InterviewSolution
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InterviewSolution
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Answer» SOLUTION :`ararrp, brarrq, r, crarrs, r,drarrq,r.` a. `f(1-X)=f(1+x)` `therefore""-f'(1-x)=f'(1+x)` Hence, graph of f(x) is SYMMETRICAL about point (1, 0) [ as if f(x) = - f(-x), then f(x) is odd and its graphs is symmetrical about (0,0). Now shift the graph at (1,0)]. b. `f(2-x)+f(x)=0` `"Replace" x by 1+x. Then f(2-(1+x))+f(1+x)=0` `"or "f(1-x)+f(1+x)=0` `"or "-f'(1-x)+f'(1+x)=0` `"or "f'(1-x)=f'(1+x)"(1)"` Therefore, graph of f'(x) is symmetrical about line x=1. Alos, put x=2 in (1). Then f'(-1)=f'(3). c. `f(x+2)+f(x)=0"(1)"` Replace `x by x + 2. then f(x+4)+g(c+2)=0"(2)"` From (1) and (2), we have f(x) = f(x+4) Hence, f(x) is periodic with period 4. Also, f'(x)=f'(x+4). Hence, f'(x) is periodic with period 4. Put `=-1 in f'(x)=f'(x+4). Then f'(-1)=f'(3)`. d. `"Putting "x=0, y = 0, we GET 2f(0)+{f(0)}^(2)=1` `"or "f(0)=sqrt(2)-1""[because f(0)gt0]` Putting `y=x, 2f(x)+{f(x)}^(2)=1` Differentiating w.r.t. x, we get `2f'(x)+2fcdotf'(x)=0` `"or "f'(x){1+f(x)}=0` `"or "f'(x)=0," because "f(x)gt0`. |
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