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QJEEM19Maa3c4: JEE Main 2019 Mathematics - Class 12 JEEM19Maa3c4 Solution: If the function f ( x ) = { a

If the function f(x)={a|π−x|+1,x≤5b|x−π|+3,x>5
is continuous at x = 5, then the value of a – b is :-

Multiple Choice Options

  • A 2π−5{2 \over {\pi - 5 }}
  • B 25−π{2 \over {5 - \pi }}
  • C −2π+5{-2 \over {\pi + 5 }}
  • D 2π+5{2 \over {\pi + 5 }}

Step-by-Step Verified Solution

As f(x) is continuous at x = 5 then

lim⁡x→5−f(x)=f(5)=lim⁡x→5+f(x)\mathop {\lim }\limits_{x \to {5^ - }} f\left( x \right) = f\left( 5 \right) = \mathop {\lim }\limits_{x \to {5^ + }} f\left( x \right)

⇒\Rightarrow lim⁡h→0f(5−h)=f(5)=lim⁡h→0f(5+h)\mathop {\lim }\limits_{h \to 0} f\left( {5 - h} \right) = f\left( 5 \right) = \mathop {\lim }\limits_{h \to 0} f\left( {5 + h} \right)

⇒\Rightarrow lim⁡h→0(a∣π−(5−h)∣+1)=a∣π−5∣+1\mathop {\lim }\limits_{h \to 0} \left( {a\left| {\pi - \left( {5 - h} \right)} \right| + 1} \right) = a\left| {\pi - 5} \right| + 1

= lim⁡h→0(b∣5+h−π∣+3)\mathop {\lim }\limits_{h \to 0} \left( {b\left| {5 + h - \pi } \right| + 3} \right)

⇒\Rightarrow a∣π−5∣+1{a\left| {\pi - 5} \right| + 1} = a∣π−5∣+1a\left| {\pi - 5} \right| + 1

= b∣5−π∣+3{b\left| {5 - \pi } \right| + 3}

Note : As π\pi = 3.14, then π\pi - 5 = 3.14 - 5 = -1.84 < 0

∴\therefore ∣π−5∣{\left| {\pi - 5} \right|} = - ( π\pi - 5) and ∣5−π∣{\left| {5 - \pi } \right|} = (5 - π\pi )

⇒\Rightarrow −a(π−5)+1- a\left( {\pi - 5} \right) + 1 = −a(π−5)+1- a\left( {\pi - 5} \right) + 1 = b(5−π)+3{b\left( {5 - \pi } \right) + 3}

⇒\Rightarrow −a(π−5)+1- a\left( {\pi - 5} \right) + 1 = b(5−π)+3{b\left( {5 - \pi } \right) + 3}

⇒\Rightarrow (a−b)(5−π)=2\left( {a - b} \right)\left( {5 - \pi } \right) = 2

⇒\Rightarrow (a−b)=2(5−π)\left( {a - b} \right) = {2 \over {\left( {5 - \pi } \right)}}

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