Question
If the function f(x) = 2x2 - kx + 5 is increasing on [1, 2], then k lies in the interval:
- $(-\infty,4)$
- $(4,\infty)$
- $(-\infty,8)$
- $(8,\infty)$
Solution:
f(x) = 2x2 - kx + 5
f'(x) = 4x - k
f(x) is increasing
4x - k < 0 on [1, 2]
k < 4x
Minimum value of k is 4.
k < 4
$\text{k}\in(-\infty,4)$
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$h(x)=\left\{\begin{array}{lll}\max & \{f(x), g(x)\} & \text { if } x \leq 0, \\ \min & \{f(x), g(x)\} & \text { if } x > 0 .\end{array}\right.$ The number of points at which $h(x)$ is not differentiable is
$(A)$ $M=I$ $(B)$ $\operatorname{det} M =1$ $(C)$ $M ^2= I$ $(D)$ $(\operatorname{adj} M)^2=I$