88733
The equations of the lines on which the perpendiculars from the origin make \(30^{\circ}\) angle with \(x\)-axis and which form a triangle of area \(\frac{50}{\sqrt{3}}\) with axes, are
1 \(x+\sqrt{3} y \pm 10=0\)
2 \(\sqrt{3} x+y \pm 10=0\)
3 \(x \pm \sqrt{3} y-10=0\)
4 None of above
Explanation:
(B) : Let \(\mathrm{AB}\) be the line which perpendicular to \(P\) drawn from the origin on the line
Given \(\quad \alpha=30^{\circ}\)
the equation of the normal form is, \(\mathrm{x} \cos 30^{\circ}+\mathrm{y} \sin 30^{\circ}=\mathrm{P}\)
\(\sqrt{3} \mathrm{x}+\mathrm{y}=2 \mathrm{P}\)
\(\sqrt{3} \mathrm{x}+\mathrm{y}-2 \mathrm{P}=0\)
Area \(=50 \sqrt{3}\)
(Given)
This meet \(\mathrm{x}\) - axis in point \(\left[\frac{2 \mathrm{P}}{\sqrt{3}}, 0\right]\) ie point \(\mathrm{A}\).
This meet \(\mathrm{y}\)-axis in point \((0,2 \mathrm{P})\) i.e point \(\mathrm{B}\)
Area of triangle \(=\frac{1}{2} \times\left(\frac{2 \mathrm{P}}{\sqrt{3}}\right) \times 2 \mathrm{P}\)
\(\frac{50}{\sqrt{3}}=\frac{2 \mathrm{P}^{2}}{\sqrt{3}}\)
\(\mathrm{P}= \pm 5\)
\(\therefore\) The equation of line is \(\sqrt{3} \mathrm{x}+\mathrm{y} \pm 10=0\)
UPSEE-2010
Straight Line
88734
If the line \(p x-q y=r\) intersects the coordinate axes at \((a, 0)\) and \((0, b)\) then the value of \((a+b)\) is equal to
(B): The equation of line,
\(p x-q y=r\)
According to question
\(p \cdot a-q \cdot 0=r\)
\(a p=r \tag{i}\)
\(p \cdot 0-q \cdot b=r\)
\(-b q=r \tag{ii}\)
Value of \(a+b=r / p+(-r / q)\)
\(=r(1 / p-1 / q)=r\left(\frac{q-p}{p q}\right)\)
Shift-II]
Straight Line
88735
if \(2 a+3 b+6 c=0\), then the equation \(a x^{2}+b x+\) \(\mathbf{c}=0\) has at least one real root in
1 \((0,1)\)
2 \(\left(0, \frac{1}{2}\right)\)
3 \(\left(\frac{1}{4}, \frac{1}{2}\right)\)
4 None of these
Explanation:
(A) : Given, \(2 \mathrm{a}+3 \mathrm{~b}+6 \mathrm{c}=0\)
Let, \(\quad f^{\prime}(x)=a x^{2}+b x+c\)
\(f(x)=\frac{a x^{3}}{3}+\frac{b x^{2}}{2}+c x+d\)
\(f(x)=\frac{2 a x^{3}+3 b x^{2}+6 c x+6 d}{6}\)
Now,
\(f(1)=\frac{2 a+3 b+6 c+6 d}{6}\)
\(=\frac{6 \mathrm{~d}}{6}=\mathrm{d} \quad\left[\begin{array}{c}\because 2 \mathrm{a}+3 \mathrm{~b}+6 \mathrm{c}=0 \\ \text { given }\end{array}\right]\)
and,
\(f(0)=\frac{6 d}{6}=d\)
\(\therefore \quad \mathrm{f}(0)=\mathrm{f}(1)\)
\(\mathrm{f}^{\prime}(\mathrm{x})\) will vanish at least once between 0 and 1 (by Relle's theorem)
Therefore, one of the roots of the equation \(\mathrm{ax}^{2}+\mathrm{bx}+\mathrm{c}\) \(=0\) lies between 0 and 1 .
Manipal UGET-2020
Straight Line
88736
The equations of the lines passing through the point
\((1,0)\) and at a distance \(\frac{\sqrt{3}}{2}\) from the origin are
(A) : Let slope of a line be \(m\).
