Question

IIT Maths problem.Friends I want this solution very fast.​

Answer 1

$\boxed{1.\ \ \sqrt{2+\sqrt{2+\sqrt{2+ \dots\dots\dots\infty}}}\ =\ ?}$

$\begin{aligned}\text{Let}\ \ &\sqrt{2+\sqrt{2+\sqrt{2+ \dots\dots\dots\infty}}}\ =\ x\\ \\ \Longrightarrow\ \ &2+\sqrt{2+\sqrt{2+ \dots\dots\dots\infty}}\ =\ x^2\\ \\ \Longrightarrow\ \ &\sqrt{2+\sqrt{2+ \dots\dots\dots\infty}}\ =\ x^2-2\\ \\ \Longrightarrow\ \ &x=x^2-2\\ \\ \Longrightarrow\ \ &x^2-x-2=0\\ \\ \Longrightarrow\ \ &x^2+x-2x-2=0\\ \\ \Longrightarrow\ \ &x(x+1)-2(x+1)=0\\ \\ \Longrightarrow\ \ &(x+1)(x-2)=0\\ \\ \Longrightarrow\ \ &x=-1\ \ \ ; \ \ \ x=2\end{aligned}$

$\text{But since x\geq 0,\ \ x\neq -1.\ \ \therefore\ \ x=2 }$

$\boxed{2.\ \ \sqrt{2\sqrt{2\sqrt{2 \dots\dots\dots\infty}}}\ =\ ?}$

$\begin{aligned}\text{Let}\ \ &\sqrt{2\sqrt{2\sqrt{2 \dots\dots\dots\infty}}}\ =\ x\\ \\ \Longrightarrow\ \ &2\sqrt{2\sqrt{2 \dots\dots\dots\infty}}\ =\ x^2\\ \\ \Longrightarrow\ \ &\sqrt{2\sqrt{2 \dots\dots\dots\infty}}}\ =\ \frac{x^2}{2}\\ \\ \Longrightarrow\ \ &x=\frac{x^2}{2}\\ \\ \Longrightarrow\ \ &x^2=2x\\ \\ \Longrightarrow\ \ &x^2-2x=0\\ \\ \Longrightarrow\ \ &x(x-2)=0\\ \\ \Longrightarrow\ \ &x=0\ \ \ ; \ \ \ x=2\end{aligned}$

$\boxed{3.\ \ \sqrt{6+\sqrt{6+\sqrt{6+ \dots\dots\dots\infty}}}\ =\ ?}$

$\begin{aligned}\text{Let}\ \ &\sqrt{6+\sqrt{6+\sqrt{6+ \dots\dots\dots\infty}}}\ =\ x\\ \\ \Longrightarrow\ \ &6+\sqrt{6+\sqrt{6+\dots\dots\dots\infty}}\ =\ x^2\\ \\ \Longrightarrow\ \ &\sqrt{6+\sqrt{6+ \dots\dots\dots\infty}}\ =\ x^2-6\\ \\ \Longrightarrow\ \ &x=x^2-6\\ \\ \Longrightarrow\ \ &x^2-x-6=0\\ \\ \Longrightarrow\ \ &x^2+2x-3x-6=0\\ \\ \Longrightarrow\ \ &x(x+2)-3(x+2)=0\\ \\ \Longrightarrow\ \ &(x+2)(x-3)=0\\ \\ \Longrightarrow\ \ &x=-2\ \ \ ; \ \ \ x=3\end{aligned}$

$\text{But since x\geq 0,\ \ x\neq -2.\ \ \therefore\ \ x=3 }$

$\boxed{4.\ \ \sqrt{6\sqrt{6\sqrt{6 \dots\dots\dots\infty}}}\ =\ ?}$

$\begin{aligned}\text{Let}\ \ &\sqrt{6\sqrt{6\sqrt{6 \dots\dots\dots\infty}}}\ =\ x\\ \\ \Longrightarrow\ \ &6\sqrt{6\sqrt{6 \dots\dots\dots\infty}}\ =\ x^2\\ \\ \Longrightarrow\ \ &\sqrt{6\sqrt{6 \dots\dots\dots\infty}}}\ =\ \frac{x^2}{6}\\ \\ \Longrightarrow\ \ &x=\frac{x^2}{6}\\ \\ \Longrightarrow\ \ &x^2=6x\\ \\ \Longrightarrow\ \ &x^2-6x=0\\ \\ \Longrightarrow\ \ &x(x-6)=0\\ \\ \Longrightarrow\ \ &x=0\ \ \ ; \ \ \ x=6\end{aligned}$

$\left[\ \text{Remember}\ \ \sqrt{x^2}\ =\ |x|\ \geq\ 0\ \right]$

Answer 2

Solution :-

(1)

$\sqrt{2 +\sqrt{2 +\sqrt{2+ ...\infty}}}$

Now let it be equal to x then

$x = \sqrt{2 +\sqrt{2 +\sqrt{2+ ...\infty}}}$

Now by squaring both sides.

$x^2 = 2 +\sqrt{2 +\sqrt{2 +\sqrt{2+ ...\infty}}}$

$x^2 = 2+x$

$x^2 - x - 2 = 0$

$(x+1)(x-2) = 0$

→ x = -1 , 2 but as x cannot be negative

→ x = 2

(2)

$\sqrt{2 \sqrt{2 \sqrt{2 ...\infty}}}$

Now let it be equal to x then

$x = \sqrt{2 \sqrt{2 \sqrt{2 ...\infty}}}$

Now by squaring both sides.

$x^2 = 2 \sqrt{2 \sqrt{2 \sqrt{2 ...\infty}}}$

$x^2 = 2x$

$x^2 - 2x = 0$

$(x)(x-2) = 0$

→ x = 0 , 2 but as x cannot be zero

→ x = 2

(3)

$\sqrt{6 +\sqrt{6 +\sqrt{6+ ...\infty}}}$

Now let it be equal to x then

$x = \sqrt{6 +\sqrt{6 +\sqrt{6+ ...\infty}}}$

Now by squaring both sides.

$x^2 = 6 +\sqrt{6 +\sqrt{6 +\sqrt{6+ ...\infty}}}$

$x^2 = 6+x$

$x^2 - x -6= 0$

$(x+2)(x-3) = 0$

→ x = -2 , 3 but as x cannot be negative

→ x = 3

(iv)

$\sqrt{6\sqrt{6\sqrt{6 ...\infty}}}$

Now let it be equal to x then

$x = \sqrt{6\sqrt{6\sqrt{6...\infty}}}$

Now by squaring both sides.

$x^2 = 6 \sqrt{6\sqrt{6\sqrt{6 ...\infty}}}$

$x^2 = 6x$

$x^2 - 6x = 0$

$(x)(x-6) = 0$

→ x = 0 , 6 but as x cannot be zero

→ x = 6

So Answer :-

1 :- r

2 :- r

3 :- p

4 :- q

Option (b)