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Let a,b,\epsilon R  be such that the function f given by f(x)=In \left | x \right |+bx^{2}+ax,x\neq 0

has extreme values at x = –1 and x =  2.

Statement 1 : f has local maximum at x = –1 and at x = 2.

Statement 2 : a= \frac{1}{2} \: and\: b=\frac{-1}{4}

 

Option: 1

Statement 1 is false, statement 2 is true


Option: 2

Statement 1 is true, statement 2 is true; statement 2 is a correct explanation for statement 1


Option: 3

Statement 1 is true, statement 2 is true; statement 2 is not a correct explanation for statement 1


Option: 4

Statement 1 is true, statement 2 is false


Answers (1)

best_answer

f(x)=1n\left | x \right |+bx^{2}+ax,x\neq 0\; has\; extreme \; values\; at

x=-1,x=2

\Rightarrow f'(x)=\frac{1}{x}+2bx+a

f'(-1)=0\; and\; f'(2)=0\; \left [ Given \right ]

\Rightarrow -1-2b+a=0\Rightarrow b=-\frac{1}{4}

and\; \frac{1}{2}+4b+a=0\Rightarrow a=\frac{1}{2}

f''(x)=-\frac{1}{x^{2}}+2b=-\frac{1}{x^{2}}-\frac{1}{2}=-\left ( \frac{1}{x^{2}} +\frac{1}{2}\right )< 0

for\; all\; x\; \varepsilon \; R-\left \{ 0 \right \}

\Rightarrow f\; has\; a\; local\; maximum\; at\; x=-1,x=2

\therefore \; \; Statement \; 1 :\; f\; has \; local\; maxima \; at\; x=-1,x=2

\therefore \; \; Statement \; 2 :\; a=\frac{1}{2},b=-\frac{1}{4}

f(x) = ln[x] + bx2 + ax

f1(x) = 1/x + 2bx + a

f1(x-1) = 0 and f1(2) = 0

=> - 1 - 2b + a = 0

=>b = -1/4

and 1/2 +4b + a = 0

=> a= 1/2

f1(x) = -1/x2 + 2b = - 1/x2 - 1/2 < 0

=> f has a local maximum of x = -1, x= 2

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