Now, the equation of a line passing through \((1,0)\) is
\(y-0=m(x-1)\)
\(\Rightarrow \quad \mathrm{mx}-\mathrm{y}-\mathrm{m}=0\)
Distance from origin \(=\frac{\sqrt{3}}{2}\)
\(\Rightarrow \quad \frac{|-\mathrm{m}|}{\sqrt{1+\mathrm{m}^{2}}}=\frac{\sqrt{3}}{2}\)
\(\Rightarrow \quad 4 \mathrm{~m}^{2}=3\left(1+\mathrm{m}^{2}\right)\)
\(\Rightarrow \quad \mathrm{m}^{2}=3\)
\(\Rightarrow \quad \mathrm{m}= \pm \sqrt{3}\)
\(\therefore\) Equation of lines are
\(\sqrt{3} x-y-\sqrt{3}=0\)
and \(\quad-\sqrt{3} x-y+\sqrt{3}=0\)
\(\Rightarrow \quad \sqrt{3} \mathrm{x}-\mathrm{y}-\sqrt{3}=0\)
and \(\quad \sqrt{3} x+y-\sqrt{3}=0\)
hence, option (a) is correct.
88733
The equations of the lines on which the perpendiculars from the origin make \(30^{\circ}\) angle with \(x\)-axis and which form a triangle of area \(\frac{50}{\sqrt{3}}\) with axes, are
1 \(x+\sqrt{3} y \pm 10=0\)
2 \(\sqrt{3} x+y \pm 10=0\)
3 \(x \pm \sqrt{3} y-10=0\)
4 None of above
Explanation:
(B) : Let \(\mathrm{AB}\) be the line which perpendicular to \(P\) drawn from the origin on the line
Given \(\quad \alpha=30^{\circ}\)
the equation of the normal form is, \(\mathrm{x} \cos 30^{\circ}+\mathrm{y} \sin 30^{\circ}=\mathrm{P}\)
\(\sqrt{3} \mathrm{x}+\mathrm{y}=2 \mathrm{P}\)
\(\sqrt{3} \mathrm{x}+\mathrm{y}-2 \mathrm{P}=0\)
Area \(=50 \sqrt{3}\)
(Given)
This meet \(\mathrm{x}\) - axis in point \(\left[\frac{2 \mathrm{P}}{\sqrt{3}}, 0\right]\) ie point \(\mathrm{A}\).
This meet \(\mathrm{y}\)-axis in point \((0,2 \mathrm{P})\) i.e point \(\mathrm{B}\)
Area of triangle \(=\frac{1}{2} \times\left(\frac{2 \mathrm{P}}{\sqrt{3}}\right) \times 2 \mathrm{P}\)
\(\frac{50}{\sqrt{3}}=\frac{2 \mathrm{P}^{2}}{\sqrt{3}}\)
\(\mathrm{P}= \pm 5\)
\(\therefore\) The equation of line is \(\sqrt{3} \mathrm{x}+\mathrm{y} \pm 10=0\)
UPSEE-2010
Straight Line
88734
If the line \(p x-q y=r\) intersects the coordinate axes at \((a, 0)\) and \((0, b)\) then the value of \((a+b)\) is equal to
(B): The equation of line,
\(p x-q y=r\)
According to question
\(p \cdot a-q \cdot 0=r\)
\(a p=r \tag{i}\)
\(p \cdot 0-q \cdot b=r\)
\(-b q=r \tag{ii}\)
Value of \(a+b=r / p+(-r / q)\)
\(=r(1 / p-1 / q)=r\left(\frac{q-p}{p q}\right)\)
Shift-II]
Straight Line
88735
if \(2 a+3 b+6 c=0\), then the equation \(a x^{2}+b x+\) \(\mathbf{c}=0\) has at least one real root in
1 \((0,1)\)
2 \(\left(0, \frac{1}{2}\right)\)
3 \(\left(\frac{1}{4}, \frac{1}{2}\right)\)
4 None of these
Explanation:
(A) : Given, \(2 \mathrm{a}+3 \mathrm{~b}+6 \mathrm{c}=0\)
Let, \(\quad f^{\prime}(x)=a x^{2}+b x+c\)
\(f(x)=\frac{a x^{3}}{3}+\frac{b x^{2}}{2}+c x+d\)
\(f(x)=\frac{2 a x^{3}+3 b x^{2}+6 c x+6 d}{6}\)
Now,
\(f(1)=\frac{2 a+3 b+6 c+6 d}{6}\)
\(=\frac{6 \mathrm{~d}}{6}=\mathrm{d} \quad\left[\begin{array}{c}\because 2 \mathrm{a}+3 \mathrm{~b}+6 \mathrm{c}=0 \\ \text { given }\end{array}\right]\)
and,
\(f(0)=\frac{6 d}{6}=d\)
\(\therefore \quad \mathrm{f}(0)=\mathrm{f}(1)\)
\(\mathrm{f}^{\prime}(\mathrm{x})\) will vanish at least once between 0 and 1 (by Relle's theorem)
Therefore, one of the roots of the equation \(\mathrm{ax}^{2}+\mathrm{bx}+\mathrm{c}\) \(=0\) lies between 0 and 1 .
