Let fx= a ; x=(π)/2 √(2x-π)√(9+2x-π)-b ; x(π)/2 . If fx is continuous at x=(π)/2, then the value of (a^2)/5b is

Question

Let fx= a ; x=(π)/2 √(2x-π)√(9+2x-π)-b ; x>(π)/2 . If fx is continuous at x=(π)/2, then the value of (a^2)/5b is

MARKS Content Team · Accepted Answer

Correct answer: 2.40. limx→π^(2+)fx=f(π)/2 ⇒limx→π^(2+)(√(2x-π))/(√(9+2x-π)-b)=a For the denominator to be zero, b=3 Now, on rationalization limx→π^(2+)(√(2x-π)√(9+2x-π)+3)/(√(9+2x-π)-9)=a ∴a=6

Let $f (x) = \{\begin{cases} a & ; x = \frac{π}{2} \\ \frac{\sqrt{2 x - π}}{\sqrt{9 + \sqrt{2 x - π}} - b} & ; x > \frac{π}{2} \end{cases}$ . If $f (x)$ is continuous at $x = \frac{π}{2}$ , then the value of $\frac{a^{2}}{5 b}$ is

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More Continuity and Differentiability Questions

Let fx=a; x=π22x-π9+2x-π-b; x>π2. If fx is continuous at x=π2, then the value of a25b is

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More Continuity and Differentiability Questions

Let $f (x) = \{\begin{cases} a & ; x = \frac{π}{2} \\ \frac{\sqrt{2 x - π}}{\sqrt{9 + \sqrt{2 x - π}} - b} & ; x > \frac{π}{2} \end{cases}$ . If $f (x)$ is continuous at $x = \frac{π}{2}$ , then the value of $\frac{a^{2}}{5 b}$ is