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Let aR and f:RR be given by f(x)=x55x+a then,

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f(x) has three real roots, if a > 4
f(x) has only one real root, if a > 4
f(x) has three real roots, if a < - 4
f(x) has three real roots, if - 4

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detailed solution

Correct option is B

Let y=x5−5x  (i) As x→∞,y→∞ and x→−∞,y→−∞(ii) Also, at x = 0, y = 0, thus the curve passes through the origin. (iii) dydx=5x4−5=5x4−1=5x2−1x2+1        =5(x−1)(x+1)x2+1Now, dydx>0 in (−∞,−1)∪(1,∞) thus f(x) is increasing in these intervalAlso, dydx<0(−1,1) thus decreasing in (-1, 1).(iv) Also , at x = -1,  dydx changes its sign from + ve to vex= - 1 is point of local maxima.Similarly, x = 1 is point of local minimaLocal maximum value,y=(−1)5−5(−1)=4Local minimum value,y=(1)5−5(1)=−4Now, let y = -aAs, evident from the graph, if −a∈(−4,4) i.e., a∈(−4,4)Then, f(x) has three real roots and if −a>4 or −a>−4 then f(x) has one real root.

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