If I = integral from 1 to 2 of dx / sqrt(2x^3 - 9x^2 + 12x + 4), then: (A) 1/6

Q: Wait, why can't we just integrate this directly? It's a cubic polynomial.

That is a great question! While you could technically attempt to integrate it, the expression $\int \frac{dx}{\sqrt{2x^3 - 9x^2 + 12x + 4}}$ does not have a simple elementary antiderivative. In JEE Advanced, when you see an integral that looks impossible to solve, it is almost always a hint to use estimation or properties of definite integrals instead of brute force.

Q: Why does the inequality flip when we take the reciprocal?

Think about simple numbers: $2 0.25$, the inequality must flip. This is a fundamental property of inequalities for positive numbers. Whenever you take the reciprocal of an inequality, you must reverse the inequality signs.

Q: How do I know if I should use strict or non-strict inequalities?

In this problem, the function $g(x)$ is strictly decreasing. This means it only reaches the values $8$ and $9$ at the exact endpoints $x=2$ and $x=1$. Since the integral is over the open interval $(1, 2)$, the value of the integral will be strictly between the values obtained by integrating the bounds. Thus, strict inequalities are appropriate.

Question

If $I = \int_1^2 \frac{dx}{\sqrt{2x^3 - 9x^2 + 12x + 4}}$, then:

Options:

(A) $\frac{1}{6} < I^2 < \frac{1}{2}$

(B) $\frac{1}{8} < I^2 < \frac{1}{4}$

(C) $\frac{1}{9} < I^2 < \frac{1}{8}$

(D) $\frac{1}{16} < I^2 < \frac{1}{9}$

Sigma Percentile · Accepted Answer

### The Art of Estimation: Taming the Impossible Integral

Welcome, future engineers! Today, we are going to face a problem that might look like a nightmare of calculus, but it is actually a beautiful exercise in strategic thinking.

Imagine you are standing before a complex integral:

$$I = \int_1^2 \frac{dx}{\sqrt{2x^3 - 9x^2 + 12x + 4}}$$

If you try to integrate this directly, you will find yourself lost in a forest of algebraic complexity. The JEE is not testing your ability to perform brute-force integration; it is testing your ability to see the structure of the function.

Let us define the denominator as $g(x) = 2x^3 - 9x^2 + 12x + 4$. Our goal is to understand how this function behaves between $x=1$ and $x=2$.

### Phase 1: Analyzing the Beast

First, we look for the slope of the curve by finding the derivative:

$$g'(x) = \frac{d}{dx}(2x^3 - 9x^2 + 12x + 4) = 6x^2 - 18x + 12$$

Factoring this, we get:

$$g'(x) = 6(x^2 - 3x + 2) = 6(x-1)(x-2)$$

Notice that for any $x$ in the interval $(1, 2)$, the term $(x-1)$ is positive and $(x-2)$ is negative. This means the derivative is negative, and our function $g(x)$ is **strictly decreasing**.

Because it is strictly decreasing, the maximum value must occur at the start of the interval, $x=1$, and the minimum value must occur at the end, $x=2$.

Calculating these, we find:

$$g(1) = 2(1)^3 - 9(1)^2 + 12(1) + 4 = 9$$

$$g(2) = 2(2)^3 - 9(2)^2 + 12(2) + 4 = 8$$

Now, we have trapped our polynomial: $8 \le g(x) \le 9$.

### Phase 2: The Reciprocal Flip

Now, we need to transform this into our integrand. Taking the square root, we get $\sqrt{8} \le \sqrt{g(x)} \le \sqrt{9}$, which simplifies to $\sqrt{8} \le \sqrt{g(x)} \le 3$.

Here is the crucial step: when we take the reciprocal, the inequality flips! We get:

$$\frac{1}{3} \le \frac{1}{\sqrt{g(x)}} \le \frac{1}{\sqrt{8}}$$

This is where many students stumble, but you now know to be careful. Finally, integrating this inequality from $1$ to $2$ gives us the bounds for $I$.

Since the integral of a constant $k$ over an interval of length $1$ is just $k$, we get:

$$\int_1^2 \frac{1}{3} dx \le I \le \int_1^2 \frac{1}{\sqrt{8}} dx$$

This simplifies to $\frac{1}{3} \le I \le \frac{1}{\sqrt{8}}$.

### Phase 3: The Final Victory

We are almost there! The question asks for the range of $I^2$. Squaring our inequality, we arrive at:

$$\left(\frac{1}{3}\right)^2 < I^2 < \left(\frac{1}{\sqrt{8}}\right)^2$$

This gives us the final result:

$$\frac{1}{9} < I^2 < \frac{1}{8}$$

This is the elegance of mathematics—we didn't need to solve the integral; we only needed to understand its bounds. Keep this technique in your toolkit, and you will find that even the most terrifying integrals can be tamed with a bit of insight.

Animated Solution for Mathematics - Definite Integration: If $I = \int_{1}^{2} \frac{d x}{2 x ^{3} - 9 x ^{2} + 12 x + 4}$ , then:

Select Answer:

Visualized Solution

Introduction to the Integral

Analyzing the Denominator $g (x)$

Finding Critical Points

Determining Monotonicity

Calculating Maximum Value

Calculating Minimum Value

Establishing the Range of $g (x)$

Applying the Square Root

Taking the Reciprocal

Integrating the Inequality

Squaring to find $I^{2}$

Final Conclusion

The Sigma Insight: Fundamental Theorem & Properties of Definite Integrals

The Art of Estimation

Phase 1

Phase 2

Phase 3

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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 I=∫12​2x3−9x2+12x+4​dx​, then: