The solution set for - -1x-2 - 6- - 1x- - 9 - 0 is

The correct option is C none of these [^ 1x]^2 6^ 1x+9≤ 0 ⇒ (^ 1x 3)^2≤ 0 but square of any no. cannot be negative , so ^ 1x 3=0 x=cot3so ans . is ...

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Standard XII

Mathematics

Nature of Roots

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Question

The solution set for ${[{cot}^{- 1} x]}^{2} - 6 [{cot}^{- 1} x] + 9 \leq 0$ is

(- \infty, cot 3]

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[cot 3, cot 2]

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[cot 3, \infty]

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none of these

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Solution

The correct option is C none of these
$[{cot}^{- 1} x]^{2} - 6 {cot}^{- 1} x + 9 \leq 0$
$\Rightarrow ({cot}^{- 1} x - 3)^{2} \leq 0$
but square of any no. cannot be negative , so
${cot}^{- 1} x - 3 = 0$
x=cot3
so ans . is none of these.

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The solution set for

[c o t^{- 1} x]^{2} - 6 [c o t^{- 1} x] + 9 \leq 0

, is

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The solution set for [cot−1x]2−6[cot−1x]+9≤0 is

The correct option is C none of these[cot−1x]2−6cot−1x+9≤0⇒(cot−1x−3)2≤0but square of any no. cannot be negative , socot−1x−3=0x=cot3so ans . is none of these.

The solution set for ${[{cot}^{- 1} x]}^{2} - 6 [{cot}^{- 1} x] + 9 \leq 0$ is

The correct option is C none of these
$[{cot}^{- 1} x]^{2} - 6 {cot}^{- 1} x + 9 \leq 0$
$\Rightarrow ({cot}^{- 1} x - 3)^{2} \leq 0$
but square of any no. cannot be negative , so
${cot}^{- 1} x - 3 = 0$
x=cot3
so ans . is none of these.