If $$\sqrt {\left( {{a^2} + {b^2} + ab} \right)} + \sqrt {\left( {{a^2} + {b^2} - ab} \right)} = 1,$$ then what is the value of (1 - a2) (1 - b2) ?
A. $$\frac{1}{4}$$
B. $$\frac{4}{7}$$
C. $$\frac{5}{4}$$
D. $$\frac{3}{4}$$
Answer: Option D
Solution(By Examveda Team)
$$\eqalign{ & \sqrt {\left( {{a^2} + {b^2} + ab} \right)} + \sqrt {\left( {{a^2} + {b^2} - ab} \right)} = 1 \cr & {\text{Squaring both sides}} \cr & {a^2} + {b^2} + ab + {a^2} + {b^2} - ab + 2\sqrt {{{\left( {{a^2} + {b^2}} \right)}^2} - {{\left( {ab} \right)}^2}} = 1 \cr & \sqrt {{{\left( {{a^2} + {b^2}} \right)}^2} - {{\left( {ab} \right)}^2}} = \frac{1}{2} - \left( {{a^2} + {b^2}} \right) \cr & {\text{Again squaring both sides}} \cr & {a^4} + {b^4} + 2{a^2}{b^2} - {a^2}{b^2} = \frac{1}{4} + {\left( {{a^2} + {b^2}} \right)^2} - \left( {{a^2} + {b^2}} \right) \cr & {a^4} + {b^4} + {a^2}{b^2} = \frac{1}{4} + {a^4} + {b^4} + 2{a^2}{b^2} - {a^2} - {b^2} \cr & {a^2} + {b^2} - {a^2}{b^2} = \frac{1}{4}........\left( {\text{i}} \right) \cr & \Rightarrow \left( {1 - {a^2}} \right)\left( {1 - {b^2}} \right) \cr & = 1 - {a^2} - {b^2} + {a^2}{b^2} \cr & = 1 - \left[ {{a^2} + {b^2} - {a^2}{b^2}} \right] \cr & = 1 - \frac{1}{4} \cr & = \frac{3}{4} \cr & \cr & {\bf{Alternate:}} \cr & {\text{Let }}b = 0 \cr & \sqrt {\left( {{a^2} + {b^2} + ab} \right)} + \sqrt {\left( {{a^2} + {b^2} - ab} \right)} = 1 \cr & \sqrt {{a^2} + 0 + 0} + \sqrt {{a^2} + 0 + 0} = 1 \cr & a + a = 1 \cr & 2a = 1 \cr & a = \frac{1}{2} \cr & \left( {1 - {a^2}} \right)\left( {1 - {b^2}} \right) \cr & = \left( {1 - \frac{1}{4}} \right)\left( {1 - 0} \right) \cr & = \frac{3}{4} \cr} $$Related Questions on Algebra
If $$p \times q = p + q + \frac{p}{q}{\text{,}}$$ then the value of 8 × 2 is?
A. 6
B. 10
C. 14
D. 16
A. $$1 + \frac{1}{{x + 4}}$$
B. x + 4
C. $$\frac{1}{x}$$
D. $$\frac{{x + 4}}{x}$$
A. $$\frac{{20}}{{27}}$$
B. $$\frac{{27}}{{20}}$$
C. $$\frac{6}{8}$$
D. $$\frac{8}{6}$$
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