Question : What is the value of $\frac{x^2-x-6}{x^2+x-12}÷\frac{x^2+5x+6}{x^2+7x+12}$?
Option 1: $1$
Option 2: $\frac{(x-3)}{(x+3)}$
Option 3: $\frac{(x+4)}{(x-3)}$
Option 4: $\frac{(x-3)}{(x+4)}$
Correct Answer: $1$
Solution : Given: $\frac{x^2-x-6}{x^2+x-12}÷\frac{x^2+5x+6}{x^2+7x+12}$ = $\frac{(x-3)(x+2)}{(x+4)(x-3)}÷\frac{(x+3)(x+2)}{(x+4)(x+3)}$ = $\frac{(x-3)(x+2)}{(x+4)(x-3)}×\frac{(x+4)(x+3)}{(x+3)(x+2)}$ = $1$ Hence, the correct answer is $1$.
Result | Eligibility | Application | Selection Process | Cutoff | Admit Card | Preparation Tips
Question : If $x^2-7x+1=0$, what is the value of $(x+\frac{1}{x})$.
Question : If for a non-zero $x$, $3x^{2}+5x+3=0,$ then the value of $x^{3}+\frac{1}{x^{3}}$ is:
Question : If $\frac{1}{x+2}=\frac{3}{y+3}=\frac{1331}{z+1331}=\frac{1}{3}$, then what is the value of $\frac{x}{x+1}+\frac{y}{y+6}+\frac{z}{z+2662}$?
Question : If $\frac{x^{2}-x+1}{x^{2}+x+1}=\frac{2}{3}$, then the value of $\left (x+\frac{1}{x} \right)$ is:
Question : If $x^2-2\sqrt{10}x+1=0$, what is the value of $(x-\frac{1}{x})$?
Regular exam updates, QnA, Predictors, College Applications & E-books now on your Mobile