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 option is 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]