Question
Find the minimum value of x which the expression x3 - 7x2 + 11x - 5 ≥ 0.
Answer: Option C
$$\eqalign{
& {x^3} - 7{x^2} + 11x - 5 \geqslant 0 \cr
& \Rightarrow {x^3} - 5{x^2} - 2{x^2} + 10x + x - 5 \geqslant 0 \cr
& \Rightarrow {x^2}\left( {x - 5} \right) - 2x\left( {x - 5} \right) + 1\left( {x - 5} \right) \geqslant 0 \cr
& \Rightarrow \left( {x - 5} \right)\left( {{x^2} - 2x + 1} \right) \geqslant 0 \cr
& \Rightarrow \left( {x - 5} \right){\left( {x - 1} \right)^2} \geqslant 0 \cr
& \Rightarrow \left( {x - 5} \right)\left( {x - 1} \right)\left( {x - 1} \right) \geqslant 0 \cr
& {\text{So, }}x = 1\& 5 \cr} $$
Equation satisfies at both the values, but the minimum value of these two
x = 1
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$$\eqalign{
& {x^3} - 7{x^2} + 11x - 5 \geqslant 0 \cr
& \Rightarrow {x^3} - 5{x^2} - 2{x^2} + 10x + x - 5 \geqslant 0 \cr
& \Rightarrow {x^2}\left( {x - 5} \right) - 2x\left( {x - 5} \right) + 1\left( {x - 5} \right) \geqslant 0 \cr
& \Rightarrow \left( {x - 5} \right)\left( {{x^2} - 2x + 1} \right) \geqslant 0 \cr
& \Rightarrow \left( {x - 5} \right){\left( {x - 1} \right)^2} \geqslant 0 \cr
& \Rightarrow \left( {x - 5} \right)\left( {x - 1} \right)\left( {x - 1} \right) \geqslant 0 \cr
& {\text{So, }}x = 1\& 5 \cr} $$
Equation satisfies at both the values, but the minimum value of these two
x = 1
Was this answer helpful ?
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