Graphs of Quadratic Equation for Different Values of D when a>0
Consider the ...
Question
Consider the function f(x)=x2−4x+17. If M and m are the maximum and minimum values of f in [0,3] respectively, then the value of 2M−m is
A
21.0
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B
21.00
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C
21
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Solution
f(x)=x2−4x+17
Here, a=1>0 D=−52<0
So, the graph of y=f(x) lies above x-axis.
The graph of y=f(x) looks like
Minimum value occurs at x=−b2a=2 ∴m=f(2)=13
Maximum value of f will occur at one of the end points of the interval [0,3] f(0)=17 and f(3)=14 ∴M=max{f(0),f(3)}=17
Hence, 2M−m=21