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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
If the function $f\left( x \right) = \left\{ \begin{array}{l} a\,\left| {\pi  - x} \right|\, + 1,\,\,x \le 5\,\\ b\,\,\left| {\pi  - x} \right|\, + 3,\,\,x > 5\,\, \end{array} \right.$ is continuous at $x = 5$, then the value of $a -b$ is
  • A
    $\frac{2}{{5 - \pi }}$
  • B
    $\frac{2}{{\pi  - 5}}$
  • C
    $\frac{2}{{\pi  + 5}}$
  • D
    $\frac{-2}{{\pi  + 5}}$
✓

Answer

$f\left( x \right) = \left\{ \begin{array}{l} a\left| {\pi  - x} \right| + 1,x \le 5\\ b\left| {\pi  - x} \right| + 3,x > 5 \end{array} \right.$Contributes at $x=5$
$\therefore \text{L.H.L. = R.H.L} = f\left( 5 \right)$
$ \Rightarrow b\left| {\pi  - 5} \right| + 3 = a\left| {\pi  - 5} \right| + 1$
$ \Rightarrow  - b\left( {\pi  - 5} \right) + 3 =  - a\left( {5 - \pi } \right) + 1$
$ \Rightarrow \left( {a - b} \right)\left( {\pi  - 5} \right) =  - 2$
$ \Rightarrow a - b = \frac{{ - 2}}{{\pi  - 5}} = \frac{2}{{5 - \pi }}$

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