Let `f(x) =ax^2+bx+c` where a,b,c are real numbers. Suppose `f(x)!=x` for any real number x. Th...
79 views · Published 11 October 2018 · 4:58 · Indexed 26 September 2026
Channel: Doubtnut · 2018 · Education
To ask Unlimited Maths doubts download Doubtnut from - https://goo.gl/9WZjCW Let `f(x) =ax^2+bx+c` where a,b,c are real numbers. Suppose `f(x)!=x` for any real number x. Then the number of solutions for f(f(x))=x in real numbers x is
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