First slide

Trigonometric Identities

Question

If $S = \{x \in N : 2 + \log_{2} \sqrt{x + 1} > 1 - \log_{1 / 2} \sqrt{4 - x^{2}}\}$ , then

Moderate

A

s={1}
B

S=Z
C

S=N
D

none of these

Solution

$\begin{array}{l} 2 + \log_{2} \sqrt{x + 1} > 1 - \log_{1 / 2} \sqrt{4 - x^{2}} \\ \Rightarrow 1 + \log_{2} \sqrt{x + 1} - \log_{2} \sqrt{4 - x^{2}} > 0 \\ or \log_{2} 2 + \log_{2} \sqrt{x + 1} - \log_{2} \sqrt{4 - x^{2}} > 0 \\ or \log_{2} \frac{2 \sqrt{x + 1}}{\sqrt{4 - x^{2}}} > 0 or \frac{2 \sqrt{x + 1}}{\sqrt{4 - x^{2}}} > 1 \\ ∴ 4 (x + 1) > 4 - x^{2} \\ ∴ x^{2} + 4 x > 0 \\ x < - 4 or x > 0 - - - - (I) \end{array}$
$\begin{array}{l} Also, x + 1 > 0 or x > - 1 \\ and 4 - x^{2} > 0 \\ or x \in (- 2, 2) \\ ∴ x \in (0, 2) \\ \Rightarrow x = 1, as x \in N \end{array}$

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