Question : If the equation k(21x2 + 24) + rx + (14x2 – 9) = 0, k(7x2 + 8) + px + (2x2 – 3) = 0 have both roots common, then the value of $\frac{p}{r}$ is:
Option 1: $\frac{1}{3}$
Option 2: $\frac{2}{5}$
Option 3: $\frac{4}{3}$
Option 4: $\frac{7}{5}$
Correct Answer: $\frac{1}{3}$
Solution : Since the roots of the equation are common, the coefficients are in proportion. The equation 1 can be written as, (21k +14)x2 + rx + 24k – 9 = 0 The equation 2 can be written as, (7k + 2)x2 + px + 8k – 3 = 0 Therefore, $\frac{21k+14}{7k+2} = \frac{r}{p} = \frac{24k-9}{8k-3}$ $⇒\frac{r}{p} = \frac{24k−9}{8k−3}$ $⇒\frac{r}{p} = \frac{3 \times (8k−3)}{(8k−3)}$ $⇒\frac{r}{p} = \frac{3}{1}$ $\therefore\frac{p}{r}=\frac{1}{3}$ Hence, the correct answer is $\frac{1}{3}$.
Application | Eligibility | Selection Process | Result | Cutoff | Admit Card | Preparation Tips
Question : What is the value of m in the quadratic equation $x^{2}+mx+24=0$, if one of its roots is $\frac{3}{2}$.
Option 1: $–\frac{45}{2}$
Option 2: $16$
Option 3: $–\frac{21}{2}$
Option 4: $–\frac{35}{2}$
Question : The value of (1+ sin4 A – cos4 A) cosec2 A is:
Option 1: –2
Option 2: 2
Option 3: 1
Option 4: –1
Question : The value of $2 \frac{1}{3} \div 2 \frac{1}{2}$ of $1 \frac{3}{5}+\left(\frac{3}{8}+\frac{1}{7} \times 1 \frac{3}{4}\right)$ is:
Option 1: $\frac{25}{24}$
Option 2: $\frac{29}{24}$
Option 3: $\frac{35}{24}$
Option 4: $\frac{5}{24}$
Question : The value of $3 \frac{1}{3} \div 2 \frac{1}{2}$ of $1 \frac{3}{5}+\left(\frac{3}{8}+\frac{1}{7} \times 1 \frac{3}{4}\right)$ is:
Option 1: $\frac{55}{24}$
Option 2: $\frac{35}{24}$
Option 3: $\frac{25}{24}$
Question : The simplified value of (0.2)3 × 400 ÷ 2000 of (0.2)2 is:
Option 1: $\frac{1}{25}$
Option 2: $\frac{3}{25}$
Option 3: $\frac{2}{25}$
Option 4: $\frac{1}{50}$
Regular exam updates, QnA, Predictors, College Applications & E-books now on your Mobile