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BITSAT 2010
Exam
BITSAT
Year
2010
Subject
Mathematics
Chapter
Application of Derivatives

BITSAT 2010 · Mathematics · Application of Derivatives

Let (f(x)= (\ array l a x²+1, x > 1 x+a, x ≤ 1 array .). Then f(x)...

Let (\mathrm{f}(\mathrm{x})=(\left\${ax2+1,x>1x+a,x≤1\begin{array}{l}a x²+1, x > 1 \\ x+a, x \leq 1\end{array}\right.). Thenf(x)isderivableatis derivable atx=1,if, if

Options

  1. Aa=2a=2
  2. Ba=1a=1
  3. Ca=0a=0
  4. Da=1/2a=1 / 2

Correct answer

D. a=1/2a=1 / 2

Step-by-step solution

$\begin{aligned}& \operatorname{Lf}^{\prime}(1)=\lim _{h \rightarrow 0} \frac{f(1-h)-f(1)}{-h}\\ & =\lim _{h \rightarrow 0} \frac{(1-h+a)-(1+a)}{-h}=\lim _{h \rightarrow 0} \frac{-h}{-h}=1 \\ & \operatorname{Rf}^{\prime}(1)=\lim _{h \rightarrow 0} \frac{f(1+h)-f(1)}{h}\\ & =\lim _{h \rightarrow 0} \frac{\left[a(1+h)^2+1\right]-(1+a)}{h}\\ & =\lim _{h \rightarrow 0}( a h+2 a)=2 a \end{aligned}$) Since f' (1) exists, (\therefore L f^{\prime}(1)=R f^{\prime}(1) \Rightarrow a=1 / 2)
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