NEET Test Series from KOTA - 10 Papers In MS WORD
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Application of the Integrals
87025
The three lies of a triangle are give by \(\left(x^{2}-y^{2}\right)(2 x+3 y-6)=0\). If the point \((-2, \lambda)\) lies inside and \((\mu, 1)\) lies outside the triangle, then
(D) : Three lines of triangle are given by \(\left(x^{2}-y^{2}\right)(2 x+3 y-6)=0\) \(\Rightarrow \quad(x-y)(x+y)(2 x+3 y-6)=0\) \(\therefore\) The three lines of triangle are \(x-y=0, x+y=0\) and \(\quad 2 x+3 y-6=0\) From given lines of triangle, the required \(\triangle \mathrm{OAB}\) is formed. \(\because(-2, \lambda)\) lines inside the triangle. \(\therefore 2(-2)+3(\lambda)-6\lt 0\) and \(-2+\lambda>0\) \(\Rightarrow \quad-4+3 \lambda-6\lt 0\) and \(\lambda>2\) \(\Rightarrow \quad 3 \lambda\lt 10\) and \(\lambda>2\) \(\Rightarrow \quad \lambda\lt \frac{10}{3}\) and \(\lambda>2\) \(\therefore \quad \lambda \in\left(2, \frac{10}{3}\right)\) Now, \((\mu, 1)\) lies outside the triangle. To find value of \(\mu\), we find the interval \([\mathrm{M}, \mathrm{N}]\) for values of \(\mathrm{x}\). \(\mathrm{x}+1 \geq 0 \text { and } \mathrm{x}-1 \leq 0\) \(\Rightarrow \mathrm{x} \in[-1,1]\) \(\because(\mu, 1)\) lies outside the triangle. \(\therefore \mu \in(-\infty,-1) \cup(1, \infty)\) \(\text { or } \mu \in R-[-1,1]\)
CG PET-2015
Application of the Integrals
87027
Let \(q\) be the maximum integral value of \(p\) in \([0\), 10] for which the roots of the equation \(x^{2}-p x+\frac{5}{4} p=0\) are rational. Then the area of the region \(\left\{(x, y): 0 \leq y \leq(x-q)^{2}, 0 \leq x \leq q\right)\) is
1 164
2 243
3 \(\frac{125}{3}\)
4 25
Explanation:
(B) : Maximum integral value of \(\mathrm{p}=\mathrm{q}\) Roots of equation \(=x^{2}-p x+\frac{5}{4} p=0\) The graphs of curve are as shown in the figure. \(\Rightarrow \quad \mathrm{x}^{2}-\mathrm{px}+\frac{5}{4} \mathrm{p}=0\) \(\mathrm{a}=1, \quad \mathrm{~b}=-\mathrm{p}, \quad \mathrm{c}=\frac{5}{4} \mathrm{p}\) \(D=b^{2}-4 a c=(-p)^{2}-4 \times 1 \times \frac{5}{4} p\) \(=\mathrm{p}^{2}-5 \mathrm{p}=\mathrm{p}(\mathrm{p}-5)\) \(q=9\) Area of shared region \(=\int_{0}^{9}(x-9)^{2} d x\) \(=\left[\frac{(x-9)^{3}}{3}\right]_{0}^{9}=\frac{(+9)^{3}}{3}=243 \text { sq. units }\)
Shift-II
Application of the Integrals
87028
The area of the region given by \(\{(x, y): x y \leq 8\), \(\left.1 \leq \mathrm{y} \leq \mathrm{x}^{2}\right\}\) is
87029
If an equilateral triangle is inscribed in the circle \(x^{2}+y^{2}=a^{2}\), the length of its each side is
1 \(\sqrt{2} \mathrm{a}\)
2 \(\sqrt{3} \mathrm{a}\)
3 \(\frac{\sqrt{3}}{2} \mathrm{a}\)
4 \(\frac{1}{\sqrt{3}} \mathrm{a}\)
Explanation:
