# Joint Entrance Examination/AIEEE 2009/Q014 calculus

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| Original Question by 164.100.170.4 Here's the question... Given P(x)=x$^4$+ax$^3$+bx$^2$+cx+d such that x=0 is the only real root of P'(x)=0 . If P(-1)<P(1) , then in the interval [-1,1], (1)P(-1) neither is the minimum nor P(1) is the maximum of P (2) P(-1) is the minimum and P(1) is the maximum of P (3) P(-1) is not minimum but P(1) is the maximum of P (4) P(-1) is the minimum but P(1) is not the maximum of P Posted by 164.100.170.4 05:42, 27 September 2009 (UTC) Post a reply! | What's Up? Latest discussions on the forum Calculus-Help User:Thewinster/HHF/Calculus-Preload Joint Entrance Examination/AIEEE 2009/Q014 calculus Mathematics-Help Category:Calculus-Help Category:Vectors-3D-Geometry-Help Category:Misc-Math-Topics-Help User:Thewinster/Sandbox/Forum-Watch User:Thewinster/Sandbox |
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Reply to the thread, posted by 164.100.170.4
Thread post
Additional notes
ans:[3]
P(x)=x$^4$+ax$^3$+bx$^2$+cx+d
Now P'(0)= 0 => c=0.
=>P(x)= P(x)=x$^4$+ax$^3$+bx$^2$+d
clearly we see that P(1) is maximum but P(-1) is not minimum.
Clearly is maximum but is not minimum.
Correct choice: (3)
These comments, along with the notes on the left, were contributed by 164.100.170.4 05:48, 27 September 2009 (UTC) :