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CUET · MATHS · PYQ PAPER 2025

For what value of \(\alpha\), the function \(f\) defined by \(f(x)=\left\{\begin{array}{ll}\alpha\left(x^2-2 x+1\right), & \text { if } x \leq 0 \\ 2 x+1, & \text { if } x>0\end{array}\right.\) is continuous at \(x=0\) ?

  1. A \(\alpha=1\)
  2. B \(\alpha=2\)
  3. C \(\alpha=-1\)
  4. D \(\alpha=0\)
Verified Solution

Answer & Solution

Correct Answer

(A) \(\alpha=1\)

Step-by-step Solution

Detailed explanation

\(\lim_{x \to 0^-} f(x) = \lim_{x \to 0^+} f(x)\) \(\alpha(0^2 - 2(0) + 1) = 2(0) + 1\) \(\alpha(1) = 1\) \(\alpha = 1\)