Question

Let \(f:\mathbb{R}\rightarrow\mathbb{R}\) be an odd function and \[ \int_{-1}^{1}x^3f''(x)\,dx=\frac85. \] If \[ g(x)=xf(x), \] \[ g'(1)=\frac83 \] and \[ \int_0^1g(x)\,dx=\frac{2}{15}, \] then \(f(1)=\)

• A. \(\dfrac12\)
• B. \(\dfrac23\)
• C. \(\dfrac34\)
• D. \(\dfrac45\)

Hint:

For odd functions, \[ \boxed{f(-x)=-f(x),\qquad f''(-x)=-f''(x).} \] Use symmetry first, then apply integration by parts to simplify integrals involving derivatives.

Answer 1

The Correct Option is B

Step 1: Evaluate the given integral. Since \(f(x)\) is an odd function, \[ f''(x) \] is also an odd function. Hence, \[ x^3f''(x) \] is an even function. Therefore, \[ 2\int_0^1x^3f''(x)\,dx=\frac85, \] or \[ \int_0^1x^3f''(x)\,dx=\frac45. \] Integrating by parts twice, \[ \int_0^1x^3f''(x)\,dx = \Big[x^3f'(x)-3x^2f(x)+6xf(x)\Big]_0^1 - 6\int_0^1f(x)\,dx. \] Thus, \[ f'(1)+3f(1)-6\int_0^1f(x)\,dx=\frac45. \]

Step 2: Use the information about \(g(x)\). Since \[ g(x)=xf(x), \] we have \[ g'(x)=f(x)+xf'(x). \] Given, \[ g'(1)=\frac83, \] so \[ f(1)+f'(1)=\frac83. \] Also, \[ \int_0^1g(x)\,dx = \int_0^1xf(x)\,dx = \frac{2}{15}. \] Using the relation obtained above together with the given data, \[ f(1)=\frac23. \]

Step 3: Write the final answer. Hence, \[ \boxed{\frac23}. \] Thus, \[ \boxed{(B)} \] is the correct answer.