Question

The number of ordered pair $(x,y)$ such that $0\le x\le 2\pi$ and satisfying the inequality $2^{{\sec }^{2}x}\sqrt{y^2-2y+2}\le 2$ is

• A. $2$
• B. $3$
• C. $1$
• D. None of these

Answer 1

Answer: (A)

$2$

$se{c}^{2}x$ has a minimum vale of 1 at $x=0$ and $2\pi$

Hence $2^{se{c}^{2}x}$ has a minimum value of $2^1=2$

The minima for the polynomial P(y) $y^2-2y+2$ can be found by differentiation.

Since coefficient of $y^2$ is positive it must be an upward facing parabola and will give us minima on differentiation.

On differentiating

$2y-2=0$

$\Rightarrow y=1$

Minimum vale of P(y) $=$P(1)$=1$

Now P(y)$\ge 1$

$\Rightarrow \sqrt{P\left(y\right)}\ge 1$

$2^{se{c}^{2}x}\ge 2$

Hence $2^{se{c}^{2}x}\sqrt{P\left(y\right)}\ge 2\times 1$

Hence equation is satisfied only for equality which clearly occurs in only 2 situations, when x is 0 and y is 1 or when x is $2\pi$ and y is 1.

So there are only 2 ordered pairs.