Question : If $\frac{(a+b+c)}{2}=16$ and $2ab+2 bc+2ca=120$ ,then what is the value of $4a^2+4b^2+4c^2$?
Option 1: 2828
Option 2: 3368
Option 3: 3616
Option 4: 3056
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Correct Answer: 3616
Solution : Given: $\frac{(a+b+c)}{2}=16$ $⇒ a+b+c=32$ Also given $2ab+2 bc+2ca=120$ We know that, $(a+b+c)^2=a^2+b^2+c^2+2ab+2bc+2ca$ Putting the values we get: ⇒ $(32)^2=a^2+b^2+c^2+120$ ⇒ $1024-120=a^2+b^2+c^2$ ⇒ $a^2+b^2+c^2=904$ $\therefore 4(a^2+b^2+c^2)=4\times904=3616$ Hence, the correct answer is 3616.
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Question : If $\frac{x+1}{x-1}=\frac{a}{b}$ and $\frac{1-y}{1+y}=\frac{b}{a}$, then the value of $\frac{x-y}{1+xy}$ is:
Option 1: $\frac{2ab}{a^{2}-b^{2}}$
Option 2: $\frac{a^{2}-b^{2}}{2ab}$
Option 3: $\frac{a^{2}+b^{2}}{2ab}$
Option 4: $\frac{a^{2}-b^{2}}{ab}$
Question : If $\left (2a-1 \right )^{2}+\left (4b-3 \right)^{2}+\left (4c+5 \right)^{2}=0$, then the value of $\frac{a^{3}+b^{3}+c^{3}-3abc}{a^{2}+b^{2}+c^{2}}$ is:
Option 1: $1\tfrac{3}{8}$
Option 2: $2\tfrac{3}{8}$
Option 3: $3\tfrac{3}{8}$
Option 4: $0$
Question : If $\frac{x}{y}=\frac{a+2}{a-2}$, then the value of $\frac{x^{2}-y^{2}}{x^{2}+y^{2}}$ is:
Option 1: $\frac{4a}{a^{2}+2}$
Option 2: $\frac{2a}{a^{2}+2}$
Option 3: $\frac{4a}{a^{2}+4}$
Option 4: $\frac{2a}{a^{2}+4}$
Question : If $a^2+b^2+c^2=ab+bc+ca$, then the value of $\frac{11a^4+13b^4+15c^4}{16a^2b^2+19b^2c^2+17c^2a^2}$ is:
Option 1: $1 \frac{1}{3}$
Option 2: $\frac{1}{4}$
Option 3: $\frac{3}{4}$
Option 4: $1 \frac{3}{4}$
Question : If $\small c+\frac{1}{c}=3$, then the value of $\left (c-3 \right )^{7}+\frac{1}{c^{7}}$ is:
Option 1: 2
Option 2: 0
Option 3: 3
Option 4: 1
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