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STD 12 - 5. continuity and differentiation — Mathematics STD 12 Science — Question

CBSE BoardEnglish MediumSTD 12 ScienceMathematicsSTD 12 - 5. continuity and differentiation1 Mark
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
    $a=b+\frac{2}{5}$
  • ✓
    $a=b+\frac{2}{3}$
✓

Answer

Correct option: D.
$a=b+\frac{2}{3}$
d
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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