The slope of the tangent to a curve C : y = y(x) at any point (x, y) on it is (2e²ˣ - 6e⁻ˣ + 9) / (2

Q: Why did we multiply by $e^{2x}$? Couldn't we just integrate as is?

You *could* attempt to integrate as is, but it would be a nightmare of chain rules and messy algebra. Multiplying by $e^{2x}$ is a strategic move to normalize the expression into a form where the denominator's derivative is visible in the numerator. It's about working smarter, not harder. In JEE, recognizing these algebraic 'normalizations' is often the difference between solving a problem in 3 minutes versus 15.

Q: I got stuck at the $\tan^{-1}$ part. How do I combine $\tan^{-1}(A) = \frac{\pi}{4} + \tan^{-1}(B)$?

Great question! Never try to solve for the angle directly by taking the inverse of both sides. Instead, take the tangent of both sides. Use the identity $\tan(A+B) = \frac{\tan A + \tan B}{1 - \tan A \tan B}$. It turns a transcendental equation into a simple algebraic one. This is a standard trick to clear inverse trigonometric functions.

Q: Why do we reject the negative root or worry about the constant $C$?

The constant $C$ is not just a number; it's the 'DNA' of your specific curve. Without finding $C$ using the given point $(0, \frac{1}{2} + \frac{\pi}{2\sqrt{2}})$, you are describing a family of curves, not the specific one passing through your points. Always respect the boundary conditions! They are the keys that unlock the specific solution from the general one.

Question

The slope of the tangent to a curve $C : y = y(x)$ at any point $(x, y)$ on it is $\frac{2e^{2x} - 6e^{-x} + 9}{2 + 9e^{-2x}}$. If $C$ passes through the points $(0, \frac{1}{2} + \frac{\pi}{2\sqrt{2}})$ and $(\alpha, \frac{1}{2}e^{2\alpha})$, then $e^{\alpha}$ is equal to :

Options:

(A) $\frac{3+\sqrt{2}}{3-\sqrt{2}}$

(B) $\frac{3}{\sqrt{2}}\left(\frac{3+\sqrt{2}}{3-\sqrt{2}}\right)$

(C) $\frac{1}{\sqrt{2}}\left(\frac{\sqrt{2}+1}{\sqrt{2}-1}\right)$

(D) $\frac{\sqrt{2}+1}{\sqrt{2}-1}$

Sigma Percentile · Accepted Answer

### The Anatomy of a Curve: A Journey Through Differential Equations

Welcome, future engineer. Today, we are going to dismantle a problem that, at first glance, looks like a chaotic mess of exponentials. You see an expression like $\frac{2e^{2x} - 6e^{-x} + 9}{2 + 9e^{-2x}}$ and your brain might instinctively want to run away.

But I want you to pause. Take a breath. In the world of JEE Advanced, complexity is often just a mask for elegance. Let's peel back that mask together.

### Phase 1: The Algebraic Surgery

We are given the slope of the tangent, which is the derivative $\frac{dy}{dx}$. Our goal is to find the curve $y(x)$. The expression we are given is:

$$
\frac{dy}{dx} = \frac{2e^{2x} - 6e^{-x} + 9}{2 + 9e^{-2x}}
$$

This is our starting point. The negative exponents are the 'noise' in the system. To clear this noise, we perform a bit of algebraic surgery.

We multiply the numerator and the denominator by $e^{2x}$. Why? Because $e^{2x} \cdot e^{-2x} = 1$, which effectively eliminates the negative powers. After this operation, our equation transforms into:

$$
\frac{dy}{dx} = \frac{2e^{4x} - 6e^x + 9e^{2x}}{2e^{2x} + 9}
$$

Now, look at the numerator. We want to match it to the denominator, $2e^{2x} + 9$. By grouping terms cleverly, we can rewrite the numerator as $e^{2x}(2e^{2x} + 9) - 6e^x$.

When we divide this by the denominator, the expression simplifies beautifully into:

$$
\frac{dy}{dx} = e^{2x} - \frac{6e^x}{2e^{2x} + 9}
$$

Suddenly, the monster is tamed. We have two simple terms ready for integration.

### Phase 2: The Art of Integration

Now we integrate both sides with respect to $x$. The first part, $\int e^{2x} dx$, is a standard integral that gives us $\frac{1}{2}e^{2x}$.

The second part, $\int \frac{6e^x}{2e^{2x} + 9} dx$, is where the real work happens. We use the substitution method. Let $t = e^x$, which implies $dt = e^x dx$. The integral becomes:

$$
I_1 = \int \frac{6 dt}{2t^2 + 9} = 3 \int \frac{dt}{t^2 + \left(\frac{3}{\sqrt{2}}ight)^2}
$$

This is the classic form $\int \frac{dx}{x^2 + a^2} = \frac{1}{a} 	an^{-1}\left(\frac{x}{a}ight)$. Applying this, we get:

$$
I_1 = \sqrt{2} 	an^{-1}\left(\frac{\sqrt{2}e^x}{3}ight)
$$

Combining these, we have our general solution for the curve $C$:

$$
y = \frac{1}{2}e^{2x} - \sqrt{2} 	an^{-1}\left(\frac{\sqrt{2}e^x}{3}ight) + C
$$

### Phase 3: The Boundary Conditions and the Finale

We are not done yet. We have a constant $C$ to find. The problem tells us the curve passes through $(0, \frac{1}{2} + \frac{\pi}{2\sqrt{2}})$.

