Question

The number of triplets (x,y,z), where $x,y,zϵ[0,2\pi ],$ satisfying the inequality $(1+co{s}^{4}x)(2+co{t}^{2}y)(4+sin4z)\le 12co{s}^{2}x$ is

• A. 16
• B. 18
• C. 20
• D. 24

Answer 1

Answer: (D)

24

GIven,

$(1+{\cos }^{4}x)(2+{\cot }^{2}y)(4+\sin 4x)\le 12{\cos }^{2}x$

$\left(\frac{1+{\cos }^{4}x}{{\cos }^{2}x}\right)(2+{\cot }^{2}y)(4+\sin 4x)\le 12$

$\Rightarrow ({\cos }^{2}x+{\sec }^{2}x)(2+{\cot }^{2}y)(4+{\sin }^{4}x)\le 12$...........(1)

Since A.M $\ge$ G.M

$\Rightarrow ({\cos }^{2}x+{\sec }^{2}x)\ge 2$

$\Rightarrow (2+{\cot }^{2}y)\ge 3$

$\Rightarrow (4+\sin 4x)\ge 3$

$\Rightarrow ({\cos }^{2}x+{\sec }^{2}x)(2+{\cot }^{2}y)(4+\sin 4x)\ge 12$...........(2)

from (1) and (2) it says that any limiting is possible. So

$\therefore {\cos }^{2}x={\sec }^{2}x=1$

$\Rightarrow x=0$ or $x=\pi$ or $x=2\pi$ which gives 3 values.

for $2+{\cot }^{2}y$

$\Rightarrow {\cot }^{2}y=0$

$\Rightarrow y=\frac{\pi }{2}$ or $y=\frac{3\pi }{2}$ which gives 2 values.

for $4+\sin 4x$

$\Rightarrow \sin 4x=-1$

$\Rightarrow x=\frac{3x}{8}$ or $x=\frac{7x}{8}$ or $x=\frac{11x}{8}$ or $x=\frac{15x}{8}$

which gives 4 values

$\therefore$ total number of triples is $3\times 2\times 4=24$

Hence answer is $24$