Manipal UGET-2020
Straight Line
88736
The equations of the lines passing through the point
\((1,0)\) and at a distance \(\frac{\sqrt{3}}{2}\) from the origin are
(A) : Let slope of a line be \(m\).
Now, the equation of a line passing through \((1,0)\) is
\(y-0=m(x-1)\)
\(\Rightarrow \quad \mathrm{mx}-\mathrm{y}-\mathrm{m}=0\)
Distance from origin \(=\frac{\sqrt{3}}{2}\)
\(\Rightarrow \quad \frac{|-\mathrm{m}|}{\sqrt{1+\mathrm{m}^{2}}}=\frac{\sqrt{3}}{2}\)
\(\Rightarrow \quad 4 \mathrm{~m}^{2}=3\left(1+\mathrm{m}^{2}\right)\)
\(\Rightarrow \quad \mathrm{m}^{2}=3\)
\(\Rightarrow \quad \mathrm{m}= \pm \sqrt{3}\)
\(\therefore\) Equation of lines are
\(\sqrt{3} x-y-\sqrt{3}=0\)
and \(\quad-\sqrt{3} x-y+\sqrt{3}=0\)
\(\Rightarrow \quad \sqrt{3} \mathrm{x}-\mathrm{y}-\sqrt{3}=0\)
and \(\quad \sqrt{3} x+y-\sqrt{3}=0\)
hence, option (a) is correct.
88733
The equations of the lines on which the perpendiculars from the origin make \(30^{\circ}\) angle with \(x\)-axis and which form a triangle of area \(\frac{50}{\sqrt{3}}\) with axes, are
1 \(x+\sqrt{3} y \pm 10=0\)
2 \(\sqrt{3} x+y \pm 10=0\)
3 \(x \pm \sqrt{3} y-10=0\)
4 None of above
Explanation:
(B) : Let \(\mathrm{AB}\) be the line which perpendicular to \(P\) drawn from the origin on the line
Given \(\quad \alpha=30^{\circ}\)
the equation of the normal form is, \(\mathrm{x} \cos 30^{\circ}+\mathrm{y} \sin 30^{\circ}=\mathrm{P}\)
\(\sqrt{3} \mathrm{x}+\mathrm{y}=2 \mathrm{P}\)
\(\sqrt{3} \mathrm{x}+\mathrm{y}-2 \mathrm{P}=0\)
Area \(=50 \sqrt{3}\)
(Given)
This meet \(\mathrm{x}\) - axis in point \(\left[\frac{2 \mathrm{P}}{\sqrt{3}}, 0\right]\) ie point \(\mathrm{A}\).
