If the x intercept of the line y=mx+2 is greater than 12, then the gradient of the line lies in the interval

for X intercept ⇒ Put y=0⇒mx+2=0⇒x=−2m −2m>12 −2m−12>0 −(4+m2m)>0⇒−4<m<0

If the x intercept of the line y=mx+2 is greater than 12, then the gradient of the line lies in the interval

for X intercept ⇒ Put y=0⇒mx+2=0⇒x=−2m −2m>12 −2m−12>0 −(4+m2m)>0⇒−4<m<0

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If the $x$ intercept of the line $y = m x + 2$ is greater than $\frac{1}{2},$ then the gradient of the line lies in the interval

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$(- 1, 0)$

$(\frac{- 1}{4}, 0)$

$(- \infty, - 4)$

$(- 4, 0)$

answer is D.

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Detailed Solution

for $X$ intercept $\Rightarrow$ Put $y = 0$

$\Rightarrow m x + 2 = 0 \Rightarrow x = \frac{- 2}{m} \frac{- 2}{m} > \frac{1}{2} \frac{- 2}{m} - \frac{1}{2} > 0 - (\frac{4 + m}{2 m}) > 0 \Rightarrow - 4 < m < 0$

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If the x intercept of the line y=mx+2 is greater than 12, then the gradient of the line lies in the interval

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If the $x$ intercept of the line $y = m x + 2$ is greater than $\frac{1}{2},$ then the gradient of the line lies in the interval