Question

What is the least value of k such that the function x${}^2$ + kx + 1 is strictly increasing on (1,2)

• A. 1
• B. -1
• C. 2
• D. -2

Answer 1

Answer: (D)

Let f(x) = x${}^2$ + kx + 1
f '(x) = 2x + k
f(x) is strictly increasing on (1, 2)
if f '(x) > 0 for x $\in$(1, 2)
$\Rightarrow$ 2x + k > 0 for x $\in$(1, 2)
$\Rightarrow$ k > -2x for x $\in$ (1, 2)
Now, 1 < x < 2 $\Rightarrow$ 2 < 2x < 4
$\Rightarrow$ -2 > -2x > -4
$\Rightarrow$ - 4 < -2x < -2
$\Rightarrow k\ge -2$
Hence least value of k = -2.