(B) : Given, equation of the circle is \(x^{2}+y^{2}=a^{2}\) Where, \(\text { radius }=\mathrm{a}\) Let \(\mathrm{ABC}\) be the inscribed equilateral triangle and length of its each sides be \(l\). Area of \(\triangle \mathrm{ABC}=\frac{\sqrt{3}}{4} l^{2}\) Also, \(\mathrm{OD}=\sqrt{\mathrm{a}^{2}-\frac{l^{2}}{4}}\) [ from \(\Delta \mathrm{ODB}]\) \(\therefore\) Area of \(\triangle \mathrm{OBC}=\frac{1}{2} \times l \times \sqrt{\mathrm{a}^{2}-\frac{l}{4}}\) Now, Area of \(\triangle \mathrm{ABC}=3 \times\) Area of \(\triangle \mathrm{OBC}\) \(\Rightarrow \frac{\sqrt{3}}{4} l^{2}=3 \times \frac{l}{2} \sqrt{\mathrm{a}^{2}-\frac{l^{2}}{4}} \Rightarrow \frac{\sqrt{3}}{2} l=3 \frac{\sqrt{4 \mathrm{a}^{2}-l^{2}}}{2}\) \(\Rightarrow \sqrt{3} l=3 \sqrt{4 \mathrm{a}^{2}-l^{2}} \quad \Rightarrow 3 l^{2}=9\left(4 \mathrm{a}^{2}-l^{2}\right)\) \(\Rightarrow l^{2}=12 \mathrm{a}^{2}-3 l^{2} \quad \Rightarrow 4 l^{2}=12 \mathrm{a}^{2}\) \(\Rightarrow l^{2}=3 \mathrm{a} \quad \Rightarrow l=\sqrt{3} \mathrm{a}\)
87025
The three lies of a triangle are give by \(\left(x^{2}-y^{2}\right)(2 x+3 y-6)=0\). If the point \((-2, \lambda)\) lies inside and \((\mu, 1)\) lies outside the triangle, then
(D) : Three lines of triangle are given by \(\left(x^{2}-y^{2}\right)(2 x+3 y-6)=0\) \(\Rightarrow \quad(x-y)(x+y)(2 x+3 y-6)=0\) \(\therefore\) The three lines of triangle are \(x-y=0, x+y=0\) and \(\quad 2 x+3 y-6=0\) From given lines of triangle, the required \(\triangle \mathrm{OAB}\) is formed. \(\because(-2, \lambda)\) lines inside the triangle. \(\therefore 2(-2)+3(\lambda)-6\lt 0\) and \(-2+\lambda>0\) \(\Rightarrow \quad-4+3 \lambda-6\lt 0\) and \(\lambda>2\) \(\Rightarrow \quad 3 \lambda\lt 10\) and \(\lambda>2\) \(\Rightarrow \quad \lambda\lt \frac{10}{3}\) and \(\lambda>2\) \(\therefore \quad \lambda \in\left(2, \frac{10}{3}\right)\) Now, \((\mu, 1)\) lies outside the triangle. To find value of \(\mu\), we find the interval \([\mathrm{M}, \mathrm{N}]\) for values of \(\mathrm{x}\). \(\mathrm{x}+1 \geq 0 \text { and } \mathrm{x}-1 \leq 0\) \(\Rightarrow \mathrm{x} \in[-1,1]\) \(\because(\mu, 1)\) lies outside the triangle. \(\therefore \mu \in(-\infty,-1) \cup(1, \infty)\) \(\text { or } \mu \in R-[-1,1]\)
CG PET-2015
Application of the Integrals
87027
Let \(q\) be the maximum integral value of \(p\) in \([0\), 10] for which the roots of the equation \(x^{2}-p x+\frac{5}{4} p=0\) are rational. Then the area of the region \(\left\{(x, y): 0 \leq y \leq(x-q)^{2}, 0 \leq x \leq q\right)\) is
1 164
2 243
3 \(\frac{125}{3}\)
4 25
Explanation:
