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CAT 2022 Slot 3 Paper

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Question : 46 of 66

Marks: +1, -0

Suppose k is any integer such that the equation $2x^2+kx+5=0$ has no real roots and the equation $x^2+(k-5)x+1=0$ has two distinct real roots for $x$ . Then, the number of possible values of $k$ is

9
7
8
13

Solution:

2 x^2+k x+5=0

has no real roots so

D<0

k^2-40<0

(k-\sqrt{40})(k+\sqrt{40})<0

k \in (-\sqrt{40}, \sqrt{40})

x^2+(k-5) x+1=0

has two distinct real roots so

D>0

(k-5)^2-4>0

k^2-10 k+21>0

(k-3)(k-7)>0

k \in (-\infty, 3) \cup (7, \infty)

Therefore possible value of

k

are

-6,-5,-4,-3,-2,-1,0,1,2

In 9 total 9 integer values of

k

are possible.

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