Integral Calculus
Integral Calculus
star_batch_jee_advanced_2025
Grade 12
Question:
Let $f : \mathbb{R} \to \mathbb{R}$ be a function as $f(x) = (x-1)(x+2)(x-3)(x-6) - 100$. If $g(x)$ is a polynomial of degree $\leq 3$ such that $\int \frac{g(x)}{f(x)}dx$ does not contain any logarithm function and $g(-2) = 10$, then:
$f(x) = 0$ has two real & two imaginary roots
$(f(x))_{\min} = -84$
$\int \frac{g(x)}{f(x)}dx = \tan^{-1}\left(\frac{x-2}{2}\right) + c$
$g(2) = -42$
Step-by-Step Solution
Key Concept: Factoring $f(x)$ using substitution $u = x^2-4x$ reduces the quartic to a product of two quadratics with known discriminants.
First, expand $f(x) = (x-1)(x+2)(x-3)(x-6)-100$ by grouping as $(x^2-4x+3)(x^2-4x-12)-100$. Substitute $u = x^2-4x$ to get $(u+3)(u-12)-100 = u^2-9u-136 = (u-17)(u+8)$, giving $f(x) = (x^2-4x-17)(x^2-4x+8)$. This shows $f(x)=0$ has two real roots and two imaginary roots. Since $A, B, C$ must be zero in the partial fraction decomposition, $g(x) = D(x^2-4x-17)$. Using $g(-2) = D(4+8-17) = -5D$, and the condition that $g(x)/f(x)$ yields specific terms, we find $D=2$. Therefore $\int\frac{g(x)}{f(x)}dx = \frac{1}{2}\ln\left|\frac{x-2}{2}\right| + \tan^{-1}\left(\frac{x-2}{2}\right) + c$.
Correct Answer: 1,2,3,4