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Differential Calculus

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23 Aug 2009 16:45:10 IST

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344

App of deriviatives

None

Ques1) Let S be the set of real values of parameter k for which te eqn f(x) = 2x ³ - 3(2+k)x ² +12kx has exactly one local minimum and exactly one local maximum. S is a subset of
(a) (-4, ∞) (b) (-3,3) (c) (3,∞) (d) (- ∞,0)

Ques2) If f(x) = a sinx +1/3 Sin3x has an extremum at x = 2pie/3 then
(a) a=2 (b) f(2pie/3) is max for a=2 (c) f(2pie/3) is min for a=2 (D) there are three critical points b/w (0,piE)

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Comments (4)

Vineet Tanna

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Joined: 18 Aug 2009

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23 Aug 2009 23:54:28 IST

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f'(x)=6x^2-6x(2+k)+12ksince it has one max and one min,6x^2-6x(2+k)+12k=0 should have 2 real roots,Therefore x^2-x(2+k)+2k=0 should have 2 real roots hence D>0(2+k)^2-8k>0Hence k>2...

kabi

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Joined: 11 Jan 2009

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25 Aug 2009 18:08:27 IST

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for first

@ vineet . U r doing a mistake here .

D>0

(2+k)²-8k >0

(k-2)² >0

which is true for every valueof k except when k= 2

a , b option contain k =2 so these are out .

so c, d are the correct option .

second-

differentiate ::

acosx +4cos³x-3cosx = 0

cos(x) [ a +4cos²x -3 ] =0

a + 4 (1/4 ) -3 = 0

a = 2 ,

Max .

three critical points if a =2 , x = pi/3 ,2pi/3 , pi/2

edison

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28 Oct 2009 16:22:46 IST

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well done kabi

Aakash Anuj

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28 Oct 2009 23:16:39 IST

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a,b, and d for 2nd qstn. Differentiate and put f'(x)=0 and you will get a,b d on solving

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