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NTA Abhyas 2020
Exam
NTA Abhyas
Year
2020
Subject
Mathematics
Chapter
Continuity and Differentiability

NTA Abhyas 2020 · Mathematics · Continuity and Differentiability

Let f (x) = \ cases a & ; x = (π)/(2) 2x - π 9 + 2x - π - b & ; x >...

Let ff\left(x\right) = \left\{{a;x=π22x−π9+2x−π−b;x>π2\begin{cases}a & ; x = \dfrac{\pi}{2} \\ \dfrac{\sqrt{2x - \pi}}{\sqrt{9 + \sqrt{2x - \pi}} - b} & ; x > \dfrac{\pi}{2}\end{cases}\right). If f(x)f\left(x\right) is continuous at x=π2x = \dfrac{\pi}{2}, then the value of a25b\dfrac{{a}^{2}}{5b} is

Correct answer

2.40

Step-by-step solution

lim⁡x→π2+f(x)=f(π2)\lim_{x \to {\dfrac{\pi}{2}}^{+}}f\left(x\right) = f\left(\dfrac{\pi}{2}\right)
⇒lim⁡x→π2+2x−π9+2x−π−b=a⇒ \lim_{x \to {\dfrac{\pi}{2}}^{+}}\dfrac{\sqrt{2x - \pi}}{\sqrt{9 + \sqrt{2x - \pi}} - b} = a
For the denominator to be zero, b=3b = 3
Now, on rationalization
lim⁡x→π2+2x−π(9+2x−π+3)(9+2x−π−9)=a\lim_{x \to {\dfrac{\pi}{2}}^{+}}\dfrac{\sqrt{2x - \pi}\left(\sqrt{9 + \sqrt{2x - \pi}} + 3\right)}{\left(\sqrt{9 + \sqrt{2x - \pi}} - 9\right)} = a
∴a=6∴ a = 6
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