If the function $f(x)=\begin{cases} 1+\sin \left(\frac{\pi x}{2}\right), & \text{for } -\infty < x \leq 1 \\…

If the function $f(x)=\begin{cases} 1+\sin \left(\frac{\pi x}{2}\right), & \text{for } -\infty < x \leq 1 \\ a x+b, & for 1 < x < 3 \\ 6 \tan \left(\frac{x \pi}{12}\right), & \text{for } 3 \leq x < 6 \end{cases}$ is continuous in the interval $(-\infty, 6)$, then the values of $a$ and $b$ are respectively $f(x)=\begin{cases} 1+\sin \left(\frac{\pi x}{2}\right), & \text{for } -\infty < x \leq 1 \\ a x+b, & \text{for } 1 < x < 3 \\ 6 \tan \left(\frac{x \pi}{12}\right), & \text{for } 3 \leq x < 6 \end{cases}$ The possible values for the ordered pair $(a,b)$ is
  1. 0,2
  2. 1,1
  3. 2,0
  4. 2,1

Solution

n function is continuous at all point in \((-\infty, 6)\) and at \(x=1, x=3\) function is continuous. If function \(f(x)\) is continuous at \(x=1\), then $\(\begin{array} \lim _{x \rightarrow 1-} f(x)=\lim _{x \rightarrow 1+} f(x) \Rightarrow 1+\sin \frac{\pi}{2}=a+b \\ \Rightarrow a+b=2 \end{array}\)$ If at \(x=3\), function is continuous, then \(\begin{array} \lim _{x \rightarrow 3^{-}} f(3)=\lim _{x \rightarrow 3^{+}} f(x) \Rightarrow 3 a+b=6 \tan \frac{3 \pi}{12} \\ \Rightarrow 3 a+b=6 \end{array}\) From Eqs. (1) and (2), a = 2, b = 0

Asked in: MHT CET 2021 (21 Sep Shift 1)

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