JEE MAIN - Mathematics (2021 - 26th February Evening Shift - No. 11)
Let f : R $$ \to $$ R be defined as
$$f(x) = \left\{ \matrix{ 2\sin \left( { - {{\pi x} \over 2}} \right),if\,x < - 1 \hfill \cr |a{x^2} + x + b|,\,if - 1 \le x \le 1 \hfill \cr \sin (\pi x),\,if\,x > 1 \hfill \cr} \right.$$ If f(x) is continuous on R, then a + b equals :
$$f(x) = \left\{ \matrix{ 2\sin \left( { - {{\pi x} \over 2}} \right),if\,x < - 1 \hfill \cr |a{x^2} + x + b|,\,if - 1 \le x \le 1 \hfill \cr \sin (\pi x),\,if\,x > 1 \hfill \cr} \right.$$ If f(x) is continuous on R, then a + b equals :
$$-$$3
3
$$-$$1
1
Explanation
$$f( - {1^ - }) = 2$$
$$f( - {1^ + }) = |a + b - 1|$$
$$|a + b - 1|\, = 2$$ ... (i)
$$f({1^ - }) = |a + b + 1|$$
$$f({1^ + }) = 0$$
$$|a + b + 1| = 0 \Rightarrow a + b + 1 = 0$$
$$ \Rightarrow a + b = - 1$$ .... (ii)
$$f( - {1^ + }) = |a + b - 1|$$
$$|a + b - 1|\, = 2$$ ... (i)
$$f({1^ - }) = |a + b + 1|$$
$$f({1^ + }) = 0$$
$$|a + b + 1| = 0 \Rightarrow a + b + 1 = 0$$
$$ \Rightarrow a + b = - 1$$ .... (ii)
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