AP EAMCET Engineering 2018 Apr 23 Shift 2 Paper

Consider the function,
f(x)=cos‌x−sin‌2‌x
Is a continuous function in the interval [−

π

2

,

π

2

] anddifferentiable in interval (−

π

2

,

π

2

) .
Now according to Lagrange mean value theorem,
c∈(−

π

2

,

π

2

)
Such that,
f′(c)=

f( π 2 )−f(− π 2 )

π 2 −(− π 2 )

This implies,
−sinc−2‌cos(2c)=0
sinc+2−4sin2c=0
4sin2c−sinc−2=0
So,
sinc=

1±√1+32

8

=

1±√33

8

c=sin−1(

1±√33

8

)