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If a,b,c,d are Real number such that $\frac{3a+2b}{c+d}+\frac{3}{2}=0$. Then the equation $ax^3+bx^2+cx+d=0$ has

(1) at least one root in [-2,0]

(2) at least one root in [0,2]

(3) at least two root in [-2,2]

(4) no root in [-2,2]

I am doing hit and trial method by using $f'(x)=0$, put x=1, we get $3a+2b+c=0$, putting $3a+2b=-c$ in $\frac{3a+2b}{c+d}+\frac{3}{2}=0$, i get relation between c & d, but not able to proceed.

Samar Imam Zaidi
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2 Answers2

2

Let $$f(x)=\frac{ax^4}{4}+\frac{bx^3}{3}+\frac{cx^2}{2}+dx.$$ Then $f0)=0$ and $f(2)=\frac{16a}{4}+\frac{8b}{3}+\frac{4c}{2}+2d=\frac{48a+32b+24c+24d}{12}=\frac{16(3a+2b)+24(c+d)}{12}=0$ (from the condition given).

Now apply Rolle's theorem. There exists a point in $[0,2]$, where the derivative of $f$ must be zero.

Anurag A
  • 41,067
1

Finding f'(x)=0 gives you a point of maxima or minima which need not be the root of the equation.

From Rolle's theorem we know that is the function in continuous and diffrentiable in [a,b] and if f(a)=f(b), then there exists a c $\epsilon$(a,b) such that f'(c)=0.

So integrate the given equation and find for which of the intervals f(a)=f(b) then that implies there is atleast a root of the given equation.

Integrating,we get

$\frac{a(x)^4}{4}$+$\frac{b(x)^3}{3}$ +$\frac{c(x)^2}{2}$+dx=F(x)

F(0)=0

F(2)=4a+$\frac{8}{3}b$+6c+6d

But by the given condition,

6a+4b+3c+3d=0

Hence F(2)=0

which implies there is at least one c between 0 and 2 such that F'(c)=f(c)=0 which implies there is atleast one root between 0 and 2.

So option B is what you are looking for.

Vinanth S Bharadwaj
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