JEE Mains Model Paper 2

f '(x )= 2(1/3)sin( x/ 6) cos( x/6) +(1/3) cos x/ 3-(1/3)
= (1/3) [2sin (x/6) cos (x/6) - 2sin2 (x/6)]
= (2/3) sin (x/6) (cos (x/6) - sin (x/6))
f'(x) = 0 ⇒ sin (x/6) = 0 or tan (x/6) = 1
⇒

x

6

= kπ , k ∊ I or

x

6

= nπ +

π

4

, n ∊ I
x2 - 10 < - 19.5x ⇒ (x+9.75)2 < 105.0625
⇒ (x - 0.5) (x + 20) < 0 ⇒ -20 < x < 0.5
So the critical points satisfying the last inequality will be 0 , -6π , -9π/2