Question

If $f(x) = x^2 - 2(4K-1)x + g(K)>0$ $\forall x \in \mathbb{R}$ and for $K \in (a,b)$, and if $g(K) = 15K^2 - 2K - 7$, then

• A. g(K) attains its maximum at the midpoint of (a,b)
• B. g(K) attains its minimum at two points in (a,b)
• C. g(K) attains its both maximum and minimum in (a,b)
• D. g(K) attains no maximum and no minimum in (a,b)

Hint:

For a quadratic to be positive everywhere, $A>0$ and $\Delta<0$. On an open interval, a strictly monotonic function does not achieve its extrema within the interval.

Answer 1

The Correct Option is D

Step 1: Condition for quadratic $>0$.
For $Ax^2+Bx+C >0$ $\forall x$, we require $A>0$ and $\Delta=B^2-4AC<0$. 
Step 2: Apply discriminant condition. 
Here, $f(x) = x^2 - 2(4K-1)x + g(K)$, so \[ \Delta = [-2(4K-1)]^2 - 4g(K)<0 \implies (4K-1)^2<g(K). \] 
Step 3: Substitute $g(K)$. 
\[ (4K-1)^2<15K^2 - 2K - 7 \implies 16K^2-8K+1<15K^2-2K-7 \implies K^2-6K+8<0. \] 
Step 4: Solve inequality. 
\[ (K-2)(K-4)<0 \implies 2<K<4 \implies (a,b)=(2,4). \] 
Step 5: Analyze $g(K)$ on $(2,4)$. 
$g(K) = 15K^2 - 2K -7$ is an upward parabola. Vertex at $K_v = -B/(2A) = 1/15 \notin (2,4)$, so $g(K)$ is strictly increasing on $(2,4)$. 
Step 6: Conclude. 
A strictly increasing function on an open interval does not attain maximum or minimum inside the interval. 
Thus, $g(K)$ has no maximum or minimum in $(a,b)$.