Question : The value of $\frac{4.669 \times 4.669–9 \times(0.777)^2}{(4.669)^2+(2.331)^2+14(0.667)(2.331)}$ is $(1-k)$, where $k = $?
Option 1: 0.666
Option 2: 0.647
Option 3: 0.467
Option 4: 0.768
Correct Answer: 0.666
Solution :
Given: The expression is $\frac{4.669 \times 4.669–9 \times(0.777)^2}{(4.669)^2+(2.331)^2+14(0.667)(2.331)}=(1 - k)$
Use the algebraic identities, $(a+b)^2=a^2+b^2+2ab$ and $a^2-b^2=(a+b)(a-b)$
⇒ $\frac{4669 \times 4669–9 \times(777)^2}{(4669)^2+(2331)^2+14(0667)(2331)}=(1 - k)$
⇒ $\frac{(4669)^2–(2331)^2}{(4669)^2+(2331)^2+2(4669)(2331)}=(1 - k)$
⇒ $\frac{(4669)^2–(2331)^2}{(4669+2331)^2}=(1 - k)$
⇒ $\frac{(4669–2331)(4669+2331)}{(4669+2331)^2}=(1 - k)$
⇒ $\frac{(4669–2331)}{(4669+2331)}=(1 - k)$
⇒ $\frac{2338}{7000}=(1 - k)$
⇒ $0.334=1 - k$
⇒ $k=1 - 0.334$
⇒ $k= 0.666$
Hence, the correct answer is 0.666.
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