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Question

If family of straight lines ax+by+c=0 always passes through a fixed point (32,1), then equation 36ax2+8bx+2c=0 has

A
at least one root in [0,1]
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B
atleast one root in [12,12]
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C
atleast one root in [1,2]
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D
atleast one root in [0, 12]
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Solution

The correct options are
A at least one root in [0,1]
B atleast one root in [12,12]
C atleast one root in [1,2]
D atleast one root in [0, 12]
ax+by+c=0 passes through (32,1)
3a+2b+2c=0
32a+b+c=0
Let f(x)=12ax3+4bx2+2cx+d
such that f(x)=36ax2+8bx+2c
f(x) is continuous and differentiable
f(0)=d,
f(12)=32a+b+c+d=d
f(0)=f(12)
From above we see that f(x) satisfies all the conditions of Rolle's theorem
So, according to Rolle's theorem there is at least one root in [0,12]

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