MCQ 11 Mark
Statement-1 (A): For $0 \leq \theta < 90^{\circ}, \sec x+\cos x \geq 2$.
Statement-2 (R): For any $x > 0, x+\frac{1}{x} \geq 2$.
Statement-2 (R): For any $x > 0, x+\frac{1}{x} \geq 2$.
- ✓Statement-1 and Statement-2 are True; Statement-2 is a correct explanation for Statement-1.
- BStatement-1 and Statement-2 are True; Statement-2 is not a correct explanation for Statement-1.
- CStatement-1 is True, Statement-2 is False.
- DStatement-1 is False, Statement-2 is True.
Answer
View full question & answer→Correct option: A.
Statement-1 and Statement-2 are True; Statement-2 is a correct explanation for Statement-1.
(A)Statement-1 and Statement-2 are True; Statement-2 is a correct explanation for Statement-1.
For any $x > 0$, we find that
$\left(\sqrt{x}-\frac{1}{\sqrt{x}}\right)^2 \geq 0 \Rightarrow x+\frac{1}{x}-2 \geq 0 \Rightarrow x+\frac{1}{2} \geq 2$
So, statement 2 is true. Since, $\sec x=\frac{1}{\cos x}$. Therefore,
$\sec x+\cos x=\cos x+\frac{1}{\cos x} \geq 2$
So, statement-1 is also true and statement-2 is the correct explanation for statement-1. Hence, option (a) is correct.
For any $x > 0$, we find that
$\left(\sqrt{x}-\frac{1}{\sqrt{x}}\right)^2 \geq 0 \Rightarrow x+\frac{1}{x}-2 \geq 0 \Rightarrow x+\frac{1}{2} \geq 2$
So, statement 2 is true. Since, $\sec x=\frac{1}{\cos x}$. Therefore,
$\sec x+\cos x=\cos x+\frac{1}{\cos x} \geq 2$
So, statement-1 is also true and statement-2 is the correct explanation for statement-1. Hence, option (a) is correct.
