Let f: R R be a function given by: 1 - (2x)x^2, for x < 0 1 - (x)x, for x 0. If f is continuous at x = 0, then ^2 + ^2 is equal to:

Let f: R R be a function given by: 1 - (2x)x^2, for x < 0 1 - (x)… — Solution | JEE Mathematics

MCQMediumJEE 2024Continuity

Let $f : \mathbb{R} \to \mathbb{R}$ be a function given by:

$1 - \frac{\cos(2x)}{x^2}$ , for $x < 0$

$\beta \cdot \frac{\sqrt{1 - \cos(x)}}{x}$ , for $x > 0$ .

If $f$ is continuous at $x = 0$ , then $\alpha^2 + \beta^2$ is equal to:

A
$48$
B
$12$
C
$4$
D
$6$

Answer

Correct answer:B

Step-by-step solution

the solution unavailable for this question

Given: the solution is for a different question about differentiability at $x = 1$ and evaluation of $\int_{-2}^{2} f(x) \, dx$ .

Find: The value of $\alpha^2 + \beta^2$ for continuity at $x = 0$ .

The supplied solution content is unrelated to the asked question, so valid working could not be extracted from it. Therefore, the answer is taken from the provided correct answer field. The mapped correct option is B.

Common mistakes

Treating the solution as applicable to this question. It discusses a different piecewise function and a definite integral, so using it here is invalid. Always check that the solution matches the question before extracting results.
Ignoring one-sided continuity at $x = 0$ . For a piecewise function, continuity must be checked using the left-hand and right-hand limits at the joining point, not by inspecting only one branch.
Misreading expressions like $\frac{1-\cos x}{x}$ or $\frac{1-\cos(2x)}{x^2}$ near $x=0$ without using standard limits or trigonometric identities. Rewrite first, then evaluate the limit carefully.

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Answer

Step-by-step solution

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