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If $x$ is real, the maximum value of $\frac{3x^{2}+9x+17}{3x^{2}+9x+7}$ is

Option 1)
$1/4$

Option 2)
$41$

Option 3)
$1$

Option 4)
$17/7$

Answers (1)

best_answer

As we learnt in

Rate Measurement -

Rate of any of variable with respect to time is rate of measurement. Means according to small change in time how much other factors change is rate measurement:

$\Rightarrow \frac{dx}{dt},\:\frac{dy}{dt},\:\frac{dR}{dt},(linear),\:\frac{da}{dt}$

$\Rightarrow \frac{dS}{dt},\:\frac{dA}{dt}(Area)$

$\Rightarrow \frac{dV}{dt}(Volume)$

$\Rightarrow \frac{dV}{V}\times 100(percentage\:change\:in\:volume)$

- wherein

Where dR / dt means Rate of change of radius.

let $y=\frac{3x^{2}+9x+17}{3x^{2}+9x+7}$

$(3x^{2}+9x+7)y=3x^{2}+9x+17$

$\therefore 3x^{2}(y-1)+9x(y-1)+(7y-17)=0$

For real x

$81(y-1)^{2}-4.3(y-1)(7y-17)\geqslant 0$

$3(y-1)\left [ 27(y-1) -4(7y-17)\right ]\geqslant 0$

$(y-1)\left [ 27y-27-28y+68 \right ]\geqslant 0$

$(y-1)\left [ -y+41 \right ]\geqslant 0$

$(y-1)(y-41)\geqslant 0$ $1\leq y\leq 41$ maximum value =41

Option 1)

$1/4$

This option is incorrect

Option 2)

$41$

This option is correct

Option 3)

$1$

This option is incorrect

Option 4)

$17/7$

This option is incorrect

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