2 2 votes If two real polynomials $f(x)$ and $g(x)$ of degrees $m\;(\geq2)$ and $n\;(\geq1)$ respectively, satisfy $f(x^{2}+1) = f(x)g(x)$ $,$ for every $x\in \mathbb{R}$ , then (A) $f$ has exactly one real root $x_{0}$ such that $f'(x_{0}) \neq 0$ (B) $f$ has exactly one real root $x_{0}$ such that $f'(x_{0}) = 0$ (C) $f$ has $m$ distinct real roots (D) $f$ has no real root. Calculus engineering-mathematics calculus userisi2015 usermod + – ankitgupta.1729 1.8k views answer comment Share Follow Print See all 3 Comments 3 3 Comments reply Kushagra Chatterjee commented Feb 21, 2019 reply Follow flag I think option D 1 1 replyShare ankitgupta.1729 commented Feb 21, 2019 reply Follow flag I also think (D) should be answer but not sure about it. 0 0 replyShare Kushagra Chatterjee commented Feb 21, 2019 reply Follow flag Solution given in the answer section 0 0 replyShare Please log in or register to add a comment.
Best answer 4 4 votes Option D Kushagra Chatterjee answered Feb 21, 2019 • selected Feb 21, 2019 by ankitgupta.1729 Kushagra Chatterjee comment Share Follow See all 2 Comments 2 2 Comments reply ankitgupta.1729 commented Feb 21, 2019 reply Follow flag you have written "if they are equal then $x_{0}$ has to be imaginary". Can you please explain this line. Complex numbers can't be compared. right ? 1 1 replyShare Kushagra Chatterjee commented Feb 21, 2019 reply Follow flag x0 will be imaginary because 2 2 replyShare Please log in or register to add a comment.