(B) : Maximum integral value of \(\mathrm{p}=\mathrm{q}\) Roots of equation \(=x^{2}-p x+\frac{5}{4} p=0\) The graphs of curve are as shown in the figure. \(\Rightarrow \quad \mathrm{x}^{2}-\mathrm{px}+\frac{5}{4} \mathrm{p}=0\) \(\mathrm{a}=1, \quad \mathrm{~b}=-\mathrm{p}, \quad \mathrm{c}=\frac{5}{4} \mathrm{p}\) \(D=b^{2}-4 a c=(-p)^{2}-4 \times 1 \times \frac{5}{4} p\) \(=\mathrm{p}^{2}-5 \mathrm{p}=\mathrm{p}(\mathrm{p}-5)\) \(q=9\) Area of shared region \(=\int_{0}^{9}(x-9)^{2} d x\) \(=\left[\frac{(x-9)^{3}}{3}\right]_{0}^{9}=\frac{(+9)^{3}}{3}=243 \text { sq. units }\)
Shift-II
Application of the Integrals
87028
The area of the region given by \(\{(x, y): x y \leq 8\), \(\left.1 \leq \mathrm{y} \leq \mathrm{x}^{2}\right\}\) is
87029
If an equilateral triangle is inscribed in the circle \(x^{2}+y^{2}=a^{2}\), the length of its each side is
1 \(\sqrt{2} \mathrm{a}\)
2 \(\sqrt{3} \mathrm{a}\)
3 \(\frac{\sqrt{3}}{2} \mathrm{a}\)
4 \(\frac{1}{\sqrt{3}} \mathrm{a}\)
Explanation:
(B) : Given, equation of the circle is \(x^{2}+y^{2}=a^{2}\) Where, \(\text { radius }=\mathrm{a}\) Let \(\mathrm{ABC}\) be the inscribed equilateral triangle and length of its each sides be \(l\). Area of \(\triangle \mathrm{ABC}=\frac{\sqrt{3}}{4} l^{2}\) Also, \(\mathrm{OD}=\sqrt{\mathrm{a}^{2}-\frac{l^{2}}{4}}\) [ from \(\Delta \mathrm{ODB}]\) \(\therefore\) Area of \(\triangle \mathrm{OBC}=\frac{1}{2} \times l \times \sqrt{\mathrm{a}^{2}-\frac{l}{4}}\) Now, Area of \(\triangle \mathrm{ABC}=3 \times\) Area of \(\triangle \mathrm{OBC}\) \(\Rightarrow \frac{\sqrt{3}}{4} l^{2}=3 \times \frac{l}{2} \sqrt{\mathrm{a}^{2}-\frac{l^{2}}{4}} \Rightarrow \frac{\sqrt{3}}{2} l=3 \frac{\sqrt{4 \mathrm{a}^{2}-l^{2}}}{2}\) \(\Rightarrow \sqrt{3} l=3 \sqrt{4 \mathrm{a}^{2}-l^{2}} \quad \Rightarrow 3 l^{2}=9\left(4 \mathrm{a}^{2}-l^{2}\right)\) \(\Rightarrow l^{2}=12 \mathrm{a}^{2}-3 l^{2} \quad \Rightarrow 4 l^{2}=12 \mathrm{a}^{2}\) \(\Rightarrow l^{2}=3 \mathrm{a} \quad \Rightarrow l=\sqrt{3} \mathrm{a}\)
87025
The three lies of a triangle are give by \(\left(x^{2}-y^{2}\right)(2 x+3 y-6)=0\). If the point \((-2, \lambda)\) lies inside and \((\mu, 1)\) lies outside the triangle, then
(D) : Three lines of triangle are given by \(\left(x^{2}-y^{2}\right)(2 x+3 y-6)=0\) \(\Rightarrow \quad(x-y)(x+y)(2 x+3 y-6)=0\) \(\therefore\) The three lines of triangle are \(x-y=0, x+y=0\) and \(\quad 2 x+3 y-6=0\) From given lines of triangle, the required \(\triangle \mathrm{OAB}\) is formed. \(\because(-2, \lambda)\) lines inside the triangle. \(\therefore 2(-2)+3(\lambda)-6\lt 0\) and \(-2+\lambda>0\) \(\Rightarrow \quad-4+3 \lambda-6\lt 0\) and \(\lambda>2\) \(\Rightarrow \quad 3 \lambda\lt 10\) and \(\lambda>2\) \(\Rightarrow \quad \lambda\lt \frac{10}{3}\) and \(\lambda>2\) \(\therefore \quad \lambda \in\left(2, \frac{10}{3}\right)\) Now, \((\mu, 1)\) lies outside the triangle. To find value of \(\mu\), we find the interval \([\mathrm{M}, \mathrm{N}]\) for values of \(\mathrm{x}\). \(\mathrm{x}+1 \geq 0 \text { and } \mathrm{x}-1 \leq 0\) \(\Rightarrow \mathrm{x} \in[-1,1]\) \(\because(\mu, 1)\) lies outside the triangle. \(\therefore \mu \in(-\infty,-1) \cup(1, \infty)\) \(\text { or } \mu \in R-[-1,1]\)
