If $x = \sqrt {1 + \sqrt {1 + \sqrt {1 + \ldots \infty } } } ,\;$ then $x$ is equal to

JEE MATHEMATICS MCQ: QUADRATIC EQUATIONS

1 Easy

1 of 1

Application +4 | -1

x = \sqrt {1 + \sqrt {1 + \sqrt {1 + \ldots \infty } } } ,\;

then

x

is equal to

View answer and explanation

Correct option: A

\frac{{1 + \sqrt 5 }}{2}

The correct answer is A (

\frac{{1 + \sqrt 5 }}{2}

) because it is the positive solution to the equation

x = \sqrt {1 + x}

which arises from the given infinite nested radical.

Practice All Chapter MCQs →

\log _7 \log _5(\sqrt{x+5}+\sqrt{x})=0

\alpha ,\beta

{x^2} - 3x + a = 0,a \in R

\alpha < 1 < \beta ,

\alpha ,\;\beta

{x^2} - px + q = 0

{\alpha _1},{\beta _1}

{x^2} - qx + p = 0,

\frac{1}{{{\alpha _1}\beta }} + \frac{1}{{\alpha {\beta _1}}}\mathrm{and}\;\frac{1}{{\alpha {\alpha _1}}} + \frac{1}{{\beta {\beta _1}}},

{x^2} + 4ax + 2 > 0

x

a

f\left( x \right) = a{x^2} + bx + c\;

\;f\left( { - 1} \right) < 1,f\left( 1 \right) > - 1,f\left( 3 \right) < - 4\;

\;a \ne 0,

9{x^2} - 18\left| x \right| + 5 = 0

f\left( x \right) = \log \left( {{x^2} - x - 2} \right)