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Question

Range of the function $$f(x)=\cfrac{x^2+x+2}{x^2+x+1}$$ is:


A
[1,2]
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B
[1,]
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C
[2,73]
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D
[1,73]
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Solution

The correct option is B $$\left[1,\cfrac{7}{3}\right]$$
$$\cfrac{(x^2+x+2)}{(x^2+x+1)}=\cfrac{(x^2+x+1)+1}{(x^2+x+1)}$$

$$=1+\cfrac{1}{(x^2+x+1)}$$

$$=1+\cfrac{1}{\left(x+\cfrac{1}{2}\right)^2+\cfrac{3}{4}}$$
Minimum value of $$\left(x+\cfrac{1}{2}\right)^2=0$$

Maximum value of $$\left(x+\cfrac{1}{2}\right)^2=\infty$$

$$\therefore$$ Range $$=\left(1,1+\cfrac{1}{3/4}\right)$$

$$\implies$$ Range $$=\left[1,\cfrac{7}{3}\right]$$

Mathematics

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