pattern recognition is the heart of calculus! Whenever you see a fraction where the numerator looks like the derivative of the denominator (or a constant multiple of it), that is a huge signal to use substitution. Here, the derivative of

y + 2^{y}

is exactly

1 + 2^{y} ln 2

. Always check the derivative of your denominator first!

Why is the interval

(1, 2)

? How do I estimate

ln 3

without a calculator?

In JEE Advanced, you won't have a calculator, so you must rely on your knowledge of constants. We know

e \approx 2.718

. Since

2.718 < 3 < 7.389

(which is

e^{2}

), it follows that

ln (e) < ln (3) < ln (e^{2})

. This simplifies to

1 < ln 3 < 2

. It is all about bounding the value between known powers of

e

Animated Solution for Mathematics - Differential Equations: If $\frac{d y}{d x} = \frac{2 ^{x} y + 2 ^{y} \cdot 2 ^{x}}{2 ^{x} + 2 ^{x + y} l o g _{e} 2}, y (0) = 0$ , then for $y = 1$ , the value of $x$ lies in the interval:

Select Answer:

Visualized Solution

The Given Differential Equation

Factoring the Numerator

Factoring the Denominator

Simplifying the Equation

Separating Variables

Integrating Both Sides

Substitution Method

Executing the Integration

Applying Initial Condition

Finding the Constant $c$

Finding $x$ for $y = 1$

Estimating the Interval

Final Conclusion

The Sigma Insight: Variable Separable Method

Analyzing the Setup

The Power of Factoring

The Detective Work of Integration

The Final Estimation

Similar Questions

If $\frac{d y}{d x} = \frac{2 ^{x + y} - 2 ^{x}}{2 ^{y}}, y (0) = 1$ , then $y (1)$ is equal to :

If $\frac{d y}{d x} + \frac{2 ^{x - y} ( 2 ^{y} - 1 )}{2 ^{x} - 1} = 0, x, y > 0, y (1) = 1$ , then $y (2)$ is equal to :

The solution of the differential equation $\frac{d y}{d x} = (x - y)^{2}$ , when $y (1) = 1$ , is:

Let $\int_{0}^{x} 1 - (y^{'} (t))^{2} d t = \int_{0}^{x} y (t) d t, 0 \leq x \leq 3, y \geq 0, y (0) = 0$ . Then at $x = 2, y^{''} + y + 1$ is equal to

If $y = y (x)$ is the solution of the differential equation, $e^{y} (\frac{d y}{d x} - 1) = e^{x}$ such that $y (0) = 0$ , then $y (1)$ is equal to

Let $x = x (y)$ be the solution of the differential equation $2 y e^{x / y^{2}} d x + (y^{2} - 4 x e^{x / y^{2}}) d y = 0$ such that $x (1) = 0$ . Then, $x (e)$ is equal to

If $(2 + sin x) \frac{d y}{d x} + (y + 1) cos x = 0$ and $y (0) = 1$ , then $y (\frac{π}{2})$ is equal to:

Let the curve $y = y (x)$ be the solution of the differential equation, $\frac{d y}{d x} = 2 (x + 1)$ . If the numerical value of area bounded by the curve $y = y (x)$ and $x$ -axis is $\frac{4 8}{3}$ , then the value of $y (1)$ is equal to ____

Let $y = y (x)$ be the solution of the differential equation $\frac{d y}{d x} = 1 + x e^{y - x}, - 2 < x < 2, y (0) = 0$ , then, the minimum value of $y (x), x \in (- 2, 2)$ is equal to :

If $y = y (x)$ is the solution of the differential equation $(1 + e^{2 x}) \frac{d y}{d x} + 2 (1 + y^{2}) e^{x} = 0$ and $y (0) = 0$ , then $6 [y^{'} (0) + (y (lo g_{e} 3))^{2}]$ is equal to:

Animated Solution for Mathematics - Differential Equations: .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } If dxdy​=2x+2x+yloge​22xy+2y⋅2x​,y(0)=0, then for y=1, the value of x lies in the interval:

Select Answer:

