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Limits Continuity and Differentiability question

2019 · 9 Apr · Shift 2 · Q27
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Limits Continuity and Differentiability question

2019 · 9 Apr · Shift 2 · Q27

JEE MainMathematicsLimits Continuity and DifferentiabilityMCQ+4 / −1
If the function f(x)={a∣π−x∣+1,x≤5b∣x−π∣+3,x>5f(x) = \left\{ {\begin{matrix} {a|\pi - x| + 1,x \le 5} \\ {b|x - \pi | + 3,x \gt 5} \\ \end{matrix} } \right.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 :-
  1. A
    2π−5{2 \over {\pi - 5 }}π−52​
  2. B
    25−π{2 \over {5 - \pi }}5−π2​
  3. C
    −2π+5{-2 \over {\pi + 5 }}π+5−2​
  4. D
    2π+5{2 \over {\pi + 5 }}π+52​
View written solutionFree

Correct answer: B

  1. Given function
f(x)={a∣π−x∣+1,x≤5b∣x−π∣+3,x>5f(x)= \begin{cases} a|\pi-x|+1, & x\le 5 \\ b|x-\pi|+3, & x>5 \end{cases}f(x)={a∣π−x∣+1,b∣x−π∣+3,​x≤5x>5​

We need continuity at x=5x=5x=5.

  1. Condition for continuity at x=5x=5x=5

For continuity,

LHL at x=5=f(5)=RHL at x=5.\text{LHL at }x=5 = f(5) = \text{RHL at }x=5.LHL at x=5=f(5)=RHL at x=5.

Since 5>π5>\pi5>π, we have

∣π−5∣=5−π,|\pi-5|=5-\pi,∣π−5∣=5−π,

and also

∣5−π∣=5−π.|5-\pi|=5-\pi.∣5−π∣=5−π.
  1. Compute left value / function value at x=5x=5x=5

Because x=5x=5x=5 falls in the first branch,

f(5)=a∣π−5∣+1=a(5−π)+1.f(5)=a|\pi-5|+1=a(5-\pi)+1.f(5)=a∣π−5∣+1=a(5−π)+1.
  1. Compute right-hand limit as x→5+x\to 5^+x→5+

Using the second branch,

lim⁡x→5+f(x)=b∣5−π∣+3=b(5−π)+3.\lim_{x\to 5^+}f(x)=b|5-\pi|+3=b(5-\pi)+3.x→5+lim​f(x)=b∣5−π∣+3=b(5−π)+3.
  1. Apply continuity
a(5−π)+1=b(5−π)+3.a(5-\pi)+1=b(5-\pi)+3.a(5−π)+1=b(5−π)+3.

So,

(a−b)(5−π)=2.(a-b)(5-\pi)=2.(a−b)(5−π)=2.

Hence,

a−b=25−π.a-b=\frac{2}{5-\pi}.a−b=5−π2​.
  1. Match with options

This is Option B.

a−b=25−π\boxed{a-b=\frac{2}{5-\pi}}a−b=5−π2​​
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