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NIMCET 2025 Paper

Section: Mathematics

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Question : 2 of 120

Marks: +1, -0

The maximum value of

\sin(x) + \sin(x+1)

\cos \frac{1}{2}

. Then the value of

k

3
1
2
None of these

Solution:

Concept:

Use the sum-to-product identity to combine the sine terms.

Formula:

\sin A + \sin B = 2 \sin\left(\frac{A+B}{2}\right) \cos\left(\frac{A-B}{2}\right)

Solution:

\sin x + \sin(x+1) = 2 \sin\left(\frac{2x+1}{2}\right) \cos\left(\frac{1}{2}\right)

= 2 \sin\left(x + \frac12\right) \cos\frac12

Maximum value of

\sin\left(x+\frac12\right)

is

1

.

Hence maximum value of the sum is

2 \cos\frac12

.

Therefore, the constant

k

is

2

.

Answer:

C. 2

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