Visualized Solution

The Given Differential Equation

Factoring the Numerator

Factoring the Denominator

Simplifying the Equation

Separating Variables

Integrating Both Sides

Substitution Method

Executing the Integration

Applying Initial Condition

Finding the Constant .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } c

Finding .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } x for .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } y=1

Estimating the Interval

Final Conclusion

The Sigma Insight: Variable Separable Method

Analyzing the Setup

The Power of Factoring

The Detective Work of Integration

The Final Estimation

Similar Questions

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } If dxdy​=2y2x+y−2x​,y(0)=1, then y(1) is equal to :

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } If dxdy​+2x−12x−y(2y−1)​=0,x,y>0,y(1)=1, then y(2) is equal to :

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } The solution of the differential equation dxdy​=(x−y)2, when y(1)=1, is:

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Let ∫0x​1−(y′(t))2​dt=∫0x​y(t)dt,0≤x≤3,y≥0,y(0)=0. Then at x=2,y′′+y+1 is equal to

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } If y=y(x) is the solution of the differential equation, ey(dxdy​−1)=ex such that y(0)=0, then y(1) is equal to

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Let x=x(y) be the solution of the differential equation 2yex/y2dx+(y2−4xex/y2)dy=0 such that x(1)=0. Then, x(e) is equal to

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } If (2+sinx)dxdy​+(y+1)cosx=0 and y(0)=1, then y(2π​) is equal to:

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Let y=y(x) be the solution of the differential equation dxdy​=1+xey−x,−2​<x<2​,y(0)=0, then, the minimum value of y(x),x∈(−2​,2​) is equal to :

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } If y=y(x) is the solution of the differential equation (1+e2x)dxdy​+2(1+y2)ex=0 and y(0)=0, then 6[y′(0)+(y(loge​3​))2] is equal to:

Animated Solution for Mathematics - Differential Equations: If $\frac{d y}{d x} = \frac{2 ^{x} y + 2 ^{y} \cdot 2 ^{x}}{2 ^{x} + 2 ^{x + y} l o g _{e} 2}, y (0) = 0$ , then for $y = 1$ , the value of $x$ lies in the interval:

Finding the Constant $c$

Finding $x$ for $y = 1$

If $\frac{d y}{d x} = \frac{2 ^{x + y} - 2 ^{x}}{2 ^{y}}, y (0) = 1$ , then $y (1)$ is equal to :

If $\frac{d y}{d x} + \frac{2 ^{x - y} ( 2 ^{y} - 1 )}{2 ^{x} - 1} = 0, x, y > 0, y (1) = 1$ , then $y (2)$ is equal to :

The solution of the differential equation $\frac{d y}{d x} = (x - y)^{2}$ , when $y (1) = 1$ , is:

Let $\int_{0}^{x} 1 - (y^{'} (t))^{2} d t = \int_{0}^{x} y (t) d t, 0 \leq x \leq 3, y \geq 0, y (0) = 0$ . Then at $x = 2, y^{''} + y + 1$ is equal to

If $y = y (x)$ is the solution of the differential equation, $e^{y} (\frac{d y}{d x} - 1) = e^{x}$ such that $y (0) = 0$ , then $y (1)$ is equal to

Let $x = x (y)$ be the solution of the differential equation $2 y e^{x / y^{2}} d x + (y^{2} - 4 x e^{x / y^{2}}) d y = 0$ such that $x (1) = 0$ . Then, $x (e)$ is equal to

If $(2 + sin x) \frac{d y}{d x} + (y + 1) cos x = 0$ and $y (0) = 1$ , then $y (\frac{π}{2})$ is equal to:

Let $y = y (x)$ be the solution of the differential equation $\frac{d y}{d x} = 1 + x e^{y - x}, - 2 < x < 2, y (0) = 0$ , then, the minimum value of $y (x), x \in (- 2, 2)$ is equal to :

If $y = y (x)$ is the solution of the differential equation $(1 + e^{2 x}) \frac{d y}{d x} + 2 (1 + y^{2}) e^{x} = 0$ and $y (0) = 0$ , then $6 [y^{'} (0) + (y (lo g_{e} 3))^{2}]$ is equal to: