Let f:[ 0,1 ] \to R (the set of all real numbers) be a function. Suppose the function f is twice differentiable, f(0) = f(1)=0 and satisfies f''( x ) - 2f'( x ) + f( x ) \ge .e^x,x \in [ 0,1 ] . If the function e^ - xf( x ) assumes its minimum in the interval [ 0,1 ] at x = 1 \over 4 , which of the following is true?

f'( x ) \lt f( x ),0 \lt x \lt 1 \over 4

f'( x ) \lt f( x ),1 \over 4 \lt x \lt 3 \over 4

f'( x ) \gt f( x ),0 \lt x \lt 1 \over 4

f'( x ) \lt f( x ),3 \over 4 \lt x \lt 1

Application of Derivatives PYQ — JEE Advanced 2013 · Shift 2 · Q27 with Solution

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Correct answer: C

Let $g(x)=e^{-x}f(x).$ Then $f(x)=e^x g(x).$

We compute derivatives: $f'(x)=e^x(g+g'),$ $f''(x)=e^x(g+2g'+g'').$

So, $f''-2f'+f=e^x(g+2g'+g'')-2e^x(g+g')+e^x g=e^x g''.$

Given $f''(x)-2f'(x)+f(x)ge e^x,$ we get $e^x g''(x)ge e^x.$ Since $e^x>0$ , this implies $g''(x)ge 1 \,\text{ for all } x\in[0,1].$

Since $f(0)=0$ and $f(1)=0$ ,

\qquad g(1)=e^{-1}f(1)=0.$$ Also, we are told that $g(x)=e^{-x}f(x)$ attains its minimum at $x=\tfrac14$. Thus, $$g\left(\tfrac14\right)\le g(0)=0, \qquad g\left(\tfrac14\right)\le g(1)=0.$$ Because $g''\ge 1>0$, the function $g$ is **strictly convex**. Hence its minimum is unique and occurs where $$g'\left(\tfrac14\right)=0.$$ --- 3. **Study the monotonicity of $g'$** Since $$g''(x)\ge 1>0,$$ $g'$ is strictly increasing on $[0,1]$. And since $$g'\left(\tfrac14\right)=0,$$ we get: - for $0<x<\tfrac14$, $$g'(x)<0,$$ - for $x>\tfrac14$, $$g'(x)>0.$$ Now relate this to $f$. Because $$g(x)=e^{-x}f(x),$$ we have $$g'(x)=e^{-x}(f'(x)-f(x)).$$ Since $e^{-x}>0$, the sign of $g'(x)$ is the sign of $f'(x)-f(x)$. Thus: - if $0<x<\tfrac14$, then $$f'(x)-f(x)<0 \implies f'(x)<f(x),$$ - if $x>\tfrac14$, then $$f'(x)-f(x)>0 \implies f'(x)>f(x).$$ --- 4. **Check each option** ### Option A Claims: $$f'(x)<f(x),\quad \tfrac14<x<\tfrac34.$$ But for $x>\tfrac14$, we found $$f'(x)>f(x).$$ So **A is false**. ### Option B Claims: $$f'(x)>f(x),\quad 0<x<\tfrac14.$$ But for $0<x<\tfrac14$, we found $$f'(x)<f(x).$$ So **B is false**. ### Option C Claims: $$f'(x)<f(x),\quad 0<x<\tfrac14.$$ This matches exactly what we derived. So **C is true**. ### Option D Claims: $$f'(x)<f(x),\quad \tfrac34<x<1.$$ But for every $x>\tfrac14$, we have $$f'(x)>f(x).$$ So **D is false**. --- 5. **Final answer** The correct option is $$\boxed{\text{C}}.$$