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Question 8 : For all real values of x, \\frac{3x^{2} - 6x + 12}{x^{2} + 2x + 4}\\) lies between 1 and k, and does not take any value above k. Then k equals

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\\frac{3x^{2} - 6x + 12}{x^{2} + 2x + 4}\\) = \\frac{3(x^{2} + 2x + 4)- 12x}{x^{2} + 2x + 4}\\) = 3 - \\frac{12x}{x^{2} + 2x + 4}\\) = 3 - \\frac{12x}{3 + (x+1)^{2}}\\)

Since \\frac{3x^{2} - 6x + 12}{x^{2} + 2x + 4}\\) lies between 1 and k,

Which means, 3 - \\frac{12x}{3 + (x+1)^{2}}\\) also lies between 1 and k

So, \\frac{12x}{3 + (x+1)^{2}}\\) lies between 3-k and 2

So the minimum value of the function \\frac{12x}{3 + (x+1)^{2}}\\) occurs at differentiation \\frac{12x}{3 + (x+1)^{2}}\\) w.r.t x = 0

differentiation(\\frac{12x}{3 + (x+1)^{2}}\\) ) = \\frac{(3+(๐ฅ+1)^{2} ).12 โ12๐ฅ.2(๐ฅ+1)}{(3+(๐ฅ+1)^{2})^{2}}\\) = 0

Numerator = 0.

(3+(๐ฅ+1)^{2} ).12 โ12๐ฅ.2(๐ฅ+1)=0

(3+(๐ฅ+1)^{2} ).12 โ12๐ฅ.2(๐ฅ+1)= 0

36 + 12x^{2} + 12 + 24x โ 24x^{2} โ 24x = 0

48 โ 12x^{2} = 0

12(4-x^{2}) = 0

12 (2+x)(2-x) = 0

x=-2 or x=2.

When x = -2 the value of the function \\frac{12x}{3 + (x+1)^{2}}\\) will be -6

When x =2 the value of the function \\frac{12x}{3 + (x+1)^{2}}\\) will be 2

So the maximum value of the function is 2 and the minimum value of the function is -6.

So the values of the function lies between -6 an 2.

We also know that the values of the function lies between 3-k and 2

So 3-k = -6

k=9.

Therefore, k equals 9.

The question is **"For all real values of x, \\frac{3x^{2} - 6x + 12}{x^{2} + 2x + 4}\\) lies between 1 and k, and does not take any value above k. Then k equals" **

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