CAT 2021 Question Paper | Quant Slot 2

CAT 2021 Quant was dominated by Arithmetic followed by Algebra. In Arithmetic, the questions were dominated by topics like Speed-time-distance, Mixture and Alligations. This year, there was a surprise. The questions from Geometry were relatively on the lower side as compared to the previous years. There were 8 TITA Qs this year. Overall this section was at a medium level of difficulty.

Question 13 : For all real values of x, the range of the function f(x) = \\frac{x^{2} + 2x + 4}{2x^{2} + 4x + 9}) is

Video Explanation

Explanatory Answer

We have a function f(x) = \\frac{x^{2} + 2x + 4}{2x^{2} + 4x + 9})
We can complete the squares in the numerator as
f(x) = \\frac{x^{2} + 2x + 1 + 3}{2x^{2} + 4x + 9})
f(x) = \\frac{(x + 1)_{2} + 3}{2x^{2} + 4x + 9})
Now we try to get (x + 1)²in the denominator as well
f(x) = \\frac{(x + 1)_{2} + 3}{2(x + 1)^{2} + 7})
From here we can take out (x+1)² + 3 from both, the numerator and the denominator and put it as ‘k’
Replacing (x+1)² + 3 with k we get
f(x) = \\frac{k}{2k + 1})
The minimum value of k = 3, as the square part is always positive and another positive number is added to the equation.
So the fraction will have the minimum value when k = 3
f(x) = \\frac{3}{2(3) + 1})= \\frac{3}{7})
And, the maximum value when k approaches its maximum value. We can observe that the fraction has a 2k + 1 in the denominator and so it will always be one more than twice the numerator. As the value of k increases the value of the addition of one gets less and less of a value to the overall denominator. The fraction approaches 1/2 , but never 1/2 itself.
Hence the range becomes [\\frac{3}{7}),\\frac{1}{2}))

The question is " For all real values of x, the range of the function f(x) = \\frac{x^{2} + 2x + 4}{2x^{2} + 4x + 9}) is "

Hence, the answer is '[ \\frac{3}{7}) , \\frac{1}{2}) )'

Choice C is the correct answer.