  Question

# Suppose, f(x) is a function satisfying the following conditions (a) f(0) = 2, f(1) = 1 (b) f has a minimum value at x=52, and (c) for all x, f′(x)=∣∣ ∣∣2ax2ax−12ax+b+1bb+1−12(ax+b)2ax+2b+12ax+b∣∣ ∣∣ where a, b are some constants. Determine the constants a, b and the function f(x).f(x)=14x2−54x+2f(x)=−x3+7516x2−7516x+2f(x)=xf(x)=−14x2−54x+2

Solution

## The correct option is A f(x)=14x2−54x+2Given, f′(x)=∣∣ ∣∣2ax2ax−12ax+b+1bb+1−12(ax+b)2ax+2b+12ax+b∣∣ ∣∣ Applying R3→R3−R1−2R2, we get f′(x)=∣∣ ∣∣2ax2ax−12ax+b+1bb+1−1001∣∣ ∣∣=∣∣∣2ax2ax−1bb+1∣∣∣=∣∣∣2ax−1b1∣∣∣[C2→C2−C1]⇒     f′(x)=2ax+b On integrating, we get f(x) = ax2 + bx + c, where c is an arbitrary constant. Since, f has maximum at x=52. ⇒  f′(52)=0  ⇒  5a+b=0      . . . (i) Also,  f(0) = 2 ⇒ c = 2 and f(1) = 1 ⇒ a + b + c = 1                    . . . (ii) On solving Eqs. (i) and (ii) for a, b, we get a=14, b=−54 Thus,      f(x)=14x2−54x+2  Suggest corrections  Similar questions
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