IIT JEE , AIEEE Forum - Mathematics , Differential Calculus

kislay

let f(x) = ax³ + bx² +cx + dsinx.find the condition that f(x) is always one - one function..

ans.

b²<3a(c-|d|)

but accord. to me it sud be
b²<3a(c+|d|).
pls solve

kishan12

here we have to take care to see that the graph doesnot take a turn...that is dy/dx is not zero.
So we get 2 possibility either greater than or less than zero.
so,
3ax^2+2bx+c+dcosx is not equal to zero.
b^2<>3a(c+dcosx)
cosx<>b^2-3ac/3ad
Put the codition that cos x lies between -1 and 1
u wud get 3equations for inequality..one for< and 2 for >
take intersection of the two and get ans:b^2,3a(c-IdI)
if u have any douts be free to nudge me..

kislay

hey kishan hoe u hav used
3ax^2+2bx+c+dcosx is not equal to zero.
b^2<>3a(c+dcosx)
cosx<>b^2-3ac/3ad

the equation u have used to derive the condition is not a polynomial in x.

kishan12

See when in any function,a graph turns,that is it may have maxima or minima,then for 2 values of x we get a value of y.
so not oneone function.
hence the graph should not have any maxima or minima..so the slope of tangent at any point cannot be zero.
therefore...............
and ya for the second part..i think u r correct yaar..its not a polynomial in x that we can apply delta...but there comes a method na that shift the trigonometric part on the right side then draw the graph of the quadratic and of the trigo part........
i will confirm it and tell u tonight.

puneet

hey

see the question is simple .. n answer is correct ..

kishan12 has solved it correctly ...

cheers

avdesh100

the both the ans is right in fact u get b^2=3ad for d=0 also we get three cases in all for it to be one-one
1 case:for a<0
2 case :a>0 and d not equal 2 zero in above 2 cases
3 case: d=0
hope its clear.....

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