MCQ
If $\sqrt2=1.4142,$ then $\sqrt{\frac{\sqrt2-1}{\sqrt2+1}}$ is equal to:
- A$0.1718$
- B$5.8282$
- ✓$0.4142$
- D$2.4142$
✓
Answer
Correct option: C.
$0.4142$
By Rationalising $\frac{\sqrt2-1}{\sqrt2+1},$ we get
$\frac{\sqrt2-1}{\sqrt2+1}\times\frac{\sqrt2-1}{\sqrt2-1}=\frac{\big(\sqrt2-1\big)^2}{\big(\sqrt2\big)^2-1^2}=\frac{\big(\sqrt2-1\big)^2}{1}$
So $\sqrt{\frac{\sqrt2-1}{\sqrt2+1}}=\sqrt{\frac{\big(\sqrt2-1\big)^2}{1}}\\ \ =\big(\sqrt2-1\big)=1.4142-1=0.4142$
Hence, correct option is $(c).$
Need a full question paper?
Generate a complete, print-ready paper with questions like this in minutes — across 16+ boards, with answer keys.
Explore more
Similar questions
If $x^2-1$ is a factor of $a x^4+b x^3+c x^2+d x+e$, then
→In a $\triangle\text{ABC}$ if $D$ and $E$ are mid-points of $BC$ and $AD$ respectively such that $\text{ar}(\triangle\text{AEC})=4\text{cm}^2, $ then $\text{ar}(\triangle\text{BEC}) =$
→If ABC is an arc of a circle and $\angle A B C=135^{\circ}$, then the ratio of arc $\widehat{A B C}$ to the circumference, is
→The number of spherical bullets each $5\ dm$ in diameter which can be cast from a rectangular block of lead $11 m$ long, $10 m$ broad and $5$ high is
→In the given figure, $AOB$ is a straight line. The value of $x$ is:
→In the given figure, $AOB$ is a diameter and $ABCD$ is a cyclic quadrilateral. If $\angle\text{ADC}=120^\circ,$ then $\angle\text{BAC}=?$
→The zero of the polynomial $p(x) = 5x - 2$ is:
→$\sqrt{12}\times\sqrt{15}=$
→If the radius of the base of a right circular cone is $3r$ and its height is equal to the radius of the base, then its volume is:
→The sides of a triangle are $56\ cm, 60\ cm$ and $52\ cm$ long. Then the area of the triangle is:
→