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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
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
  • ✓
    $\frac{2}{{5 - \pi }}$
  • B
    $\frac{2}{{\pi  - 5}}$
  • C
    $\frac{2}{{\pi  + 5}}$
  • D
    $\frac{-2}{{\pi  + 5}}$
✓

Answer

Correct option: A.
$\frac{2}{{5 - \pi }}$
a
$f\left( x \right) = \left\{ \begin{array}{l}
a\left| {\pi  - x} \right| + 1,x \le 5\\
b\left| {\pi  - x} \right| + 3,x > 5
\end{array} \right.$

Contributes at $x=5$

$\therefore L.H.L. = R.H.L = f\left( 5 \right)$

$ \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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