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)

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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