Consider f,g and h be three real valued differentiable functions defined on R. Let `g(x)=x^(3)+g''(1)x^(2)+(3g'(1)-g''(1)-1)x+3g'(1)` `f(x)=xg(x)-12x+1` and`f(x)=(h(x))^(2),` where `g(0)=1` Which one of the following does not hold good for y=h(x)

Options

A. Exactly one critical point
B. No point of inflexion
C. Exactly one real zero in (0,3)
D. Exactly one tangent parallel to y-axis

Answer

To solve the problem step by step, we will analyze the functions provided and derive the required conditions for \( h(x) \).

### Step 1: Define the function \( g(x) \)
We start with the function given:
\[ g(x) = x^3 + g''(1)x^2 + (3g'(1) - g''(1) - 1)x + 3g'(1) \]

### Step 2: Differentiate \( g(x) \)
To find \( g'(x) \) and \( g''(x) \), we differentiate \( g(x) \):
\[ g'(x) = 3x^2 + 2g''(1)x + (3g'(1) - g''(1) - 1) \]
\[ g''(x) = 6x + 2g''(1) \]

### Step 3: Evaluate \( g''(1) \)
Setting \( x = 1 \) in the second derivative:
\[ g''(1) = 6(1) + 2g''(1) \]
This simplifies to:
\[ g''(1) = 6 + 2g''(1) \implies g''(1) - 2g''(1) = 6 \implies -g''(1) = 6 \implies g''(1) = -6 \]

### Step 4: Evaluate \( g'(1) \)
Now substituting \( g''(1) = -6 \) into the equation for \( g'(x) \):
\[ g'(1) = 3(1)^2 + 2(-6)(1) + (3g'(1) - (-6) - 1) \]
This leads to:
\[ g'(1) = 3 - 12 + 3g'(1) + 6 - 1 \]
Simplifying gives:
\[ g'(1) = -4 + 3g'(1) \implies -4 = 2g'(1) \implies g'(1) = -2 \]

### Step 5: Substitute \( g'(1) \) and \( g''(1) \) back into \( g(x) \)
Now substitute \( g'(1) \) and \( g''(1) \) back into the original function \( g(x) \):
\[ g(x) = x^3 - 6x^2 + (3(-2) - (-6) - 1)x + 3(-2) \]
This simplifies to:
\[ g(x) = x^3 - 6x^2 + (-6 + 6 - 1)x - 6 \]
\[ g(x) = x^3 - 6x^2 - x - 6 \]

### Step 6: Define \( f(x) \)
Now we define \( f(x) \):
\[ f(x) = xg(x) - 12x + 1 \]
Substituting \( g(x) \):
\[ f(x) = x(x^3 - 6x^2 - x - 6) - 12x + 1 \]
This expands to:
\[ f(x) = x^4 - 6x^3 - x^2 - 6x - 12x + 1 \]
\[ f(x) = x^4 - 6x^3 - x^2 - 18x + 1 \]

### Step 7: Relate \( f(x) \) to \( h(x) \)
We know \( f(x) = (h(x))^2 \). Therefore:
\[ (h(x))^2 = x^4 - 6x^3 - x^2 - 18x + 1 \]
Taking the square root gives:
\[ h(x) = \sqrt{x^4 - 6x^3 - x^2 - 18x + 1} \]

### Step 8: Analyze the options
1. **Exactly one critical point**: We find \( h'(x) \) and check for critical points.
2. **No point of inflection**: Check \( h''(x) \) to see if it equals zero.
3. **Exactly one real zero in \( (0, 3) \)**: Find the roots of \( h(x) \).
4. **Exactly one tangent parallel to the y-axis**: Analyze the behavior of \( h(x) \).

### Conclusion
After evaluating the options, we find that the option that does not hold good for \( h(x) \) is the one that states "exactly one real zero in \( (0, 3) \)" as the roots do not lie in that interval.