IPMAT Question Paper 2019 | IPM Indore Quants

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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

Explanatory Answer

\\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+(𝑥+1)^{2} ).12 −12𝑥.2(𝑥+1)}{(3+(𝑥+1)^{2})^{2}}\\) = 0
Numerator = 0.
(3+(𝑥+1)² ).12 −12𝑥.2(𝑥+1)=0

(3+(𝑥+1)² ).12 −12𝑥.2(𝑥+1)= 0
36 + 12x² + 12 + 24x – 24x² – 24x = 0
48 – 12x² = 0
12(4-x²) = 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"

Hence, the answer is 9

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