Question : If $a+\frac{1}{a}=2$ and $b+\frac{1}{b}=-2$, then what is the value of $a^2+\frac{1}{a^2}+b^2+\frac{1}{b^2} ?$
Option 1: 0
Option 2: 2
Option 3: 8
Option 4: 4
Recommended: How to crack SSC CHSL | SSC CHSL exam guide
Don't Miss: Month-wise Current Affairs | Upcoming government exams
New: Unlock 10% OFF on PTE Academic. Use Code: 'C360SPL10'
Correct Answer: 4
Solution : $a+\frac{1}{a}=2$ ⇒ $a^2+1 = 2a$ ⇒ $a^2-2a+1 = 0$ ⇒ $(a-1)^2 = 0$ ⇒ $a = 1$ Similarly, $b+\frac{1}{b}=-2$ ⇒ $b^2+1 = -2b$ ⇒ $b^2+2b+1 = 0$ ⇒ $(b+1)^2 = 0$ ⇒ $b = -1$ $a^2+\frac{1}{a^2}+b^2+\frac{1}{b^2} ?$ $\therefore 1^2+\frac{1}{1^2}+(-1)^2+\frac{1}{(-1)^2}$ = $1+1+1+1 = 4$ Hence, the correct answer is 4.
Candidates can download this e-book to give a boost to thier preparation.
Application | Eligibility | Admit Card | Answer Key | Preparation Tips | Result | Cutoff
Question : If $\frac{1}{a}–\frac{1}{b}=\frac{1}{a–b}$, then the value of $a^{3}+b^{3}$ is:
Option 2: –1
Option 3: 1
Option 4: 2
Question : If $\cos^{4}\theta-\sin^{4}\theta=\frac{2}{3}$, then the value of $1-2\sin^{2}\theta$ is:
Option 1: $\frac{2}{3}$
Option 2: $\frac{3}{2}$
Option 3: $1$
Option 4: $0$
Question : If $a+\frac{1}{a}=8$, then what is the value of $\frac{1}{a^4}+a^4 ?$
Option 1: 4098
Option 2: 3846
Option 3: 3842
Option 4: 4094
Question : If $\left (2a-1 \right )^{2}+\left (4b-3 \right)^{2}+\left (4c+5 \right)^{2}=0$, then the value of $\frac{a^{3}+b^{3}+c^{3}-3abc}{a^{2}+b^{2}+c^{2}}$ is:
Option 1: $1\tfrac{3}{8}$
Option 2: $2\tfrac{3}{8}$
Option 3: $3\tfrac{3}{8}$
Question : If $a^2+\frac{1}{a^2}=\frac{7}{3}$, then what is the value of $\left(a^3-\frac{1}{a^3}\right)?$
Option 1: $\frac{5}{3 \sqrt{3}}$
Option 2: $\frac{10}{3 \sqrt{3}}$
Option 3: $\frac{7}{3 \sqrt{3}}$
Option 4: $\frac{8}{3 \sqrt{3}}$
Regular exam updates, QnA, Predictors, College Applications & E-books now on your Mobile