If φ(x) = (1/√x) ∫_(π/4)ˣ (4√2 sin t - 3φ'(t)) dt, x 0, then φ'(π/4) is equal to : (1) 8/√π, (2) 4

Q: Why can't I just differentiate the integral directly using the quotient rule?

You *could*, but it's a path to algebraic misery. The quotient rule would introduce $(\sqrt{x})^2$ in the denominator and make the derivative of the integral term much more complex. Multiplying by $\sqrt{x}$ first is the hallmark of a smart JEE aspirant—it keeps the algebra clean and minimizes the chance of silly errors.

Q: Wait, how do I find $\phi(\pi/4)$? It's not given in the problem!

This is a classic 'hidden' condition. Look at the integral definition: $\int_{\pi/4}^{x} ... dt$. If you set $x = \pi/4$, the integral becomes $\int_{\pi/4}^{\pi/4} ... dt$. By the fundamental properties of definite integrals, any integral from $a$ to $a$ is zero. This gives you $\phi(\pi/4) = 0$ instantly. Always check the limits of integration for these 'free' values!

Q: Is the Newton-Leibniz rule always just replacing $t$ with $x$?

Almost! The rule states $\frac{d}{dx} \int_{a}^{x} f(t) dt = f(x)$. It works perfectly here because the upper limit is just $x$. If the upper limit were $x^2$, you would need to use the Chain Rule: $f(x^2) \cdot \frac{d}{dx}(x^2)$. Always keep an eye on that upper limit!

Question

If $\phi(x) = \frac{1}{\sqrt{x}} \int_{\pi/4}^{x} (4\sqrt{2} \sin t - 3\phi'(t)) dt, x > 0$, then $\phi'(\pi/4)$ is equal to :

Options:

(A) $\frac{8}{\sqrt{\pi}}$

(B) $\frac{4}{6+\sqrt{\pi}}$

(C) $\frac{8}{6+\sqrt{\pi}}$

(D) $\frac{4}{6-\sqrt{\pi}}$

Sigma Percentile · Accepted Answer

### The Symphony of Calculus: Solving the Integral Equation

Welcome, future engineers! Today, we are going to dissect a problem that perfectly captures the elegance of JEE Advanced mathematics. It is not just about crunching numbers; it is about seeing the structure of the equation and choosing the most graceful path to the solution.

### Phase 1: The Algebraic Setup

We are given the equation:

$$\phi(x) = \frac{1}{\sqrt{x}} \int_{\pi/4}^{x} (4\sqrt{2} \sin t - 3\phi'(t)) dt$$

At first glance, it looks intimidating. We have a function $\phi(x)$ defined by an integral, and we need to find its derivative at a specific point, $\phi'(\pi/4)$.

The immediate instinct for many students is to jump straight into differentiation. But stop! If you differentiate immediately, you will be forced to use the Quotient Rule on the right-hand side, which will lead to a messy expression involving $\sqrt{x}$ in the denominator.

Instead, let us be strategic. We multiply both sides by $\sqrt{x}$ to obtain:

$$\sqrt{x}\phi(x) = \int_{\pi/4}^{x} (4\sqrt{2} \sin t - 3\phi'(t)) dt$$

Now, the equation is linear and clean. This simple algebraic maneuver is the difference between a chaotic calculation and a smooth derivation.

### Phase 2: The Calculus Symphony

Now that we have cleared the denominator, we are ready to differentiate both sides with respect to $x$. On the left-hand side, we have a product of two functions: $\sqrt{x}$ and $\phi(x)$.

We apply the Product Rule:

$$\frac{d}{dx}(\sqrt{x}\phi(x)) = \frac{1}{2\sqrt{x}}\phi(x) + \sqrt{x}\phi'(x)$$

On the right-hand side, we invoke the powerful Newton-Leibniz Rule. This rule is a bridge between integration and differentiation, stating that the derivative of an integral with a variable upper limit $x$ is simply the integrand evaluated at $x$.

Thus, the right-hand side becomes: $4\sqrt{2}\sin x - 3\phi'(x)$. Equating the two sides, we get our master differential equation:

$$\frac{\phi(x)}{2\sqrt{x}} + \sqrt{x}\phi'(x) = 4\sqrt{2}\sin x - 3\phi'(x)$$

### Phase 3: The Boundary Condition

We are almost there, but we have a problem. We need to find $\phi'(\pi/4)$, but our equation contains $\phi(x)$. If we plug in $x = \pi/4$ right now, we will need the value of $\phi(\pi/4)$.

We return to the original integral equation:

$$\phi(x) = \frac{1}{\sqrt{x}} \int_{\pi/4}^{x} (4\sqrt{2} \sin t - 3\phi'(t)) dt$$

If we substitute $x = \pi/4$, the integral becomes $\int_{\pi/4}^{\pi/4} ... dt$. By the fundamental properties of definite integrals, an integral from a point to itself is always zero. Therefore, $\phi(\pi/4) = 0$.

This is our 'Aha!' moment. The term $\phi(x)$ vanishes at our point of interest!

### Phase 4: The Final Tally

With $\phi(\pi/4) = 0$, our master equation at $x = \pi/4$ simplifies beautifully:

$$0 + \sqrt{\frac{\pi}{4}}\phi'\left(\frac{\pi}{4}\right) = 4\sqrt{2}\sin\left(\frac{\pi}{4}\right) - 3\phi'\left(\frac{\pi}{4}\right)$$

We know that $\sin(\pi/4) = 1/\sqrt{2}$ and $\sqrt{\pi/4} = \sqrt{\pi}/2$. Substituting these values, we get:

$$\frac{\sqrt{\pi}}{2}\phi'\left(\frac{\pi}{4}\right) = 4 - 3\phi'\left(\frac{\pi}{4}\right)$$

Now, it is just simple algebra. Group the $\phi'(\pi/4)$ terms:

$$\phi'\left(\frac{\pi}{4}\right) \left( \frac{\sqrt{\pi}}{2} + 3 \right) = 4$$

Simplifying the bracket gives us $(\sqrt{\pi} + 6)/2$. Finally, isolating $\phi'(\pi/4)$, we arrive at the **final answer**:

$$\phi'\left(\frac{\pi}{4}\right) = \frac{8}{6 + \sqrt{\pi}}$$

And there you have it! By staying calm, simplifying the algebra first, and using the Newton-Leibniz rule effectively, we turned a complex problem into a series of elegant steps. Keep practicing, and remember: every problem is just a story waiting to be solved.

Animated Solution for Mathematics - Definite Integration: If $ϕ (x) = \frac{1}{x} \int_{π /4}^{x} (42 sin t - 3 ϕ^{'} (t)) d t, x > 0$ , then $ϕ^{'} (π /4)$ is equal to :

Select Answer:

Visualized Solution

Analyze the Given Equation

Simplify the Equation

Prepare for Differentiation

Differentiate the Left Hand Side

Differentiate the Right Hand Side

Equate the Derivatives

Find the Boundary Condition $ϕ (\frac{π}{4})$

Substitute $x = \frac{π}{4}$ into the Differential Equation

Evaluate Trigonometric and Root Terms

Group the $ϕ^{'} (\frac{π}{4})$ Terms

Simplify the Bracket and Final Calculation

The Sigma Insight: Newton-Leibniz & Reduction Formulas

The Symphony of Calculus

Phase 1

Phase 2

Phase 3

Phase 4

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Animated Solution for Mathematics - Definite Integration: .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } If ϕ(x)=x​1​∫π/4x​(42​sint−3ϕ′(t))dt,x>0, then ϕ′(π/4) is equal to :