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Question

Let the graph of f(x)=ax2+bx+c passes through origin and makes an intercept of 10 units on xaxis. If the maximum value of f(x) is 25, then the least value of |a+b+c| is

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Solution

f(x) passes through origin
f(0)=0c=0
f(x) takes the maximum value
a<0
Now, x inetercept is 10, so
The other root is either 10 or 10.

f(x)=kx(x+10) or kx(x10)f(x)=kx2+10kx or kx210kx
The maximum of the parabola is,
D4a=254acb24a=25100k24k=25k=1
Therefore,
f(x)=x210x or x2+10x
a+b+c=1+10+0 or 110+0a+b+c=9,11

Hence, the least value of |a+b+c| is 9.

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