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Question #40655

Difficulty: 8/10 | 🟢 Free

The Question

Let $f: \mathbb{R} \to \mathbb{R}$ be a function defined as: $f(x) = \begin{cases} \prod_{r=1}^6 \left(1 + \binom{6}{r} x\right) - 1, & x \ge 0 \\ \lambda |x| + \frac{\mu}{2} x^2, & x < 0 \end{cases}$ If $f(x)$ is twice differentiable at $x=0$ , then the value of $\mu + \lambda$ is:

Options

A.
2983
B.
3109
C.
1460
D.
1586

Check Your Answer

Correct Answer

Step-by-Step Solution

Explanation

How the Examiner Thinks

Monster

Ultra-high difficulty questions designed to test your limit, stamina, and deep conceptual understanding.

Concepts Behind This Question

Expansion of Products of Binomials|General Term Formulation|Polynomial Continuity/Differentiability Link|Differentiability vs Continuity

Why This Question Is Tricky

Common Mistakes

This question tests the expansion of a product of binomials to extract coefficients, combined with the calculus concepts of piecewise differentiability.

The primary trap lies in the absolute value function $|x|$ for $x < 0$ .

Students often mechanically differentiate $\lambda|x|$ as $\lambda$ or ignore the domain $x < 0$ , leading to $f'(0^-) = \lambda$ instead of $-\lambda$ , which yields the incorrect answer 3109.

A secondary trap is forgetting the $-1$ when calculating the sum of squares of binomial coefficients $\sum_{r=1}^n \binom{n}{r}^2 = \binom{2n}{n} - 1$ , or forgetting to multiply $S_2$ by 2 for the second derivative.

Quick Facts About This Question

Question Information

Subject	Mathematics
Chapter	Binomial Theorem - Class 11
Concept	Expansion of Products of Binomials
Difficulty	8
Cognitive Load	7
Target Solve Time	160 sec
Access	Free