AP EAMCET · Maths · Continuity and Differentiability

If \(f(x)= \begin{cases}\frac{\cos a x-\cos b x}{x^2}, & x \neq 0 \\ \frac{1}{2}\left(a^2-b^2\right) & , x=0\end{cases}\)
where \(a, b\) are real and district constants, then

A \(f\) is discontinuous at \(x=0\)
B \(f\) is continuous at \(x=0\)
C \(\lim _{x \rightarrow 0} f(x)\) does not exist
D \(f(0)\) is not defined

Verified Solution

Answer & Solution

Correct Answer

(A) \(f\) is discontinuous at \(x=0\)

Step-by-step Solution

Detailed explanation

No solution. Refer to answer key.

Apply the relevant identity from the chapter and substitute the values established in the previous step, keeping the original constraint in view.

Use the simplified expression to isolate the unknown, then plug the result back into the question setup to confirm the working is internally consistent.

Cross-check against a boundary or limiting case. Both the dimensional balance and the qualitative behaviour at the extreme should agree with the candidate answer.

Combine the steps above to identify the correct option. The remaining choices each fail at least one of the consistency, dimensional, or boundary checks.

Same subject

Explore more questions on app

From AP EAMCET

Explore more questions on app

If \(f(x)= \begin{cases}\frac{\cos a x-\cos b x}{x^2}, & x \neq 0 \\ \frac{1}{2}\left(a^2-b^2\right) & , x=0\end{cases}\)where \(a, b\) are real and district constants, then

Step-by-step Solution

If \(f(x)= \begin{cases}\frac{\cos a x-\cos b x}{x^2}, & x \neq 0 \\ \frac{1}{2}\left(a^2-b^2\right) & , x=0\end{cases}\)
where \(a, b\) are real and district constants, then