Question 133 of 157MediumMCQ
The integer 'k', for which the inequality x2 - 2(3k - 1)x + 8k2 - 7 > 0 is valid for every x in R, is :
Solutions · 1
Official solution
x2−2(3k−1)x+8k2−7>0
Now, D < 0
\[ \Rightarrow 4{(3k - 1)^2} - 4 \times 1 \times (8{k^2} - 7) < 0\]
\[ \Rightarrow 9{k^2} - 6k + 1 - 8{k^2} + 7 < 0\]
\[ \Rightarrow {k^2} - 6k + 8 < 0\]
\[ \Rightarrow (k - 4)(k - 2) < 0\]
2 < k < 4
then k = 3 Answer the question to unlock