If the normal to the curve y(x) = ∫ (2t² - 15t + 10) dt from 0 to x at a point (a, b) is parallel to

Q: Wait, why do we reject the root $a = \frac{1}{2}$? Is it mathematically wrong?

It is not mathematically wrong, but it is contextually invalid. The problem explicitly states $a > 1$. In JEE problems, constraints are often the 'trap' designed to test if you are reading carefully. Always check your final roots against the domain constraints provided in the question statement.

Q: Why is the slope of the normal $-\frac{1}{m_t}$? Can't we just use $m_t$?

Great question! The tangent and the normal are perpendicular to each other at the point of contact. Two lines with slopes $m_1$ and $m_2$ are perpendicular if and only if $m_1 \cdot m_2 = -1$. Therefore, the slope of the normal must be the negative reciprocal of the tangent's slope. If you use $m_t$ instead of $-\frac{1}{m_t}$, you are finding the slope of the tangent, not the normal.

Q: Do I really need the Newton-Leibniz rule? Can't I just integrate the function first and then differentiate?

You absolutely could! You could evaluate $\int_0^x (2t^2 - 15t + 10) dt$ to get $\frac{2x^3}{3} - \frac{15x^2}{2} + 10x$, and then differentiate that. You would get the same result: $2x^2 - 15x + 10$. However, the Newton-Leibniz rule is a powerful shortcut that saves time and reduces the chance of algebraic errors during integration. It is a vital tool for your JEE toolkit.

Question

If the normal to the curve $y(x) = \int_0^x (2t^2 - 15t + 10) dt$ at a point $(a, b)$ is parallel to the line $x + 3y = -5, a > 1$, then the value of $|a + 6b|$ is equal to

Sigma Percentile · Accepted Answer

### The Geometry of Calculus: A Journey Through the Curve

Welcome, fellow traveler on the road to JEE excellence. Today, we are not just solving a problem; we are peeling back the layers of a beautiful geometric puzzle.

We have a curve defined by an integral, a normal line dancing in parallel to a reference line, and a final value that hides behind a curtain of algebra. Let us begin.

### Phase 1: The Geometric Setup

Imagine standing on a coordinate plane. You have a curve defined by the function $y(x) = \int_0^x (2t^2 - 15t + 10) dt$.

This isn't just a static line; it is an area function, accumulating value as $x$ increases. We are interested in a specific point $(a, b)$ on this curve.

At this point, a normal line is drawn. The problem whispers a secret: this normal is parallel to the line $x + 3y = -5$.

Geometrically, this means the normal and the line share the same slope. Our mission is to find $a$ and $b$ and then calculate the value of $|a + 6b|$.

### Phase 2: The Power of Newton-Leibniz

To find the slope of the tangent at $(a, b)$, we need the derivative $\frac{dy}{dx}$. Many students panic when they see an integral in a function definition, but this is where the **Newton-Leibniz Rule** becomes your best friend.

It tells us that the derivative of an integral with respect to its upper limit is simply the integrand evaluated at that limit. So, for our function $y(x) = \int_0^x (2t^2 - 15t + 10) dt$, the derivative is:

$$y'(x) = 2x^2 - 15x + 10$$

At our point of interest, $x = a$, the slope of the tangent is $m_t = 2a^2 - 15a + 10$.

### Phase 3: The Normal Connection

Now, we must pivot. The problem discusses the *normal*, not the tangent.

Recall that the normal is the line perpendicular to the tangent at the point of contact. If the tangent has slope $m_t$, the normal must have slope $m_n = -\frac{1}{m_t}$.

Substituting our expression, we get:

$$m_n = -\frac{1}{2a^2 - 15a + 10}$$

### Phase 4: The Reference Line and the Quadratic Fork

Look at the reference line: $x + 3y = -5$. To find its slope, we rewrite it in the slope-intercept form $y = mx + c$.

Rearranging, we get $3y = -x - 5$, or $y = -\frac{1}{3}x - \frac{5}{3}$. The slope is clearly $m = -\frac{1}{3}$.

Since the normal is parallel to this line, their slopes must be equal:

$$-\frac{1}{2a^2 - 15a + 10} = -\frac{1}{3}$$

Canceling the negatives and taking the reciprocal, we arrive at the quadratic equation: $2a^2 - 15a + 10 = 3$, which simplifies to $2a^2 - 15a + 7 = 0$.

Factoring this, we find $(2a - 1)(a - 7) = 0$. This gives us two potential paths: $a = \frac{1}{2}$ or $a = 7$.

But wait—the problem constraints are the gatekeepers of the correct answer. Since $a > 1$, we must reject $a = \frac{1}{2}$ and embrace $a = 7$.

### Phase 5: The Final Calculation

With $a = 7$ in hand, we find $b$ by evaluating the original integral at $x = 7$:

$$b = \int_0^7 (2t^2 - 15t + 10) dt$$

Integrating term by term, we get:

$$b = \left[ \frac{2t^3}{3} - \frac{15t^2}{2} + 10t \right]_0^7$$

Substituting the limits, we calculate:

$$b = \frac{2(343)}{3} - \frac{15(49)}{2} + 10(7) = \frac{686}{3} - \frac{735}{2} + 70$$

Finding the common denominator of $6$, we get:

$$b = \frac{1372 - 2205 + 420}{6} = -\frac{413}{6}$$

Finally, we compute $|a + 6b| = |7 + 6(-\frac{413}{6})|$.

The $6$ cancels out beautifully, leaving us with $|7 - 413| = |-406| = 406$.

And there it is—the elegance of the final result. You have navigated the calculus, respected the constraints, and mastered the algebra. The final answer is **406**.

Animated Solution for Mathematics - Differentiation: If the normal to the curve $y (x) = \int_{0}^{x} (2 t^{2} - 15 t + 10) d t$ at a point $(a, b)$ is parallel to the line $x + 3 y = - 5, a > 1$ , then the value of $∣ a + 6 b ∣$ is equal to

Enter Numerical Value:

Visualized Solution

The Geometric Setup

Differentiating the Integral

Slope of the Tangent

Slope of the Normal

The Reference Line

Equating the Slopes

Solving the Quadratic Equation

Applying the Constraint

Setting up the Integral for 'b'

Evaluating the Integral

Substituting the Limits

Calculating the Value of 'b'

Finding $∣ a + 6 b ∣$

The Sigma Insight: Tangents, Normals and Rate Measure

The Geometry of Calculus

Phase 1

Phase 2

Phase 3

Phase 4

Phase 5

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