IIT JEE , AIEEE Forum - Mathematics , Algebra

Earth_saver

Let mode of p be less than or equal to 1. show dat eqn

4x(cube)-3x-p=0 has a unique root in interval [1/2,1] & identify it.

sandeepramesh

Its very easy to show the existence of a root :)

Note that f(1/2) = -1-p <=0 and f(1) = 1-p >= 0 hence the result :)

akhil_o

the function is continuous everywhere
now at x=1/2

f(x)=4(1/8)-3(1/2)-p=-(p+1)
since |p|<1
f(1/2)<0

now at x=1
f(x)=4-3-p
=1-p

but |p|<1
so f(x)>0

since it is a continuous fx, it takes all values from f(1/2) to f(1)
and since f(1)>0, f(1/2)<0
atleast 1 root exists between 1/2 and 1

akhil_o

sorry sandeepramesh posted already

Earth_saver

But u have to FIND the value of the root too.

Earth_saver

Hey! someone please explain.

The ans is cos{1/3cos^{-1 _{(p)}. How can we know if no options r given???}}

akhil_o

writeit as
p=4x(cube)-3x

now x=cos A
4 (cosA)^3-3cos A=cos 3A
so
RHS= cos 3A
p= cos 3A
cos inv p= 3A
A=1/3 cos inv erse p

but x=cos A
=cos(1/3(cos inverse p))

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