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Question #41635

Difficulty: 8/10 | 🟢 Free

The Question

Let $T_m$ be the numerically greatest term in the expansion of $(3 + 2x)^{10}$ at $x = 1$ . Let $m$ arithmetic means $A_1, A_2, \dots, A_m$ be inserted between $5$ and $23$ . The points $P(A_1, A_2, A_1+A_2)$ , $Q(A_3, \beta, A_3+\beta)$ , and $R(A_5, \alpha, A_5+\alpha)$ are collinear in 3D space.

Consider the polynomial $g(x) = \frac{x^4}{4} - \frac{13 x^3}{3} + \frac{c x^2}{2} - q x + r$ . If $g(x)$ has a point of inflection at $x = \alpha$ (where $g'(\alpha)=0$ ) and a local minimum at $x = \beta$ , find the value of $c + q$ .

Options

A.
114
B.
84
C.
102
D.
96

Check Your Answer

Correct Answer

Step-by-Step Solution

Explanation

How the Examiner Thinks

Specialist

Targeted practice for niche, high-difficulty concepts that frequently appear in JEE Advanced exams.

Concepts Behind This Question

Collinearity vs Coplanarity|Insertion of Means|Numerically Greatest Term|Even Multiplicity Roots

Why This Question Is Tricky

Common Mistakes

This question engineers a 'Lethal Collision' between Algebra (Binomial NGT, Arithmetic Progressions) and Calculus/Vectors.

The primary trap lies in the root multiplicity of the derivative $g'(x)$ .

Students often correctly identify the sum of the roots as 13 but fail to map the 'point of inflection' to the even multiplicity root ( $\alpha$ ) and the 'local minimum' to the odd multiplicity root ( $\beta$ ).

If a student blindly swaps their roles, assuming $g'(x) = (x-\beta)^2(x-\alpha)$ , they obtain the system $\alpha+2\beta=13$ and $2\beta-\alpha=11$ , which yields perfectly clean integer traps ( $\alpha=1, \beta=6$ ) leading to the incorrect option 84.

The dynamic constraint of the NGT forces an iterative verification that consumes time, heavily penalizing superficial derivative checks.

Quick Facts About This Question

Question Information

Subject	Mathematics
Chapter	Three-Dimensional Geometry - Class 12
Concept	Collinearity vs Coplanarity
Difficulty	8
Cognitive Load	7
Target Solve Time	160 sec
Access	Free