Question

least value of the function $f\left(x\right)=4{\cos }^{2}x+3{\sec }^{2}x$ is equal to:

• A. $4$
• B. $4\sqrt{3}$
• C. $2\sqrt{3}$
• D. $6$

Answer 1

Answer: (B)

$4\sqrt{3}$

$f\left(x\right)=4{\cos }^{2}x+2{\sec }^{2}x$

As we know that,

A.M. of two no. a and b $=\frac{a+b}{2}$

G.M. of two no. a and b $=\sqrt{ab}$

Let $4{\cos }^{2}x\&3{\sec }^{2}x$ be two numbers.

Since it is clear that $4{\cos }^{2}x\&3{\sec }^{2}x$ are positive numbers.

$\therefore A.M.=\frac{4{\cos }^{2}x+3{\sec }^{2}x}{2}$

$G.M.=\sqrt{4{\cos }^{2}x.3{\sec }^{2}x}$

$\therefore$ By the theorem of arithmetic and geometric mean, i.e.,

A.M. $\ge$ G.M.

$\frac{4{\cos }^{2}x+3{\sec }^{2}x}{2}\ge \sqrt{4{\cos }^{2}x.3{\sec }^{2}x}$

$\Rightarrow 4{\cos }^{2}x+3{\sec }^{2}x\ge 2\sqrt{12{\cos }^{2}x\left(\frac{1}{{\cos }^{2}x}\right)}$

$\Rightarrow 4{\cos }^{2}x+3{\sec }^{2}x\ge 2\sqrt{12}$

$\Rightarrow 4{\cos }^{2}x+3{\sec }^{2}x\ge 4\sqrt{3}$

$\Rightarrow f\left(x\right)\ge 4\sqrt{3}$

Hence, the least value of the function $f\left(x\right)=4{\cos }^{2}x+2{\sec }^{2}x$ is equal to $4\sqrt{3}$.