This meet \(\mathrm{y}\)-axis in point \((0,2 \mathrm{P})\) i.e point \(\mathrm{B}\)
Area of triangle \(=\frac{1}{2} \times\left(\frac{2 \mathrm{P}}{\sqrt{3}}\right) \times 2 \mathrm{P}\)
\(\frac{50}{\sqrt{3}}=\frac{2 \mathrm{P}^{2}}{\sqrt{3}}\)
\(\mathrm{P}= \pm 5\)
\(\therefore\) The equation of line is \(\sqrt{3} \mathrm{x}+\mathrm{y} \pm 10=0\)
UPSEE-2010
Straight Line
88734
If the line \(p x-q y=r\) intersects the coordinate axes at \((a, 0)\) and \((0, b)\) then the value of \((a+b)\) is equal to
(B): The equation of line,
\(p x-q y=r\)
According to question
\(p \cdot a-q \cdot 0=r\)
\(a p=r \tag{i}\)
\(p \cdot 0-q \cdot b=r\)
\(-b q=r \tag{ii}\)
Value of \(a+b=r / p+(-r / q)\)
\(=r(1 / p-1 / q)=r\left(\frac{q-p}{p q}\right)\)
Shift-II]
Straight Line
88735
if \(2 a+3 b+6 c=0\), then the equation \(a x^{2}+b x+\) \(\mathbf{c}=0\) has at least one real root in
1 \((0,1)\)
2 \(\left(0, \frac{1}{2}\right)\)
3 \(\left(\frac{1}{4}, \frac{1}{2}\right)\)
4 None of these
Explanation:
(A) : Given, \(2 \mathrm{a}+3 \mathrm{~b}+6 \mathrm{c}=0\)
Let, \(\quad f^{\prime}(x)=a x^{2}+b x+c\)
\(f(x)=\frac{a x^{3}}{3}+\frac{b x^{2}}{2}+c x+d\)
\(f(x)=\frac{2 a x^{3}+3 b x^{2}+6 c x+6 d}{6}\)
Now,
\(f(1)=\frac{2 a+3 b+6 c+6 d}{6}\)
\(=\frac{6 \mathrm{~d}}{6}=\mathrm{d} \quad\left[\begin{array}{c}\because 2 \mathrm{a}+3 \mathrm{~b}+6 \mathrm{c}=0 \\ \text { given }\end{array}\right]\)
and,
\(f(0)=\frac{6 d}{6}=d\)
\(\therefore \quad \mathrm{f}(0)=\mathrm{f}(1)\)
\(\mathrm{f}^{\prime}(\mathrm{x})\) will vanish at least once between 0 and 1 (by Relle's theorem)
Therefore, one of the roots of the equation \(\mathrm{ax}^{2}+\mathrm{bx}+\mathrm{c}\) \(=0\) lies between 0 and 1 .
Manipal UGET-2020
Straight Line
88736
The equations of the lines passing through the point
\((1,0)\) and at a distance \(\frac{\sqrt{3}}{2}\) from the origin are
(A) : Let slope of a line be \(m\).
Now, the equation of a line passing through \((1,0)\) is
\(y-0=m(x-1)\)
\(\Rightarrow \quad \mathrm{mx}-\mathrm{y}-\mathrm{m}=0\)
Distance from origin \(=\frac{\sqrt{3}}{2}\)
\(\Rightarrow \quad \frac{|-\mathrm{m}|}{\sqrt{1+\mathrm{m}^{2}}}=\frac{\sqrt{3}}{2}\)
\(\Rightarrow \quad 4 \mathrm{~m}^{2}=3\left(1+\mathrm{m}^{2}\right)\)
\(\Rightarrow \quad \mathrm{m}^{2}=3\)
\(\Rightarrow \quad \mathrm{m}= \pm \sqrt{3}\)
\(\therefore\) Equation of lines are
\(\sqrt{3} x-y-\sqrt{3}=0\)
and \(\quad-\sqrt{3} x-y+\sqrt{3}=0\)
\(\Rightarrow \quad \sqrt{3} \mathrm{x}-\mathrm{y}-\sqrt{3}=0\)
and \(\quad \sqrt{3} x+y-\sqrt{3}=0\)
hence, option (a) is correct.
88733
The equations of the lines on which the perpendiculars from the origin make \(30^{\circ}\) angle with \(x\)-axis and which form a triangle of area \(\frac{50}{\sqrt{3}}\) with axes, are
1 \(x+\sqrt{3} y \pm 10=0\)
2 \(\sqrt{3} x+y \pm 10=0\)
3 \(x \pm \sqrt{3} y-10=0\)
4 None of above
Explanation:
(B) : Let \(\mathrm{AB}\) be the line which perpendicular to \(P\) drawn from the origin on the line
Given \(\quad \alpha=30^{\circ}\)
the equation of the normal form is, \(\mathrm{x} \cos 30^{\circ}+\mathrm{y} \sin 30^{\circ}=\mathrm{P}\)
\(\sqrt{3} \mathrm{x}+\mathrm{y}=2 \mathrm{P}\)
\(\sqrt{3} \mathrm{x}+\mathrm{y}-2 \mathrm{P}=0\)
Area \(=50 \sqrt{3}\)
(Given)
This meet \(\mathrm{x}\) - axis in point \(\left[\frac{2 \mathrm{P}}{\sqrt{3}}, 0\right]\) ie point \(\mathrm{A}\).
