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4-2.Quadratic Equations and Inequations
hard
જો $x$ કોઇ વાસ્તવિક સંખ્યા હોય તો $\frac{{3{x^2} + 9x + 17}}{{3{x^2} + 9x + 7}}$ ની મહતમ કિંમત . . . હોય . .
A
$\frac{1}{4}$
B
$1$
C
$41$
D
$\frac{{17}}{7}$
(AIEEE-2006)
Solution
$y=\frac{3 x^{2}+9 x+17}{3 x^{2}+9 x+7}$
$3 x^{2}(y-1)+9 x(y-1)+7 y-17=0$
$D \geq 0$ as $x$ is real
$81(y-1)^{2}-4 \times 3(y-1)(7 y-17) \geq 0$
$\Rightarrow(y-1)(y-41) \leq 0$
$\Rightarrow 1 \leq y \leq 41$
$\therefore$ Max value of $y$ is $41$
Standard 11
Mathematics
