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NDA 2011
Exam
NDA
Year
2011
Subject
Mathematics
Chapter
Continuity and Differentiability

NDA 2011 · Mathematics · Continuity and Differentiability

What is the value of k for which the following function f(x) is...

What is the value of kk for which the following function f(x)f(x) is continuous for all xx ? f(x)=\left\{x3−3x+2(x−1)2 for x≠1k for x=1\begin{array}{cc} \frac{x^3-3 x+2}{(x-1)^2} & \text{ for } x \neq 1 \\ k & \text{ for } x=1 \end{array}\right.

Options

  1. A3
  2. B2
  3. C1
  4. D-1

Correct answer

A. 3

Step-by-step solution

Since the given function is continuous so that f(1+h)=f(1−h)=f(h) Now f(1+h)=lth→1(1+h)3−3(1+h)+2(1+h−1)2⇒lt⁡h→01+h3+3h2+3h−3−3h+2(h)2∴f(x)=3f(1+h)=f(1-h)=f(h)\text{ Now } \quad f(1+h)=lt_{h \rightarrow 1} \frac{(1+h)^3-3(1+h)+2}{(1+h-1)^2}\Rightarrow \quad \operatorname{lt}_{h \rightarrow 0} \frac{1+h^3+3 h^2+3 h-3-3 h+2}{(h)^2}\therefore \quad f(x)=3.
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