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Question

If 3(a+2c)=4(b+3d), then the equation ax3+bx2+cx+d=0 will have

A
no real solution
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B
atleast one real root in (1,0)
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C
atleast one real root in (0,1)
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D
none of these
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Solution

The correct option is B atleast one real root in (1,0)
Let f(x)=ax44+bx33+cx22+dx
Clearly, f(x) is continuous on [-1,0] and differentiable in (-1,0)
Now, f(0)=0
and f(1)=a4b3+c2d

=a+2c4b+3d3
=0
f(0)=f(1)=0
Hence, all the condition of Rolle's theorem are satisfied, so there exists at least one root of f(x)=0 in [1,0]
and f(x)=ax3+bx2+cx+d
So, it has at least one root in [1,0]

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