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STD 12 - 5. continuity and differentiation — JEE STD 11 Science — Question

Gujarat BoardEnglish MediumSTD 11 ScienceJEESTD 12 - 5. continuity and differentiation4 Marks
MCQ
Find the relationship be $a$ and $b$ so that the function $f$ defined by $f(x) = \left\{ {\begin{array}{*{20}{l}}
{ax + 1,{\rm{ if }}\,x\, \le \,3}\\
{bx + 3,{\rm{ if }}\,x\, > \,3}
\end{array}} \right.$ is continous at $x=3.$ 
  • A
    $a=b+\frac{1}{3}$
  • B
    $a=b-\frac{2}{3}$
  • C
    $a=b+\frac{2}{5}$
  • D
    $a=b+\frac{2}{3}$
✓

Answer

The given function $f$ is $f(x)=\left\{\begin{array}{l} a x+1, \text { if } x \leq 3 \\ b x+3, \text { if } x>3
\end{array}\right.$If $f$ is continuous at $x=3,$ then
$\mathop {\lim }\limits_{x \to {3^ - }} f(x) = \mathop {\lim }\limits_{x \to {3^ - }} f(x) = f(3)............. (1)$
Also, $\mathop {\lim }\limits_{x \to {3^ - }} f(x) = \mathop {\lim }\limits_{x \to {3^ - }} f(ax + 1) = 3a + 1$
$\mathop {\lim }\limits_{x \to {3^ + }} f(x) = \mathop {\lim }\limits_{x \to {3^ + }} f(bx + 1) = 3b + 3$
$f(3)=3 a+1$
Therefore, from $(1),$ we obtain
$3 a+1=3 b+3=3 a+1$
$\Rightarrow 3 a+1=3 b+3$
$\Rightarrow 3 a=3 b+2$
$\Rightarrow a=b+\frac{2}{3}$
Therefore, the required relationship is given by, $a=b+\frac{2}{3}$

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