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BITSAT 2012
Exam
BITSAT
Year
2012
Subject
Mathematics
Chapter
Continuity and Differentiability

BITSAT 2012 · Mathematics · Continuity and Differentiability

If the function f(x)= cases A x-B, & x ≤ 1 3 x & , 1< x< 2 B x^ 2...

If the function f(x)=f(x)=\begin{cases}A x-B, & x \leq 1 \\ 3 x & , 1< x< 2 \\ B x^{2}-A, & x \geq 2\end{cases}$iscontinuousatis continuous atx=2andand4,thenthevaluesofaandthen the values of a andb$ are

Options

  1. A3,5
  2. B3,-5
  3. C0,3
  4. D0,5

Correct answer

B. 3,-5

Step-by-step solution

Since f(x)\mathrm{f}(\mathrm{x}) is continuous at x=2∴f(2)=lim⁡x→2+f(x)⇒1=lim⁡x→2+(ax+b)∴1=2a+b\mathrm{x}=2\therefore \mathrm{f}(2)=\lim _{\mathrm{x} \rightarrow 2^{+}} \mathrm{f}(\mathrm{x}) \Rightarrow 1=\lim _{\mathrm{x} \rightarrow 2^{+}}(\mathrm{ax}+\mathrm{b})\therefore 1=2 \mathrm{a}+\mathrm{b} Again f(x)\mathrm{f}(\mathrm{x}) is continuous at x=4∴f(4)=lim⁡x→4−f(x)⇒7=lim⁡x→4−(ax+b)∴7=4a+b\mathrm{x}=4\therefore \mathrm{f}(4)=\lim _{x \rightarrow 4^{-}} \mathrm{f}(\mathrm{x}) \Rightarrow 7=\lim _{\mathrm{x} \rightarrow 4^{-}}(\mathrm{ax}+\mathrm{b})\therefore 7=4 \mathrm{a}+\mathrm{b} Solving (1)and(2),and (2), we get a=3,b=−5a=3, b=-5
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