If f:R→R and g:R→R are two functions such that - f(x)+f′′(x)=−xg(x)f′(x) , g(x)>0 and f(x)∈R,f′(0)≠0 . Then the function f2(x)+(f′(x))2 has
A
maxima at x=0
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B
minima at x=0
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C
point of inflexion at x=0
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D
none of the above
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Solution
The correct option is A maxima at x=0 Let S=f2(x)+(f′(x))2 S′=2f(x).f′(x)+2f′(x)f′′(x)=2f′(x)[f(x)+f′′(x)]=−2xg(x)(f′(x))2 at x=0 For x→0− ⇒x<0⇒S′=−2xg(x)(f′(x))2>0 For x→0+ ⇒x>0⇒S′=−2xg(x)(f′(x))2<0 S′ changes its sign from positive to negative ⇒f2(x)+(f′(x))2 has maxima at x=0