This meet \(\mathrm{y}\)-axis in point \((0,2 \mathrm{P})\) i.e point \(\mathrm{B}\)
Area of triangle \(=\frac{1}{2} \times\left(\frac{2 \mathrm{P}}{\sqrt{3}}\right) \times 2 \mathrm{P}\)
\(\frac{50}{\sqrt{3}}=\frac{2 \mathrm{P}^{2}}{\sqrt{3}}\)
\(\mathrm{P}= \pm 5\)
\(\therefore\) The equation of line is \(\sqrt{3} \mathrm{x}+\mathrm{y} \pm 10=0\)
UPSEE-2010
Straight Line
88734
If the line \(p x-q y=r\) intersects the coordinate axes at \((a, 0)\) and \((0, b)\) then the value of \((a+b)\) is equal to
(B): The equation of line,
\(p x-q y=r\)
According to question
\(p \cdot a-q \cdot 0=r\)
\(a p=r \tag{i}\)
\(p \cdot 0-q \cdot b=r\)
\(-b q=r \tag{ii}\)
Value of \(a+b=r / p+(-r / q)\)
\(=r(1 / p-1 / q)=r\left(\frac{q-p}{p q}\right)\)
Shift-II]
Straight Line
88735
if \(2 a+3 b+6 c=0\), then the equation \(a x^{2}+b x+\) \(\mathbf{c}=0\) has at least one real root in
1 \((0,1)\)
2 \(\left(0, \frac{1}{2}\right)\)
3 \(\left(\frac{1}{4}, \frac{1}{2}\right)\)
4 None of these
Explanation:
(A) : Given, \(2 \mathrm{a}+3 \mathrm{~b}+6 \mathrm{c}=0\)
Let, \(\quad f^{\prime}(x)=a x^{2}+b x+c\)
\(f(x)=\frac{a x^{3}}{3}+\frac{b x^{2}}{2}+c x+d\)
\(f(x)=\frac{2 a x^{3}+3 b x^{2}+6 c x+6 d}{6}\)
Now,
\(f(1)=\frac{2 a+3 b+6 c+6 d}{6}\)
\(=\frac{6 \mathrm{~d}}{6}=\mathrm{d} \quad\left[\begin{array}{c}\because 2 \mathrm{a}+3 \mathrm{~b}+6 \mathrm{c}=0 \\ \text { given }\end{array}\right]\)
and,
\(f(0)=\frac{6 d}{6}=d\)
\(\therefore \quad \mathrm{f}(0)=\mathrm{f}(1)\)
\(\mathrm{f}^{\prime}(\mathrm{x})\) will vanish at least once between 0 and 1 (by Relle's theorem)
Therefore, one of the roots of the equation \(\mathrm{ax}^{2}+\mathrm{bx}+\mathrm{c}\) \(=0\) lies between 0 and 1 .
Manipal UGET-2020
Straight Line
88736
The equations of the lines passing through the point
\((1,0)\) and at a distance \(\frac{\sqrt{3}}{2}\) from the origin are
(A) : Let slope of a line be \(m\).
Now, the equation of a line passing through \((1,0)\) is
\(y-0=m(x-1)\)
\(\Rightarrow \quad \mathrm{mx}-\mathrm{y}-\mathrm{m}=0\)
Distance from origin \(=\frac{\sqrt{3}}{2}\)
\(\Rightarrow \quad \frac{|-\mathrm{m}|}{\sqrt{1+\mathrm{m}^{2}}}=\frac{\sqrt{3}}{2}\)
\(\Rightarrow \quad 4 \mathrm{~m}^{2}=3\left(1+\mathrm{m}^{2}\right)\)
\(\Rightarrow \quad \mathrm{m}^{2}=3\)
\(\Rightarrow \quad \mathrm{m}= \pm \sqrt{3}\)
\(\therefore\) Equation of lines are
\(\sqrt{3} x-y-\sqrt{3}=0\)
and \(\quad-\sqrt{3} x-y+\sqrt{3}=0\)
\(\Rightarrow \quad \sqrt{3} \mathrm{x}-\mathrm{y}-\sqrt{3}=0\)
and \(\quad \sqrt{3} x+y-\sqrt{3}=0\)
hence, option (a) is correct.
