Question : If $a^2+b^2+c^2+84 = 4(a - 2b + 4c)$, then $\sqrt{ab - bc + ca}$ is equal to:
Option 1: $4 \sqrt{10}$
Option 2: $\sqrt{10}$
Option 3: $5\sqrt{10}$
Option 4: $2\sqrt{10}$
Correct Answer: $2\sqrt{10}$
Solution : Given: $a^2+b^2+c^2+84 = 4(a - 2b + 4c)$ ⇒ $a^2+ b^2+ c^2− 4a + 8b − 16c + 84 = 0$ ⇒ $(a^2− 4a + 4) + (b^2+ 8b + 16) + (c^2− 16c + 64) = 0$ ⇒ $(a − 2)^2+(b + 4)^2+ (c − 8)^2= 0$ ⇒ $a = 2, b = − 4$, and $c = 8$ Now, $\sqrt{ab - bc + ca}$ = $\sqrt{(2× -4) – (- 4 × 8 ) + (8 × 2)}$ = $\sqrt{-8 +32 + 16}$ = $\sqrt{40}$ = $2\sqrt{10}$ Hence, the correct answer is $2\sqrt{10}$.
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Question : If $a=\sqrt{6}+\sqrt{5},b=\sqrt{6}-\sqrt{5}$, then $2a^{2}-5\; ab+2b^{2} =?$
Option 1: 38
Option 2: 39
Option 3: 40
Option 4: 41
Question : If $a^{2}+b^{2}+c^{2}=ab+bc+ca,$ then the value of $\frac{a+c}{b}$ is:
Option 1: 3
Option 2: 2
Option 3: 0
Option 4: 1
Question : If $a+b+c=1, ab+bc+ca=-1$ and $abc=-1$, then the value of $a^{3}+b^{3}+c^{3}$ is:
Option 1: 1
Option 2: – 1
Option 3: 2
Option 4: – 2
Question : In $\triangle ABC, \angle B=90^{\circ}$ and AB : BC = 1 : 2. The value of $\cos A+\tan C$ is:
Option 1: $\frac{5+\sqrt{5}}{2 \sqrt{5}}$
Option 2: $\frac{1+\sqrt{5}}{2 \sqrt{5}}$
Option 3: $\frac{2 \sqrt{5}}{2+\sqrt{5}}$
Option 4: $\frac{2+\sqrt{5}}{2 \sqrt{5}}$
Question : If A : B = 3 : 4 and B : C = 6 : 5, then A : ( A + C) is equal to:
Option 1: 9 : 10
Option 2: 10 : 9
Option 3: 9 : 19
Option 4: 19 : 9
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