AP EAMCET · Maths · Continuity and Differentiability

Let \(f(x)=\left\{\begin{array}{cc}0, & x=0 \\ 2-x, & \text { for } 0 \lt x \lt 1 \\ 2, & \text { for } x=1 \\ \frac{1}{2}-x, & \text { for } 1 \lt x \lt 2 \\ \frac{-3}{2}, & \text { for } x \geq 2\end{array}\right.\)
then which of the following is true

A \(f\) is right continuous at \(x=0\)
B \(f\) is left continuous at \(x=1\)
C \(f\) is right continuous at \(x=1\)
D \(f\) is continuous at \(x=2\)

Verified Solution

Answer & Solution

Correct Answer

(D) \(f\) is continuous at \(x=2\)

Step-by-step Solution

Detailed explanation

For \(x=0\) \(\begin{aligned} & \lim _{x \rightarrow 0^{-}} f(x)=\text { does not exist as } f \text { is not defined for } x \lt 0 \\ & \lim _{x \rightarrow 0^{+}} f(x)=2, f(0)=0 \end{aligned}\) \(f\) is not continuous for \(x=0\) For…

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.

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Let \(f(x)=\left\{\begin{array}{cc}0, & x=0 \\ 2-x, & \text { for } 0 \lt x \lt 1 \\ 2, & \text { for } x=1 \\ \frac{1}{2}-x, & \text { for } 1 \lt x \lt 2 \\ \frac{-3}{2}, & \text { for } x \geq 2\end{array}\right.\)then which of the following is true

Step-by-step Solution

Let \(f(x)=\left\{\begin{array}{cc}0, & x=0 \\ 2-x, & \text { for } 0 \lt x \lt 1 \\ 2, & \text { for } x=1 \\ \frac{1}{2}-x, & \text { for } 1 \lt x \lt 2 \\ \frac{-3}{2}, & \text { for } x \geq 2\end{array}\right.\)
then which of the following is true