The set of values of x- which satisfy x 2-13 x - x -36-0 is- -5-7B- 6-7- -6-7D- 5-7

The correct option is D (5,7)[x]=x {x} x^2 13x+ [x]+36 = 0 ⇒ nbsp;x^2 13x+x {x}+ 36= 0 ⇒ nbsp;x^2 12x+ 36 = nbsp;{x} nbsp; ⇒ nbsp;(x 6)^2={x} ⇒ (x ...

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Standard XI

Mathematics

Binomial Expression

The set of va...

Question

The set of values of $x,$ which satisfy $x^{2} - 13 x + [x] + 36 = 0$ is

[5, 7)

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(6, 7)

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[6, 7)

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(5, 7)

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Solution

The correct option is D $(5, 7)$
$[x] = x - {x}$
$x^{2} - 13 x + [x] + 36 = 0$
$\Rightarrow x^{2} - 13 x + x - {x} + 36 = 0$
$\Rightarrow x^{2} - 12 x + 36 = {x} \Rightarrow (x - 6)^{2} = {x}$
$\Rightarrow (x - 6)^{2} \in [0, 1)$
$\Rightarrow x \in (5, 7)$

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The set of values of x, which satisfy x2−13x+[x]+36=0 is

The correct option is D (5,7)[x]=x−{x} x2−13x+[x]+36=0 ⇒x2−13x+x−{x}+36=0 ⇒x2−12x+36={x}⇒(x−6)2={x} ⇒(x−6)2∈[0,1) ⇒x∈(5,7)

The set of values of $x,$ which satisfy $x^{2} - 13 x + [x] + 36 = 0$ is

The correct option is D $(5, 7)$
$[x] = x - {x}$
$x^{2} - 13 x + [x] + 36 = 0$
$\Rightarrow x^{2} - 13 x + x - {x} + 36 = 0$
$\Rightarrow x^{2} - 12 x + 36 = {x} \Rightarrow (x - 6)^{2} = {x}$
$\Rightarrow (x - 6)^{2} \in [0, 1)$
$\Rightarrow x \in (5, 7)$