f(x) =4 log (x -1) -2x² + 4x +5, x>1, which one of the following is not correct? (a) ƒ is increasing in (1, 2) and decreasing in (2, ∞) (b) f(x)=- 1 has exactly two solutions (c) f" (e)- f"(2)< 0 (d) f(x)=0 has a root in the interval (e, e + 1)

1. Understanding the Function

The function is
$f(x)=4\log(x-1)-2x^2+4x+5,\qquad x>1$

• The part $4\log(x-1)$ blows up to $-\infty$ as $x \to 1^{+}$ and grows slowly afterwards.
• The quadratic part $-2x^2+4x+5$ dominates for large $x$ and heads to $-\infty$.

2. First Derivative – Monotonicity

To know where the graph rises or falls we differentiate:

Think of the graph as a roller-coaster

1. Start at the gate ($x=1$) – you are not allowed to go left of the gate because the track does not exist there (the log needs $x-1>0$).
2. Climb up the hill – from $x=1$ to $x=2$ the track rises.
3. Reach the highest point – exactly at $x=2$ the roller-coaster is at its tallest tower.
4. Go downhill forever – after $x=2$ the track keeps going down as you move right.
5. Checking heights – by plugging a few $x$ values you can see where the coaster is above or below certain horizontal lines (like $y=-1$ or $y=0$).
6. Curvature test – the second derivative tells you how sharply the track curves; plugging different $x$ values lets you compare curvatures.

The problem asks which of the four claims about this roller-coaster is wrong.

Step 1: First Derivative

Set $f'(x)=0$:

Sign test:

• Pick $x=1.5$ (i.e., $x-1=0.5$): $f'(1.5)=\dfrac{4}{0.5}-4(0.5)=8-2>0$.
• Pick $x=3$: $f'(3)=\dfrac{4}{2}-4(2)=2-8<0$.
  Hence $f$ increases on $(1,2)$ and decreases on $(2,\infty)$. Statement (a) is correct.

Step 2: Values Needed for Root-Counting

• As $x \to 1^{+}$, $\log(x-1)\to-\infty\;\Rightarrow\;f(x)\to-\infty$.
• $f(2)=4\log1-2(2)^2+4(2)+5 = 0-8+8+5 = 5$.
• As $x \to \infty$, $-2x^2$ dominates $\Rightarrow f(x)\to-\infty$.

Because the graph rises from $-\infty$ to 5 and falls back to $-\infty$, every horizontal line $y=k$ with $k<5$ cuts the graph exactly twice. Choosing $k=-1$ gives two intersections. Statement (b) is correct.

Step 3: Second Derivative Comparison

$f''(x)=-\frac{4}{(x-1)^2}-4.$

Compute at $x=2$ and $x=e$:

$f''(2)=-\frac{4}{1^2}-4=-8,$

$f''(e)=-\frac{4}{(e-1)^2}-4\approx-\frac{4}{(1.718)^2}-4\approx-1.355-4=-5.355.$

Difference:

$f''(e)-f''(2)\approx-5.355-(-8)=+2.645>0.$
But the option claims $f''(e)-f''(2)<0$. Hence Statement (c) is NOT correct.

Step 4: Existence of a Root in $(e,\,e+1)$

Evaluate:

$f(e)\;\approx\;4\log(1.718)-2e^2+4e+5 \approx 2.165 -14.778 +10.873 +5 = 3.26>0,$

$f(e+1)\;\approx\;4\log(2.718)-2(3.718)^2+4(3.718)+5 \approx 4 -27.658 +14.872 +5 = -3.786<0.$

A sign change over a continuous interval guarantees a root by IVT. Statement (d) is correct.

Conclusion

The only incorrect statement is (c).