Question : What is the value of $a^2+4 b^2+9 c^2-4 a b+12 b c-6 c a$?
Option 1: $(a+2 b-3 c)^2$
Option 2: $(2a- b-3 c)^2$
Option 3: $(2b-a-3 c)^2$
Option 4: $(a-2 b-3 c)^2$
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Correct Answer: $(a-2 b-3 c)^2$
Solution : $a^2+4 b^2+9 c^2-4 a b+12 b c-6 c a$ = $a^2+(-2 b)^2+(-3 c)^2+2(-2 a b+6 b c-3 c a)$ = $(a-2b-3c)^2$ Hence, the correct answer is $(a-2b-3c)^2$.
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Question : If $9 a+9 b+9 c=81$ and $4 a b+4 b c+4 c a=160$, then what is the value of $6 a^2+6 b^2+6 c^2$?
Option 1: 1
Option 2: 3
Option 3: 4
Option 4: 6
Question : $a$% of $b$ + $b$% of $a$ = ?
Option 1: $2a$% of $b$
Option 2: $2a$% of $2b$
Option 3: $2a$% of $2a$
Option 4: $2b$% of $2b$
Question : What is the value of (a + b + c) {(a - b)2 + (b - c)2 + (c - a)2}?
Option 1: $2 a^3+2 b^3+2 c^3$
Option 2: $2a^3+2b^3+2c^3-6abc$
Option 3: $3abc$
Option 4: $6abc$
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 $\sec\theta+\tan\theta=a$, then what is the value of $\sec\theta$?
Option 1: $\frac{a+1}{2a^2}$
Option 2: $\frac{(a-1)}{2a^2}$
Option 3: $\frac{a^2-1}{2a}$
Option 4: $\frac{a^2+1}{2a}$
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