If a is minimum value of ^2 - + 1 2 and b = _ x ( ( x + 1 ) ( x +… - Vidyadip

MCQ

If $a$ is minimum value of ${\sin ^2}\theta - \sin \theta + \frac{1}{2}$ and $b = \mathop {\lim }\limits_{x \to \infty } \left( {\sqrt {\left( {x + 1} \right)\left( {x + 2} \right)} - x} \right)$ then $\left| {2a + b} \right| = $

A
$3$
B
$-2$
C
$4$
✓
$2$

✓

Answer

Correct option: D.

$2$

d
${a_{\sin }} = {\sin ^2}\theta - \sin \theta + \frac{1}{2}$

$=\left(\sin \theta-\frac{1}{2}\right)^{2}+\frac{1}{4}$

$a_{1}=\frac{1}{4}$

$b = \mathop {\lim }\limits_{x \to \infty } \left( {\sqrt {(x + 1)(x + 2)} - x} \right)$

$ = \mathop {\lim }\limits_{x \to \infty } \frac{{3X + 2}}{{\sqrt {(x + 1)(x + 2)} + x}} = \frac{3}{2}$

So, $|2 a+b|=\left|\frac{1}{2}+\frac{3}{2}\right|=2$

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