Question 51

Let k be the largest integer such that the equation $$(x-1)^{2}+2kx+11=0$$ has no real roots. If y is a positive real number, then the least possible value of $$\frac{k}{4y}+9y$$ is

Correct Answer: 6


It is given that $$\left(x-1\right)^2+2kx+11=0$$ has no real roots. (Where k is the largest integer)

$$\left(x-1\right)^2+2kx+11=0$$, which can be written as:

=> $$x^2-2x+1+2kx+11=0$$

=> $$x^2-2\left(k-1\right)x+12=0$$

We know that for no real roots, D < 0 => b^2 -4ac < 0

Hence, $$\left\{2\left(k-1\right)\right\}^2-4\cdot1\cdot12\ <0$$

=> $$4\left(k-1\right)^2<48$$

=> $$\left(k-1\right)^2<12$$

Since k is an integer, it implies (k-1) is also an integer. 

Therefore, from the above inequality, we can say that the largest possible value of (k-1) = 3 => The largest possible value of k is 4.

Now we need to calculate the least possible value of $$\frac{k}{4y}+9y$$. 

$$\frac{k}{4y}+9y$$ can be written as $$\frac{4}{4y}+9y\ =\ \frac{1}{y}+9y$$

The least possible value of $$9y\ +\frac{1}{y}$$ can be calculated using A.M-G.M inequality.

Using A.M-G.M inequality, we get:

$$\ \frac{\ 9y+\frac{1}{y}}{2}\ge\ \sqrt{\ 9y\times\ \frac{1}{y}}$$

=> $$\ \frac{\ 9y+\frac{1}{y}}{2}\ge\ \sqrt{\ 9}$$

=> $$\ \frac{\ 9y+\frac{1}{y}}{2}\ge3$$

=> $$\ \ 9y+\frac{1}{y}\ge6$$

Hence, the least possible value is 6

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