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1) SHow that the roots of the eqn x4+4x3-8x2+k=0 are real for k belonging to [0,3].
2) Show that the eqn log(x2+2ax) = log(8x-6a-3) have only one solution for a=1 or a belonging to [-1/2 , -3/22].
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3 Answers
62second quesiton is simple.. take equality.. then check the domain...
661) Let
f(x)=x^4+4x^3-8x^2+k
This function attains an extremum where
f^\prime(x)=4x^3+12x^2-16x=0
i.e. 4x(x^2+3x-4)=0\ \Rightarrow \ x=0, 1, -4
The second derivative
f^{\prime\prime}(x)=4(3x^2+6x-4)
So,
f^{\prime\prime}(0)<0,\ f^{\prime\prime}(1)>0, \ f^{\prime\prime}(-4)>0
Therefore, x=0 is a local maximum while x=1 and x=-4 are local minimum. Further,
f(0)=k,\quad f(1)=k-3,\quad f(-4)=k-128
The graph of y=f(x) looks something like this
We see that for the graph to cut the real axis at four distinct points (meaning four real distinct roots), the local maximum must be positive and the greater of the local minimum must remain negative. That means
f(0)>0, \quad f(1)<0
which gives k > 0 and k < 3. So that k\in(0,3)
(There won't be closed interval as given in the problem)