AP EAMCET Engineering 21 Apr 2019 Shift 1 Paper

Consider the given function.
f(x)=tan(x+‌

2π

3

)−tan(x+‌

π

6

)+cos(x+‌

π

6

)
So,
f(x)=‌

sin‌ π 2

cos(x+‌ 2π 3 )‌cos(x+‌ π 6 )

+cos(x+‌

π

6

)
=‌

1×2

2‌cos(x+‌ 2π 3 )‌cos(x+‌ π 6 )

+cos(x+‌

π

6

)
=cos(x+30∘)−‌

2

cos(2x−30∘)

Differentiate the above function with respect to x.
f′(x)=−sin(x+30∘)−‌

4‌sin(2x−30∘)

cos2(2x−30∘)

f′(x)>0 for all −75∘<x<−60∘
fmax at x=−60∘

Find the value of the function.

f(−60∘)‌‌=−tan(60∘)−tan(−30∘)+cos(−30∘)

‌‌=√3+‌

√3

2

+‌

√3

2

‌‌=‌

6√3+2√3+3√3

6

‌‌=‌

11√3

6