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Let f: R\rightarrow R be defined as f(x)= \begin{Bmatrix} 2 \sin- \left ( \frac{\pi x}{2} \right ) &if &x < -1 \\ \left | ax^2 + x +b \right | &if &-1 \leq x\leq \leq 1 \\ \sin (\pi x) &if & x>1 \end{Bmatrix} If f(x) is continous on R, then a +b equals:
Option: 1 -1
Option: 2 -3
Option: 3 1
Option: 4 3

Answers (1)

best_answer

\begin{aligned} &f(\mathrm{x}) \text { is continuous on } \mathbb{R}\\ &\Rightarrow f\left(1^{-}\right)=f(1)=f\left(1^{+}\right)\\ &|\mathrm{a}+1+\mathrm{b}|=\lim _{\mathrm{x} \rightarrow 1} \sin (\pi \mathrm{x}) \end{aligned}

\\|\mathrm{a}+1+\mathrm{b}|=0 \Rightarrow \mathrm{a}+\mathrm{b}=-1 \;\;\;\;\;\;\;\;\;\;\;\;\;\;\ldots(1)\\ \Rightarrow \text { Also } f\left(-1^{-}\right)=f(-1)=f\left(-1^{+}\right) \\ \lim _{x \rightarrow-1} 2 \sin \left(\frac{-\pi \mathrm{x}}{2}\right)=|\mathrm{a}-1+\mathrm{b}|=2

Either a– 1 + b = 2 or a – 1 + b = –2

a + b = 3     ...(2) or a + b = –1     ...(3) 

from (1) and (2)

a + b = 3 = – 1(reject)

from (1) and (3)  

a + b = –1

Posted by

himanshu.meshram

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