Let `f (x)= int _(x^(2))^(x ^(3))(dt)/(ln t) ` for `x gt 1 and g (x) = int _(1) ^(x) (2t ^(2) -lnt ) f(t) dt(x gt 1),` then: (a) g is increasing on `(1,oo)` (b) g is decreasing on `(1,oo)` (c) g is increasing on `(1,2)` and decreasing on `(2,oo)` (d) g is decreasing on `(1,2)` and increasing on `(2,oo)`

Options

A. g is increasing on `(1,oo)`
B. g is decreasing on `(1,oo)`
C. g is increasing on `(1,20` and decreasing on `(2,00)`
D. g is decreasing on `(1,2)` and increasing on `(2,oo)`

Answer

To solve the problem, we need to analyze the functions \( f(x) \) and \( g(x) \) given in the question.

### Step 1: Define the function \( f(x) \)
The function \( f(x) \) is defined as:
\[ f(x) = \int_{x^2}^{x^3} \frac{dt}{\ln t} \]
for \( x > 1 \).

### Step 2: Differentiate \( g(x) \)
The function \( g(x) \) is defined as:
\[ g(x) = \int_{1}^{x} (2t^2 - \ln t) f(t) \, dt \]
for \( x > 1 \).

To analyze whether \( g(x) \) is increasing or decreasing, we need to find its derivative \( g'(x) \):
\[ g'(x) = (2x^2 - \ln x) f(x) \]

### Step 3: Analyze \( f(x) \)
We need to check if \( f(x) \) is positive for \( x > 1 \). Since \( f(x) \) is defined as an integral of a positive function over a positive interval, we can conclude that \( f(x) > 0 \) for \( x > 1 \).

### Step 4: Analyze \( 2x^2 - \ln x \)
Now, we need to analyze the term \( 2x^2 - \ln x \):
- As \( x \) increases, \( 2x^2 \) grows much faster than \( \ln x \).
- At \( x = 1 \):
\[ 2(1)^2 - \ln(1) = 2 - 0 = 2 > 0 \]
- For \( x > 1 \), since \( 2x^2 \) increases and \( \ln x \) increases much slower, we can conclude that \( 2x^2 - \ln x > 0 \) for \( x > 1 \).

### Step 5: Conclusion about \( g'(x) \)
Since both \( f(x) > 0 \) and \( 2x^2 - \ln x > 0 \) for \( x > 1 \), we conclude that:
\[ g'(x) > 0 \quad \text{for } x > 1 \]
This means that \( g(x) \) is an increasing function on the interval \( (1, \infty) \).

### Final Answer
Thus, the correct option is:
(a) \( g \) is increasing on \( (1, \infty) \).

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