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MHTCET 22 Apr 2025 Shift 1 Paper

Question : 120 of 150

Marks: +1, -0

If the function

f(x)=\left\{\begin{array}{ccc}x+a \sqrt{2} \sin x & \text { if } & 0 \leq x \leq \frac{\pi}{4} \\ 2 x \cot x+b & \text { if } & \frac{\pi}{4} < x \leq \frac{\pi}{2} \\ a \cos 2 x-b \sin x & \text { if } & \frac{\pi}{2} < x \leq \pi\end{array}\right.

[0, \pi]

a-b=

Solution:

Concept:

For a function to be continuous at a point, its left-hand limit, right-hand limit, and the value of the function at that point must all be equal.

Explanation:

Since

f(x)

is continuous in

[0, \pi]

, it is continuous at the junction points

x = \frac{\pi}{4}

and

x = \frac{\pi}{2}

x = \frac{\pi}{4}

\lim_{x \to \frac{\pi}{4}^-} (x + a\sqrt{2}\sin x) = \lim_{x \to \frac{\pi}{4}^+} (2x\cot x + b)

\Rightarrow \frac{\pi}{4} + a\sqrt{2}\left(\frac{1}{\sqrt{2}}\right) = 2\left(\frac{\pi}{4}\right)(1) + b

\Rightarrow \frac{\pi}{4} + a = \frac{\pi}{2} + b

\Rightarrow a - b = \frac{\pi}{2} - \frac{\pi}{4} = \frac{\pi}{4}

To verify consistency at

x = \frac{\pi}{2}

b = -a - b \Rightarrow a + 2b = 0

Solving

a - b = \frac{\pi}{4}

and

a + 2b = 0

gives

a = \frac{\pi}{6}

b = -\frac{\pi}{12}

Hence

a - b = \frac{\pi}{6} - \left(-\frac{\pi}{12}\right) = \frac{\pi}{4}

, confirming the result.

Answer:

a - b = \frac{\pi}{4}

Correct option: A.

\frac{\pi}{4}

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