Let f(x) is a polynomial of degree 3 sucha that f(-1) = -5 ; f(1)=3 , f(2)= -2 , f(4)=1 then the number of real roots of f(x)=0 will be
1
3
5
2
As we have learned
Properties of Continuous function -
If f is a continuous function defined on [a,b] such that f(a) and f(b) are of opposite signs then there exists at least one solution of the equation f(x) = 0 in the open interval (a, b)
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f(x) is a polynoimial so it will be continous everywhere , now f(-1) and f(1) are of opposite sign so , atleast one rootb in (-1,1) , similarly at least one in (1,2) and atleast one in (2,4) so atleast three roots will be there But as f(x) is 3rd degree polynomial so it can't have more than 3 roots
so exactly three roots will be there
Option 1)
1
Option 2)
3
Option 3)
5
Option 4)
2
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