Question

If $f(x)=x^3-x^2 f^{\prime}(1)+x f^{\prime \prime}(2)-f^{\prime \prime \prime}(3), x \in R$, then

• A. $2 f(0)-f(1)+f(3)=f(2)$
• B. $f(3)-f(2)=f(1)$
• C. $3 f(1)+f(2)=f(3)$
• D. $f(1)+f(2)+f(3)=f(0)$

Answer 1

The Correct Option is A

The correct answer is (A) : $2 f(0)-f(1)+f(3)=f(2)$
$f(x)=x^3-x^2{f}^{\prime }(1)+x{f}^{\prime \prime }(2)-f^{\prime \prime \prime }(3),x\in R$
Let $f^{\prime }(1)=a,f^{\prime \prime }(2)=b,f^{\prime \prime \prime }(3)=c$
$f(x)=x^3-a{x}^2+bx-c$
$f^{\prime }(x)=3x^2-2ax+b$
$f^{\prime \prime }(x)=6x-2a$
$f^{\prime \prime \prime }(x)=6$
$c=6,a=3,b=6$
$f(x)=x^3-3x^2+6x-6$
$f(1)=-2,f(2)=2,f(3)=12,f(0)=-6$
$2f(0)-f(1)+f(3)=2=f(2)$

Answer 2

Step 1: Given function

The given function is: 

\[ f(x) = x^3 - x^2 f'(1) + x f''(2) - f'''(3). \]

Let \( f'(1) = a \), \( f''(2) = b \), and \( f'''(3) = c \).

Substituting into the equation:

 

\[ f(x) = x^3 - ax^2 + bx - c = (1 - a)x^3 + bx - c. \] 

Step 2: Differentiation 
We differentiate the equation to find \( f'(x) \), \( f''(x) \), and \( f'''(x) \):

 

\[ f'(x) = 2(1 - a)x + b, \quad f''(x) = 2(1 - a), \quad f'''(x) = 0. \] 

Step 3: Substituting the given values

We are given that \( c = 6 \), \( a = 3 \), and \( b = 6 \). Substituting these values into \( f(x) \), we get:

\[ f(x) = x^3 - 3x^2 + 6x - 6 = -2x^2 + 6x - 6. \] 

Step 4: Evaluate \( f(0) \), \( f(1) \), \( f(2) \), and \( f(3) \)

Now we evaluate \( f(x) \) at different values of \( x \):

\[ f(0) = -6, \quad f(1) = -2, \quad f(2) = 2, \quad f(3) = 12. \] 

Step 5: Verify the given options

For Option (3), we verify if \( 2f(0) - f(1) + f(3) = f(2) \):

\[ 2f(0) - f(1) + f(3) = 2(-6) - (-2) + 12 = -12 + 2 + 12 = 2 = f(2). \] 

Conclusion:

The final answer is \( 2f(0) - f(1) + f(3) = f(2) \).