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Question

# f(x) is cubic polynomial with f(2)=18 and f(1)=−1. Also f(x) has local maxima at x=−1 and f'(x) has local minima at x=0, then

A
the distance between (1,2) and (a,f(a)), where x=a is the point of local minima is 25
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B
f(x) is increasing for x[1,25]
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C
f(x) has local minima at x=1
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D
the value of f(0)=15
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Solution

## The correct options are B f(x) is increasing for x∈[1,2√5] C f(x) has local minima at x=1Let f(x)=ax3+bx2+cx+d f(2)=18⇒8a+4b+2c+d=18⋯(1) f(1)=−1⇒a+b+c+d=−1⋯(2) f(x) has local maxima at x=−1, f′(−1)=0⇒3a−2b+c=0⋯(3) f′(x) has local minima at x=0, f′′(0)=0⇒b=0⋯(4) Solving (1),(2),(3) and (4), we get f(x)=14(19x3−57x+34) or f(0)=172 f′(x)=574(x2−1)>0 ∀ x>1 f′(x)=0⇒x=1,−1 f′′(−1)<0,f′′(1)>0 Thus, x=−1 is a point of local maximum and x=1 is a point of local minimum. So a=1, (a,f(a))=(1,f(1))=(1,−1) The distance between (−1,2) and (1,f(1)), i.e., (1,−1) is =√22+32=√13≠2√5

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