Answer (Detailed Solution Below)

Option 2 : 7
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Detailed Solution

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Calculation:

Let f(θ) = 7cos2θ + 5sin2θ 

= 2cos2θ + 5cos2θ + 5sin2θ 

= 2cos2θ + 5(cos2θ + sin2θ)

= 2cos2θ + 5                        [∵ cos2θ + sin2θ = 1]

As we know, 0 ≤ cos2θ ≤ 1

⇒  0 ≤ 2cos2θ ≤ 2

⇒  0 + 5 ≤ 2cos2θ + 5 ≤ 2 + 5

⇒ 5 ≤ 2cos2θ + 5 ≤ 7

⇒ 5 ≤ f(θ) ≤ 7

So, Maximum value of function is 7.

 

Alternate Method

Concept:

If acos2θ + bsin2θ  where, a > b

a will be the maximum value and b will be the minimum value

Calculation:

7cos2θ + 5sin2θ 

According to concept 

acos2θ + bsin2θ  where, a > b

On comparing

⇒ a = 7, b = 5 where, 7 > 5

∴ maximum value of 7cos2θ + 5sin2θ is 7

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