Question : If $p+q+r=0$, then what is the simplified value of the expression $(\frac{p^2}{p^2-q r}+\frac{q^2}{q^2-p r}+\frac{r^2}{r^2-p q})$?
Option 1: 0
Option 2: 2
Option 3: 1
Option 4: – 1
New: SSC CHSL Tier 2 answer key released | SSC CHSL 2024 Notification PDF
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: 2
Solution : Given: $p+q+r=0$ To find: $(\frac{p^2}{p^2-q r}+\frac{q^2}{q^2-p r}+\frac{r^2}{r^2-p q})$ By putting the values p = 1, q = –1, and r = 0, the equation $p+q+r=0$ satisfies. $(\frac{p^2}{p^2-q r}+\frac{q^2}{q^2-p r}+\frac{r^2}{r^2-p q})$ = $(\frac{1^2}{1^2-(-1)× 0}+\frac{(-1)^2}{(-1)^2-1×0}+\frac{0^2}{0^2-1×(-1)})$ = 1 + 1 + 0 = 2 Hence, the correct answer is 2.
Candidates can download this e-book to give a boost to thier preparation.
Result | Eligibility | Application | Admit Card | Answer Key | Preparation Tips | Cutoff
Question : If $\frac{1}{a}–\frac{1}{b}=\frac{1}{a–b}$, then the value of $a^{3}+b^{3}$ is:
Question : If $(p-q)=6,(r-q)=5$ and $(r-p)=3$, then find the value of $\frac{p^3+q^3+r^3-3 p q r}{p+q+r}$.
Question : If $x^{\frac{1}{4}}+x^{\frac{-1}{4}}=2$, then what is the value of $x^{81}+\frac{1}{x^{81}}$?
Question : If $x+\frac{1}{x}=0$, then the value of $x^{5}+\frac{1}{x^{5}}$ is:
Question : If in a right-angled $\triangle P Q R$, $\tan Q=\frac{5}{12}$, then what is the value of $\cos Q$?
Regular exam updates, QnA, Predictors, College Applications & E-books now on your Mobile