Question : If the sum of $\frac{a}{b}$ and its reciprocal is 1 and $a\neq 0,b\neq 0$, then the value of $a^{3}+b^{3}$ is:
Option 1: 2
Option 2: –1
Option 3: 0
Option 4: 1
Correct Answer: 0
Solution : Given: $\frac{a}{b}$+$\frac{b}{a}= 1$ $⇒\frac{a^2+b^2}{ab}= 1$ $⇒a^2+b^2-ab= 0$---------(i) Now, $a^{3}+b^{3}$ $=(a+b)(a^2+b^2-ab)$ $= (a+b)×0$ $=0$ Hence, the correct answer is 0.
Application | Eligibility | Selection Process | Result | Cutoff | Admit Card | Preparation Tips
Question : The numerical value of $\frac{(a–b)^{2}}{(b–c)(c–a)}+\frac{(b–c)^{2}}{(c–a)(a–b)}+\frac{(c–a)^{2}}{(a–b)(b–c)}$ is: $(a\neq b\neq c)$
Option 1: $0$
Option 2: $1$
Option 3: $\frac{1}{3}$
Option 4: $3$
Question : If $x=\frac{8ab}{a+b}(a\neq b),$ then the value of $\frac{x+4a}{x–4a}+\frac{x+4b}{x–4b}$ is:
Option 1: 0
Option 2: 1
Option 3: 2
Option 4: 4
Question : If $a+\frac{1}{b}=b+\frac{1}{c}=c+\frac{1}{a}(a\neq b\neq c)$, then the value of $abc$ is:
Option 1: $\pm 1$
Option 2: $\pm 2$
Option 3: $0$
Option 4: $\pm \frac{1}{2}$
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 : If $a+b+c=0$, then the value of $\frac{a^2}{b c}+\frac{b^2}{c a}+\frac{c^2}{a b}$ is:
Option 1: 1
Option 2: 3
Option 3: - 1
Option 4: 0
Regular exam updates, QnA, Predictors, College Applications & E-books now on your Mobile