Question

If the real valued function \[ f(x)=\log(2x-3)-2x^2+6x-4 \] then the interval in which \(f(x)\) is increasing is

• A. \((-\infty,2)\)
• B. \(\left(\frac32,2\right)\)
• C. \((2,\infty)\)
• D. \(\left(\frac32,\infty\right)\)

Hint:

For logarithmic functions, always check domain first before solving increasing or decreasing intervals.

Answer 1

The Correct Option is B

Concept: A function is increasing where its derivative is positive. Thus we check \[ f'(x)& gt;0 \] Also first determine domain because logarithm requires positive argument.

Step 1: Find domain of function.
Since \[ \log(2x-3) \] exists only when \[ 2x-3& gt;0 \] Thus \[ x& gt;\frac32 \] Hence domain is \[ \left(\frac32,\infty\right) \]

Step 2: Differentiate function.
Differentiating term by term, \[ f'(x) = \frac{2}{2x-3}-4x+6 \] Factorize the algebraic part. \[ = \frac{2}{2x-3}-2(2x-3) \] Taking LCM, \[ = \frac{2-(2x-3)^2}{2x-3} \]

Step 3: Apply increasing condition.
Since denominator positive for domain, \[ 2-(2x-3)^2& gt;0 \] \[ (2x-3)^2& lt;2 \] \[ -\sqrt2& lt;2x-3& lt;\sqrt2 \] Since domain restricts \[ 2x-3& gt;0 \] therefore \[ 0& lt;2x-3& lt;\sqrt2 \] \[ \frac32& lt;x& lt;\frac{3+\sqrt2}{2} \] Matching nearest option: \[ \boxed{\left(\frac32,2\right)} \]