Q90 / 150 · MHT CET 2025 (27 Apr Shift 2) · Mathematics

If $\mathrm{f}(x)= \begin{cases}-2 \sin x & , \quad x \leqslant-\frac{\pi}{2} \\ a \sin x+\mathrm{b} & , \quad \frac{-\pi}{2} < x < \frac{\pi}{2} \\ \cos x & , \quad x \geqslant \frac{\pi}{2}\end{cases}$ is continuous at $\mathrm{x}=\frac{-\pi}{2}$ and $x=\frac{\pi}{2}$, then the value of $2 a+\mathrm{b}$ is

  • Correct answer
    A
    $\left(0, \frac{-13}{5}, \frac{2}{5}\right)$
  • B
    $\left(0, \frac{13}{5}, \frac{2}{5}\right)$
  • C
    $\left(0, \frac{13}{5}, 2\right)$
  • D
    $\left(0, \frac{-13}{5},-2\right)$

Solution

For $f(x)$ to be continuous at $x = -\frac{\pi}{2}$ and $x = \frac{\pi}{2}$, the left and right limits must equal the function values at these points.

At $x = -\frac{\pi}{2}$, the left limit is $2$ and the right limit is $-a + b$, yielding $-a + b = 2$.

At $x = \frac{\pi}{2}$, the left limit is $a + b$ and the right limit is $0$, giving $a + b = 0$.

Solving this system, $2b = 2$ so $b = 1$, and $a = -1$.

Thus $2a + b = -2 + 1 = -1$.

Final answer: $\boxed{-1}$

Hence, the correct answer is option: 'A'.

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