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Find the minimum value of x which the expression x³ - 7x² + 11x - 5 ≥ 0.

10
25
31
4-3

Answer:- 1

Explanation:-

Solution:

\begin{aligned} x^{3} - 7 x^{2} + 11 x - 5 ⩾ 0 \\ \Rightarrow x^{3} - 5 x^{2} - 2 x^{2} + 10 x + x - 5 ⩾ 0 \\ \Rightarrow x^{2} (x - 5) - 2 x (x - 5) + 1 (x - 5) ⩾ 0 \\ \Rightarrow (x - 5) (x^{2} - 2 x + 1) ⩾ 0 \\ \Rightarrow (x - 5) {(x - 1)}^{2} ⩾ 0 \\ \Rightarrow (x - 5) (x - 1) (x - 1) ⩾ 0 \\ So, x = 1 & 5 \end{aligned}

$\eqalign{ & {x^3} - 7{x^2} + 11x - 5 \geqslant 0 \cr & \Rightarrow {x^3} - 5{x^2} - 2{x^2} + 10x + x - 5 \geqslant 0 \cr & \Rightarrow {x^2}\left( {x - 5} \right) - 2x\left( {x - 5} \right) + 1\left( {x - 5} \right) \geqslant 0 \cr & \Rightarrow \left( {x - 5} \right)\left( {{x^2} - 2x + 1} \right) \geqslant 0 \cr & \Rightarrow \left( {x - 5} \right){\left( {x - 1} \right)^2} \geqslant 0 \cr & \Rightarrow \left( {x - 5} \right)\left( {x - 1} \right)\left( {x - 1} \right) \geqslant 0 \cr & {\text{So, }}x = 1\& 5 \cr}$
Equation satisfies at both the values, but the minimum value of these two
x = 1

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Equation satisfies at both the values, but the minimum value of these two
x = 1 ", "dateCreated": "7/24/2019 10:09:12 AM", "author": { "@type": "Person", "name": "Nitin Sir" } }, "suggestedAnswer": { "@type": "Answer", "text": "

Solution:

\begin{aligned} x^{3} - 7 x^{2} + 11 x - 5 ⩾ 0 \\ \Rightarrow x^{3} - 5 x^{2} - 2 x^{2} + 10 x + x - 5 ⩾ 0 \\ \Rightarrow x^{2} (x - 5) - 2 x (x - 5) + 1 (x - 5) ⩾ 0 \\ \Rightarrow (x - 5) (x^{2} - 2 x + 1) ⩾ 0 \\ \Rightarrow (x - 5) {(x - 1)}^{2} ⩾ 0 \\ \Rightarrow (x - 5) (x - 1) (x - 1) ⩾ 0 \\ So, x = 1 & 5 \end{aligned}

$\eqalign{ & {x^3} - 7{x^2} + 11x - 5 \geqslant 0 \cr & \Rightarrow {x^3} - 5{x^2} - 2{x^2} + 10x + x - 5 \geqslant 0 \cr & \Rightarrow {x^2}\left( {x - 5} \right) - 2x\left( {x - 5} \right) + 1\left( {x - 5} \right) \geqslant 0 \cr & \Rightarrow \left( {x - 5} \right)\left( {{x^2} - 2x + 1} \right) \geqslant 0 \cr & \Rightarrow \left( {x - 5} \right){\left( {x - 1} \right)^2} \geqslant 0 \cr & \Rightarrow \left( {x - 5} \right)\left( {x - 1} \right)\left( {x - 1} \right) \geqslant 0 \cr & {\text{So, }}x = 1\& 5 \cr}$
Equation satisfies at both the values, but the minimum value of these two
x = 1 ", "dateCreated": "7/24/2019 10:09:12 AM" } }

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