Question : If $x$ = ${\frac{1}{\sqrt{2}+1}}$, then the value of $(x^{2}+2x–1)$ is:
Option 1: $\sqrt[2]{2}$
Option 2: 4
Option 3: 0
Option 4: 2
Latest: SSC CGL preparation tips to crack the exam
Don't Miss: SSC CGL complete guide
New: Unlock 10% OFF on PTE Academic. Use Code: 'C360SPL10'
Correct Answer: 0
Solution : Given: $x= {\frac{1}{\sqrt{2}+1}}$ Rationalising the denominator, we get, ⇒ $x ={\frac{{\sqrt{2}-1}}{(\sqrt{2}+1)×{(\sqrt{2}-1)}}}=\frac{\sqrt{2}-1}{2-1} = {\sqrt{2}-1}$ Putting this value in the expression $(x^{2}+2x-1)$, we get, $=[(\sqrt{2}-1)^{2}+2(\sqrt{2}-1)-1]$ $=2-2\sqrt{2}+1+2\sqrt{2}-2-1 $ $= 0$ Hence, the correct answer is 0.
Candidates can download this ebook to know all about SSC CGL.
Answer Key | Eligibility | Application | Selection Process | Preparation Tips | Result | Admit Card
Question : If $2x=\sqrt{a}+\frac{1}{\sqrt{a}}, a>0$, then the value of $\frac{\sqrt{x^{2}-1}}{x-\sqrt{x^{2}-1}}$ is:
Question : If $\cos27^{\circ}$ = $x$, then the value of $\tan 63°$ is:
Question : If $\frac{3}{(x+2)(2x+1)}=\frac{a}{2x+1}+\frac{b}{x+2}$ is an identity, the value of $b$ is:
Question : If $x=\frac{1}{x-3},(x>0)$, then the value of $x+\frac{1}{x}$ is:
Question : If $2x^2+5 x+1=0$, then one of the values of $x-\frac{1}{2 x}$ is:
Regular exam updates, QnA, Predictors, College Applications & E-books now on your Mobile