IPMAT Question Paper 2019 | IPM Indore Quants

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Question 36 : Given that cos x + cos y = 1, the range of sin x - sin y is

1. [-1, 1]
2. [-2, 2]
3. [0, √3]
4. [-√3, √3]

Explanatory Answer

Let’s say sin x – sin y = k. So we intend to find the range of k.
We have cos x + cos y = 1

So, k² + 1² = (sin x - sin y )² + (cos x + cosy )²
k² + 1² = (sin x)² + (sin y)² – 2(sin x)(sin y) + (cos x)² + (cos y)² + 2(cos x)(cos y)
k² + 1²= sin² x + cos² x + sin² y + cos² y + 2(cos x)(cos y) – 2(sin x)(sin y)
k² + 1² = 1 + 1 + 2(cos x)(cos y) – 2(sin x)(sin y) {Because, sin² x + cos² x = 1}
k² + 1²= 2 + 2[(cos x)(cos y) – (sin x)(sin y)] {Because, (cos x)(cos y) – (sin x)(sin y) = cos(x + y)}
k² = 1 + 2(cos(x + y))

The maximum value that a cos function can take is 1, and the minimum value is -1.

Therefore the maximum value that k² can take is 1+2(1) = 3,
and the minimum value that k² can take is 1+2(-1) = -1,
This is impossible, k² is always non-negative.
The minimum value that k² can take is 0.

Therefore (k²)_max = 3 and (k²)_min = 0.

To get the range of k, we concentrate on (k²)_max.

(k²)_max = 3, implies that the extreme values of k are:

k = +√3 or -√3
So, the values of k range between +√3 and -√3, both +√3 and -√3 inclusive.

k = sin x – sin y, Remember!!

Therefore, Range of sin x – sin y = [-√3, √3]

The question is "Given that cos x + cos y = 1, the range of sin x - sin y is"

Hence, the answer is [-√3, √3]

Choice D is the correct answer

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