f ( x ) = ( 2 x 2 + 3 x − 2 ) ( 1 4 x − 4 5 − x 2 ) ( x 2 − 2 x ) ( 2 x 2 − 5 x − 3 )
If the number of points where the above function is not continuous is A (also count the real numbers where the function is undefined), the number of local maxima is B , the number of local minima is C and number of integers not included in range is D . All A , B , C and D are whole numbers.
Enter your answer as A + B 2 + C 3 + D 4 .
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How D=0 D=4
B must be 1 not 2 as by drawing graph there are only three points where dy/dx=0 and only one of them is local maximum .Points where dy/dx=0 are between (-1/2,0);(2,3);(5,9). Please check and tell me if I am getting wrong.
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This was an argument in report by some senior , he made me realise that there is one more point in graph where the graph is taking a turn . I don recall it now but if you check the function values at some points you will get that exceptional point. (There is no way other than Checking! )
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Can you add a more comprehensive solution, so that those who were unable to solve this problem can learn how to do so?
D =0 is not understood!!! What do you mean by points not included in rnge? and what is the range it is (- inf,+inf)
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Yeah! That means it's Range is R , means every point on Real axis is in its Range
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A = 4 , B = 2 , C = 2 , D = 0