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Question

If $$x$$ is real, then the value of the expression $$\dfrac{x^2+14x+9}{x^2+2x+3}$$ lies between


A
5 and 4
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B
4 and 5
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C
5 and 4
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D
none of these
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Solution

The correct option is B $$-5$$ and $$4$$

$$y=\cfrac { { x }^{ 2 }+14x+9 }{ { x }^{ 2 }+2x+3 } =1+\cfrac { 12x+6 }{ { x }^{ 2 }+2x+3 } \\ y'=\cfrac { ({ x }^{ 2 }+2x+3)(12)-(12x+6)(2x+2) }{ { ({ x }^{ 2 }+2x+3) }^{ 2 } } \\ y'=\cfrac { -12{ x }^{ 2 }-12x+24 }{ { ({ x }^{ 2 }+2x+3) }^{ 2 } } $$
for max-min, $$y'=0$$
so, $$x^2+x-2=0\\ \implies x=1,-2$$
at $$x=1,y=4$$
at $$x=-2, y=-5$$
and $$\lim_{x\to \infty}y=1$$
and $$y(x=0)=3$$
concluding all these graph of $$y$$ is shown above.
So, we can say that $$y$$ lies between $$-5$$ and $$4$$.

1044204_1004253_ans_586767e8991c46bb8657077cf5828a0f.PNG

Mathematics

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