JEE MAIN - Mathematics (2016 - 10th April Morning Slot - No. 6)
If x is a solution of the equation, $$\sqrt {2x + 1} $$ $$ - \sqrt {2x - 1} = 1,$$ $$\,\,\left( {x \ge {1 \over 2}} \right),$$ then $$\sqrt {4{x^2} - 1} $$ is equal to :
$${3 \over 4}$$
$${1 \over 2}$$
2
$$2\sqrt 2 $$
Explanation
Given,
$$\sqrt {2x + 1} - \sqrt {2x - 1} = 1$$
$$ \Rightarrow $$ $$\sqrt {2x + 1} = 1 + \sqrt {2x - 1} $$
Squaring both sides, we get
2x + 1 $$=$$ 1 + 2x $$-$$ 1 + 2$$\sqrt {2x - 1} $$
$$ \Rightarrow $$ 1 $$=$$ 2$$\sqrt {2x - 1} $$
$$ \Rightarrow $$ 1 $$=$$ 4(2x $$-$$ 1)
$$ \Rightarrow $$ 8x $$-$$ 4 $$=$$ 1
$$ \Rightarrow $$ x $$=$$ $${5 \over 8}$$
So, $$\sqrt {4{x^2} - 1} $$
$$ = \sqrt {4\left( {{{25} \over {64}}} \right) - 1} $$
$$ = \sqrt {{{36} \over {64}}} $$
$$ = {6 \over 8}$$
$$ = {3 \over 4}$$
$$\sqrt {2x + 1} - \sqrt {2x - 1} = 1$$
$$ \Rightarrow $$ $$\sqrt {2x + 1} = 1 + \sqrt {2x - 1} $$
Squaring both sides, we get
2x + 1 $$=$$ 1 + 2x $$-$$ 1 + 2$$\sqrt {2x - 1} $$
$$ \Rightarrow $$ 1 $$=$$ 2$$\sqrt {2x - 1} $$
$$ \Rightarrow $$ 1 $$=$$ 4(2x $$-$$ 1)
$$ \Rightarrow $$ 8x $$-$$ 4 $$=$$ 1
$$ \Rightarrow $$ x $$=$$ $${5 \over 8}$$
So, $$\sqrt {4{x^2} - 1} $$
$$ = \sqrt {4\left( {{{25} \over {64}}} \right) - 1} $$
$$ = \sqrt {{{36} \over {64}}} $$
$$ = {6 \over 8}$$
$$ = {3 \over 4}$$
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