MCQ
$\sqrt {(3 + \sqrt 5 )} $ is equal to
- A$\sqrt 5 + 1$
- B$\sqrt 3 + \sqrt 2 $
- ✓$(\sqrt 5 + 1)/\sqrt 2 $
- D${1 \over 2}(\sqrt 5 + 1)$
$3 + \sqrt 5 = \,x + y + 2\sqrt {xy} $. Obviously $x + y = 3$
and $4xy = 5$. So ${(x - y)^2} = 9 - 5 = 4$ or $(x - y) = 2$
After solving $x = {5 \over 2},y = {1 \over 2}$.
Hence, $\sqrt {3 + \sqrt 5 } = \sqrt {{5 \over 2}} + \sqrt {{1 \over 2}} = {{\sqrt 5 + 1} \over {\sqrt 2 }}$.
Generate a complete, print-ready paper with questions like this in minutes — across 16+ boards, with answer keys.
($1$) Let $E_1, E_2$ and $F_1 F_2$ be the chords of $S$ passing through the point $P_0(1,1)$ and parallel to the $x$-axis and the $y$-axis, respectively. Let $G _1 G _2$ be the chord of $S$ passing through $P _0$ and having slope -$1$ . Let the tangents to $S$ at $E_1$ and $E_2$ meet at $E_3$, the tangents to $S$ at $F_1$ and $F_2$ meet at $F_3$, and the tangents to $S$ at $G_1$ and $G_2$ meet at $G_3$. Then, the points $E_3, F_3$, and $G _3$ lie on the curve
$(A)$ $x+y=4$ $(B)$ $(x-4)^2+(y-4)^2=16$ $(C)$ $(x-4)(y-4)=4$ $(D)$ $x y=4$
($2$) Let $P$ be a point on the circle $S$ with both coordinates being positive. Let the tangent to $S$ at $P$ intersect the coordinate axes at the points $M$ and $N$. Then, the mid-point of the line segment MN must lie on the curve
$(A)$ $(x+y)^2=3 x y$ $(B)$ $x^{2 / 3}+y^{2 / 3}=2^{4 / 3}$ $(C)$ $x^2+y^2=2 x y$ $(D)$ $x^2+y^2=x^2 y^2$
Give the answer or quetion ($1$) and ($2$)