Question

Let \(g:\mathbb{R}\to\mathbb{R}\) be a function defined by

• A. \(g\) has a local minimum at \(x=1\)
• B. \(g\) has a point of inflection at \(x=0\)
• C. \(x=0,\ x=1,\ x=2\) are critical points of \(g\)
• D. \(g''(x)=0\) has at least two distinct solutions, where \(g''\) denotes the second order derivative of \(g\)

Hint:

To test local maxima and minima, check the sign change of the first derivative. If \(g'(x)\) changes from positive to negative, the point is a local maximum; if it changes from negative to positive, the point is a local minimum.

Answer 1

The Correct Option is A

Step 1: Differentiate the given function.
Given,
\[ g(x)=\int_{0}^{x^3}(t^2-9t+8)\,dt \]
Using the Fundamental Theorem of Calculus and chain rule,
\[ g'(x)=(x^6-9x^3+8)\cdot 3x^2 \]
\[ g'(x)=3x^2(x^6-9x^3+8) \]
Factorize the expression:
\[ x^6-9x^3+8=(x^3-1)(x^3-8) \]
Therefore,
\[ g'(x)=3x^2(x^3-1)(x^3-8) \]

Step 2: Find the critical points.
Critical points occur where
\[ g'(x)=0 \]
So,
\[ 3x^2(x^3-1)(x^3-8)=0 \]
This gives
\[ x=0,\quad x^3=1,\quad x^3=8 \]
Hence,
\[ x=0,\quad x=1,\quad x=2 \]
Thus, option (C) is true.

Step 3: Check the nature of \(x=1\).
Since
\[ g'(x)=3x^2(x^3-1)(x^3-8) \]
For \(x<1\), both \((x^3-1)\) and \((x^3-8)\) are negative, so
\[ g'(x)>0 \]
For \(1<x<2\), \((x^3-1)>0\) but \((x^3-8)<0\), so
\[ g'(x)<0 \]
Thus, \(g'(x)\) changes from positive to negative at \(x=1\).
Therefore, \(g\) has a local maximum at \(x=1\), not a local minimum.
Hence, option (A) is NOT true.

Step 4: Check the point of inflection condition at \(x=0\).
Differentiate \(g'(x)\):
\[ g''(x)=\frac{d}{dx}\left(3x^2(x^6-9x^3+8)\right) \]
\[ g''(x)=24x^7-135x^4+48x \]
\[ g''(x)=3x(8x^6-45x^3+16) \]
Near \(x=0\), the sign of \(g''(x)\) changes because of the factor \(x\).
So, \(x=0\) is a point of inflection.
Thus, option (B) is true.

Step 5: Check the number of solutions of \(g''(x)=0\).
\[ g''(x)=3x(8x^6-45x^3+16) \]
Clearly,
\[ x=0 \] is one solution.
Also,
\[ 8x^6-45x^3+16=0 \]
Let
\[ y=x^3 \]
Then,
\[ 8y^2-45y+16=0 \]
This quadratic has discriminant
\[ D=(-45)^2-4(8)(16) \] \[ D=2025-512=1513>0 \]
So, it gives two distinct real values of \(y\), and therefore at least two distinct real values of \(x\).
Thus, option (D) is true.

Step 6: Final conclusion.
The only statement which is NOT true is
\[ \boxed{(A)} \]