If `A""=""sin^2x""+""cos^4x` , then for all real x : (1) `3/4lt=Alt=1` (2) `(13)/(16)lt=Alt=1` (3) `1lt=Alt=2` (4) `3/4lt=Alt=(13)/(16)`

Given A=sin^2theta+cos^4theta, then for all real theta, (a) 1lt=Alt=2 (b) 3/4lt=Alt=1 (c) (13)/(16)lt=Alt=1 (d) 3/4lt=Alt=(13)/(16)

if A=sin^2x-4sinx+2 then, for x in R a. -1lt=Alt=8 b. -1lt=Alt=7 c. -1lt=Alt=9 d. None of these

3(1 - x ) lt 2 (x + 4)

The equation sin^4x+cos^4x+sin2x+alpha=0 is solvable for -5/2lt=alphalt=1/2 (b) -3lt=alpha<1 -3/2lt=alphalt=1/2 (d) -1lt=alphalt=1

The equation sin^4x+cos^4x+sin2x+alpha=0 is solvable for -5/2lt=alphalt=1/2 (b) -3lt=alpha<1 -3/2lt=alphalt=1/2 (d) -1lt=alphalt=1

Solve -11lt=4x-3lt=13

Solve: 5lt=(2-3x)/4lt=9

If y=(sin^4x-cos^4x+sin^2xcos^2x)/(sin^4x+cos^4x+sin^2xcos^2x),x in (0,pi/2) , then (a) -3/2lt=ylt=1/2 (b) 1lt=ylt=1/2 (c) -5/3lt=ylt=1 (d) none of these

Solve the inequalities : -3lt=4-(7x)/2lt=18

The equation sin^4x-2cos^2x+a^2=0 can be solved if (a) -sqrt(3)lt=alt=sqrt(3) (b) sqrt(2)lt=alt=""sqrt(2) (c) -""1lt=alt=a (d) none of these