Question : If $16 y^2-k=\left(4 y+\frac{3}{2}\right)\left(4 y-\frac{3}{2}\right)$, then the value of $k$ is:
Option 1: $\frac{9}{4}$
Option 2: $\frac{11}{4}$
Option 3: $\frac{6}{4}$
Option 4: $\frac{7}{4}$
Correct Answer: $\frac{9}{4}$
Solution : $16 y^2-k=\left(4 y+\frac{3}{2}\right)\left(4 y-\frac{3}{2}\right)$ $⇒(4y)^2-k=(4y)^2-(\frac{3}{2})^2$ $⇒k=(\frac{3}{2})^2$ $\therefore k = \frac{9}{4}$ Hence, the correct answer is $\frac{9}{4}$.
Result | Eligibility | Application | Selection Process | Cutoff | Admit Card | Preparation Tips
Question : The value of $\left(5 \frac{1}{4} \div \frac{3}{7}\right.$ of $\left.\frac{1}{2}\right) \div\left(5 \frac{1}{9}-7 \frac{7}{8} \div 9 \frac{9}{20}\right) \times \frac{11}{21}+\left(2 \div 2\right.$ of $\left.\frac{1}{2}\right)$ is:
Question : The value of $\left(5 \frac{1}{4} \div \frac{3}{7}\right.$ of $\left.\frac{1}{2}\right) \div\left(5 \frac{1}{9}-7 \frac{7}{8} \div 9 \frac{9}{20}\right) \times \frac{11}{21}-\left(5 \div 2\right.$ of $\left.\frac{1}{2}\right)$ is:
Question : The value of $\left(5 \div 2\right.$ of $\left.\frac{1}{2}\right)+\left(5 \frac{1}{4} \div \frac{3}{7}\right.$ of $\left.\frac{1}{2}\right) \div\left(5 \frac{1}{9}-7 \frac{7}{8} \div 9 \frac{9}{20}\right) \times \frac{11}{21}$ is:
Question : If $\frac{x}{4 y}=\frac{3}{4}$ then, the value of $\frac{2 x+3 y}{x–2 y}$ is:
Question : If $x^2-3 x+1=0$, then the value of $\left(x^4+\frac{1}{x^2}\right) \div\left(x^2+1\right)$ is:
Regular exam updates, QnA, Predictors, College Applications & E-books now on your Mobile