Substituting $x=0$ and $y$ into our equation, we find $C = \frac{\pi}{2\sqrt{2}} + \sqrt{2} 	an^{-1}\left(\frac{\sqrt{2}}{3}ight)$.

Finally, we use the second point $(\alpha, \frac{1}{2}e^{2\alpha})$. When we plug this into our equation, the $\frac{1}{2}e^{2\alpha}$ terms on both sides cancel out! This is the 'aha!' moment.

We are left with:

$$
\sqrt{2} 	an^{-1}\left(\frac{\sqrt{2}e^\alpha}{3}ight) = \frac{\pi}{2\sqrt{2}} + \sqrt{2} 	an^{-1}\left(\frac{\sqrt{2}}{3}ight)
$$

Dividing by $\sqrt{2}$ and taking the tangent of both sides, we use the identity $	an(A+B) = \frac{	an A + 	an B}{1 - 	an A 	an B}$. The calculation simplifies to:

$$
\frac{\sqrt{2}e^\alpha}{3} = \frac{3+\sqrt{2}}{3-\sqrt{2}}
$$

Solving for $e^\alpha$, we arrive at our final answer. You see? The complexity was just a path to a very elegant result. Keep practicing, and soon, you will see these patterns before you even pick up your pen.

Select Answer:

Visualized Solution

The Differential Equation

Simplifying the Expression

Rearranging the Numerator

Setting up the Integration

Solving the Second Integral

Applying Standard Integral Formula

The General Solution

Finding the Constant C

Applying the Second Point

Simplifying the Equation

Final Calculation

The Sigma Insight: Variable Separable Method

The Anatomy of a Curve

Phase 1

Phase 2

Phase 3

Similar Questions

Let a curve $y = y (x)$ be given by the solution of the differential equation $cos (\frac{1}{2} cos^{- 1} (e^{- x})) d x = e^{2 x} - 1 d y$ . If it intersects $y$ -axis at $y = - 1$ , and the intersection point of the curve with $x$ -axis is $(α, 0)$ , then $e^{α}$ is equal to

If the solution curve $y = y (x)$ of the differential equation $(1 + y^{2}) (1 + lo g_{e} x) d x + x d y = 0, x > 0$ passes through the point $(1, 1)$ and $y (e) = \frac{α - t a n ( 3/2 )}{β + t a n ( 3/2 )}$ , then $α + 2 β$ is

Given that the slope of the tangent to a curve $y = y (x)$ at any point $(x, y)$ is $2 y / x^{2}$ . If the curve passes through the centre of the circle $x^{2} + y^{2} - 2 x - 2 y = 0$ , then its equation is :

Let $f$ be a differentiable function satisfying $f (x) = \frac{2}{3} \int_{0}^{3} f (\frac{λ ^{2} x}{3}) d λ, x > 0$ and $f (1) = 3$ . If $y = f (x)$ passes through the point $(α, 6)$ , then $α$ is equal to \_\_\_\_.

Let a smooth curve $y = f (x)$ be such that the slope of the tangent at any point $(x, y)$ on it is directly proportional to $\frac{- y}{x}$ . If the curve passes through the point $(1, 2)$ and $(8, 1)$ , then $∣ y (\frac{1}{8}) ∣$ is equal to

Let a curve $y = f (x)$ pass through the point $(2, (lo g_{e} 2)^{2})$ and have slope $\frac{2 y}{x l o g _{e} x}$ for all positive real value of $x$ . Then the value of $f (e)$ is equal to ___

The slope of normal at any point $(x, y), x > 0, y > 0$ on the curve $y = y (x)$ is given by $\frac{x ^{2}}{x y - x ^{2} y ^{2} - 1}$ . If the curve passes through the point $(1, 1)$ , then $e \cdot y (e)$ is equal to

Let $Γ$ denote a curve $y = y (x)$ which is in the first quadrant and let the point $(1, 0)$ lie on it. Let the tangent to $Γ$ at a point $P$ intersect the y-axis at $Y_{P}$ . If $P Y_{P}$ has length 1 for each point $P$ on $Γ$ , then which of the following options is/are correct?

Let $y = y (x)$ be solution of the differential equation $lo g_{e} (\frac{d y}{d x}) = 3 x + 4 y$ , with $y (0) = 0$ . If $y (- \frac{2}{3} lo g_{e} 2) = α lo g_{e} 2$ , then the value of $α$ is equal to:

Let $y = y (x)$ be the solution of the differential equation $d y = e^{α x + y} d x; α \in N$ . If $y (lo g_{e} 2) = lo g_{e} 2$ and $y (0) = lo g_{e} (1/2)$ , then the value of $α$ is equal to