Question

Let \(f:[-2a,2a]\to\mathbb{R}\) be a thrice differentiable function and define \[ g(x)=f(a+x)+f(a-x). \] If \(m\) is the minimum number of roots of \(g'(x)=0\) in the interval \((-a,a)\) and \(n\) is the minimum number of roots of \(g''(x)=0\) in the interval \((-a,a)\), then \(m+n\) is equal to:

• A. \(1\)
• B. \(2\)
• C. \(4\)
• D. \(5\)

Hint:

If a function is even, its derivative is odd and must vanish at the origin. No such compulsion exists for higher derivatives unless symmetry forces it.

Answer 1

The Correct Option is A

Step 1: Parity of the functions \[ g(x)=f(a+x)+f(a-x) \] Clearly, \[ g(-x)=f(a-x)+f(a+x)=g(x) \] Hence, \(g(x)\) is an even function. Step 2: First derivative \[ g'(x)=f'(a+x)-f'(a-x) \] Then, \[ g'(-x)=-g'(x) \] So \(g'(x)\) is an odd function. Thus, \[ g'(0)=0 \] Therefore, the minimum number of roots of \(g'(x)=0\) in \((-a,a)\) is: \[ m=1 \] Step 3: Second derivative \[ g''(x)=f''(a+x)+f''(a-x) \] Hence, \[ g''(-x)=g''(x) \] So \(g''(x)\) is an even function. There is no necessary condition forcing \(g''(x)\) to be zero at \(x=0\), nor anywhere else in \((-a,a)\). Thus, the minimum number of roots of \(g''(x)=0\) in \((-a,a)\) is: \[ n=0 \] Step 4: Compute \(m+n\) \[ m+n=1+0=1 \] \[ \boxed{1} \]