AP EAMCET · Maths · Continuity and Differentiability
Define \(f(x)=\left\{\begin{array}{cc}x^2+b x+c, & x < 1 \\ x, & x \geq 1\end{array}\right.\). If \(f(x)\) is differentiable at \(x=1\), then \((b-c)\) is equal to
- A \(-2\)
- B \(0\)
- C \(1\)
- D \(2\)
Answer & Solution
Correct Answer
(A) \(-2\)
Step-by-step Solution
Detailed explanation
Given function, \(\begin{aligned} & f(x)=\left\{\begin{array}{cc} x^2+b x+c, & x < 1 \\ x, & x \geq 1 \end{array}\right. \\ & f^{\prime}(x)=\left\{\begin{array}{cc} 2 x+b, & x < 1 \\ 1, & x \geq 1 \end{array}\right. \end{aligned}\) Since, \(f(x)\) is differentiable at \(x=1\).…
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