Differential Calculus
Differential Calculus
star_batch_jee_advanced_2025
Grade 12
Question:
Let $f(x) = x^5 + ax + b, x \in \mathbb{R}, f(0) > 0$ & $f(x)$ has integral roots. Tangent at $\left(\frac{5}{2}, p\right)$ to $y = f(x)$ is parallel to $x$-axis & $g(x) = f(x+1)$.
Column 1:
(A) $(a + b)$ can be
(B) Value of $[p]$ can be (where $[.]$ represents greatest integer function)
(C) Number of points where $g(|x|)$ is non differentiable can be
(D) Number of points where $|g(|x|)|$ is non differentiable can be
Column 2:
(p) $-1$
(q) $1$
(r) $3$
(s) $-3$
(t) $5$
Step-by-Step Solution
Key Concept: Use derivative conditions and Vieta's formulas to constrain the roots and coefficients of a quadratic.
Given $f(x) = x^2 + ax + b$ with $f(x)|_{x=5/(2p)} = 0$ implies $2x + a|_{x=5/(2p)} = 0$, giving $a = -5$. Thus $f(x) = x^2 - 5x + b$. Using conditions $f(0) > 0$ (so $b > 0$) and $\alpha + \beta = 5$, $\alpha\beta = b > 0$ where $\alpha, \beta$ are roots, we find both roots are positive. The pairs $(\alpha, \beta)$ can be $(1,4)$ or $(2,3)$, giving $b = 4$ or $b = 6$. Therefore $f(x) = x^2 - 5x + 4$ or $f(x) = x^2 - 5x + 6$.
Correct Answer: p]