Let a,b∈R be such that the function f given by f(x)=log⁡|x|+bx2+ax,x≠0 has extreme values at x=–1 and x=2 .Statement I : f has local maximum at x=–1 and x=2.Statement II : a=12 and b=–14

f(x)=log⁡|x|+bx2+ax,x≠0 f'(x)=1x+2bx+af1(−1)=0⇒-1-2b+a=0----(1) f1(2)=0⇒12+4b+a=0-----(2) solving 1 & 2 a=12, b=-14

Let a,b∈R be such that the function f given by f(x)=log⁡|x|+bx2+ax,x≠0 has extreme values at x=–1 and x=2 .Statement I : f has local maximum at x=–1 and x=2.Statement II : a=12 and b=–14

f(x)=log⁡|x|+bx2+ax,x≠0 f'(x)=1x+2bx+af1(−1)=0⇒-1-2b+a=0----(1) f1(2)=0⇒12+4b+a=0-----(2) solving 1 & 2 a=12, b=-14

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Let $a, b \in R$ be such that the function f given by $f (x) = \log | x | + b x^{2} + a x, x \neq 0$ has extreme values at $x = - 1$ and $x = 2$ .
Statement I : $f$ has local maximum at $x = - 1$ and $x = 2$ .
Statement II : $a = \frac{1}{2}$ and $b = - \frac{1}{4}$

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I is false and II is true

Both I and II are true and II is the correct explanation of I

I is true and II is false

Both I and II are true and II is not the correct explanation of I

answer is C.

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Detailed Solution

$f (x) = \log | x | + b x^{2} + a x, x \neq 0 f' (x) = \frac{1}{x} + 2 b x + a$

$f^{1} (- 1) = 0 \Rightarrow - 1 - 2 b + a = 0 - - - - (1) f^{1} (2) = 0 \Rightarrow \frac{1}{2} + 4 b + a = 0 - - - - - (2) s o l v i n g 1 & 2 a = \frac{1}{2}, b = - \frac{1}{4}$

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Let a,b∈R be such that the function f given by f(x)=log⁡|x|+bx2+ax,x≠0 has extreme values at x=–1 and x=2 .Statement I : f has local maximum at x=–1 and x=2.Statement II : a=12 and b=–14

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Let $a, b \in R$ be such that the function f given by $f (x) = \log | x | + b x^{2} + a x, x \neq 0$ has extreme values at $x = - 1$ and $x = 2$ .
Statement I : $f$ has local maximum at $x = - 1$ and $x = 2$ .
Statement II : $a = \frac{1}{2}$ and $b = - \frac{1}{4}$