Question : If $x=p+\frac{1}{p}$ and $y=p-\frac{1}{p}$, then the value of $x^{4}-2x^{2}y^{2}+y^{4}$ is:
Option 1: 24
Option 2: 4
Option 3: 16
Option 4: 8
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: 16
Solution : $x = p+\frac{1}{p}$ $y = p-\frac{1}{p}$ Now, $x^{4}-2x^{2}y^{2}+y^{4}$ = $(x^{2}-y^{2})^{2}$ = $((x+y)(x-y))^{2}$ = $[(p+\frac{1}{p}+p-\frac{1}{p})(p+\frac{1}{p}-(p-\frac{1}{p})]^{2}$ = $[2×p×2×\frac{1}{p}]^{2}$ = 16 Hence, the correct answer is 16.
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 $x=\sqrt{3}-\frac{1}{\sqrt{3}}, y=\sqrt{3}+\frac{1}{\sqrt{3}}$, then the value of $\frac{x^2}{y}+\frac{y^2}{x}$ is:
Question : If $\sin (x - y) = \frac{1}2$ and $\cos (x + y) = \frac{1}2$, then what is the value of $\sin x \cos x + 2\sin^2x + cos^3x \sec x$?
Question : If $x+\frac{1}{x}=\mathrm{\frac{K}{2}}$, then what is the value of $\frac{x^8+1}{x^4} ?$
Question : What is the value of $\frac{4x^2+9y^2+12xy}{144}$?
Question : If $x+\frac{1}{x}=2 K$, then what is the value of $x^4+\frac{1}{x^4}$?
Regular exam updates, QnA, Predictors, College Applications & E-books now on your Mobile