(C) Given $x = \sqrt{6 + \sqrt{6 + \sqrt{6 + \dots \infty}}}$.Since the expression is infinite,we can write $x = \sqrt{6 + x}$.Squaring both sides,we

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(C) Given $x = \sqrt{6 + \sqrt{6 + \sqrt{6 + \dots \infty}}}$.Since the expression is infinite,we can write $x = \sqrt{6 + x}$.Squaring both sides,we get $x^2 = 6 + x$,where $x > 0$.Rearranging the terms,we get the quadratic equation $x^2 - x - 6 = 0$.Factoring the quadratic equation: $(x - 3)(x + 2) = 0$.This gives $x = 3$ or $x = -2$.Since $x$ is a square root,$x$ must be positive,so $x = 3$.

Class 11 Mathematics 2 Complex Numbers and Quadratic Equations: Quadratic Equations Solution of quadratic equations and Nature of roots

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If $x = \sqrt{6 + \sqrt{6 + \sqrt{6 + \dots \text{to } \infty}}}$,then

A
$x$ is an irrational number
B
$2 < x < 3$
C
$x = 3$
D
None of these

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2 Complex Numbers and Quadratic Equations: Quadratic Equations

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Solution of quadratic equations and Nature of roots

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Assertion $(A)$: The maximum value of $-x^2+3x+1$ is $\frac{13}{4}$.Reason $(R)$: If $a < 0$,the maximum value of $ax^2+bx+c$ exists at $x = -\frac{b}{2a}$.The correct option among the following is

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Assertion $(A)$: The maximum value of $-x^2+3x+1$ is $\frac{13}{4}$.
Reason $(R)$: If $a < 0$,the maximum value of $ax^2+bx+c$ exists at $x = -\frac{b}{2a}$.
The correct option among the following is