    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]  Suggest Corrections  0      Similar questions  Explore more