CG PET-2015
Application of the Integrals
87027
Let \(q\) be the maximum integral value of \(p\) in \([0\), 10] for which the roots of the equation \(x^{2}-p x+\frac{5}{4} p=0\) are rational. Then the area of the region \(\left\{(x, y): 0 \leq y \leq(x-q)^{2}, 0 \leq x \leq q\right)\) is
1 164
2 243
3 \(\frac{125}{3}\)
4 25
Explanation:
(B) : Maximum integral value of \(\mathrm{p}=\mathrm{q}\) Roots of equation \(=x^{2}-p x+\frac{5}{4} p=0\) The graphs of curve are as shown in the figure. \(\Rightarrow \quad \mathrm{x}^{2}-\mathrm{px}+\frac{5}{4} \mathrm{p}=0\) \(\mathrm{a}=1, \quad \mathrm{~b}=-\mathrm{p}, \quad \mathrm{c}=\frac{5}{4} \mathrm{p}\) \(D=b^{2}-4 a c=(-p)^{2}-4 \times 1 \times \frac{5}{4} p\) \(=\mathrm{p}^{2}-5 \mathrm{p}=\mathrm{p}(\mathrm{p}-5)\) \(q=9\) Area of shared region \(=\int_{0}^{9}(x-9)^{2} d x\) \(=\left[\frac{(x-9)^{3}}{3}\right]_{0}^{9}=\frac{(+9)^{3}}{3}=243 \text { sq. units }\)
Shift-II
Application of the Integrals
87028
The area of the region given by \(\{(x, y): x y \leq 8\), \(\left.1 \leq \mathrm{y} \leq \mathrm{x}^{2}\right\}\) is
87029
If an equilateral triangle is inscribed in the circle \(x^{2}+y^{2}=a^{2}\), the length of its each side is
1 \(\sqrt{2} \mathrm{a}\)
2 \(\sqrt{3} \mathrm{a}\)
3 \(\frac{\sqrt{3}}{2} \mathrm{a}\)
4 \(\frac{1}{\sqrt{3}} \mathrm{a}\)
Explanation:
(B) : Given, equation of the circle is \(x^{2}+y^{2}=a^{2}\) Where, \(\text { radius }=\mathrm{a}\) Let \(\mathrm{ABC}\) be the inscribed equilateral triangle and length of its each sides be \(l\). Area of \(\triangle \mathrm{ABC}=\frac{\sqrt{3}}{4} l^{2}\) Also, \(\mathrm{OD}=\sqrt{\mathrm{a}^{2}-\frac{l^{2}}{4}}\) [ from \(\Delta \mathrm{ODB}]\) \(\therefore\) Area of \(\triangle \mathrm{OBC}=\frac{1}{2} \times l \times \sqrt{\mathrm{a}^{2}-\frac{l}{4}}\) Now, Area of \(\triangle \mathrm{ABC}=3 \times\) Area of \(\triangle \mathrm{OBC}\) \(\Rightarrow \frac{\sqrt{3}}{4} l^{2}=3 \times \frac{l}{2} \sqrt{\mathrm{a}^{2}-\frac{l^{2}}{4}} \Rightarrow \frac{\sqrt{3}}{2} l=3 \frac{\sqrt{4 \mathrm{a}^{2}-l^{2}}}{2}\) \(\Rightarrow \sqrt{3} l=3 \sqrt{4 \mathrm{a}^{2}-l^{2}} \quad \Rightarrow 3 l^{2}=9\left(4 \mathrm{a}^{2}-l^{2}\right)\) \(\Rightarrow l^{2}=12 \mathrm{a}^{2}-3 l^{2} \quad \Rightarrow 4 l^{2}=12 \mathrm{a}^{2}\) \(\Rightarrow l^{2}=3 \mathrm{a} \quad \Rightarrow l=\sqrt{3} \mathrm{a}\)
NEET Test Series from KOTA - 10 Papers In MS WORD
WhatsApp Here
Application of the Integrals
87025
The three lies of a triangle are give by \(\left(x^{2}-y^{2}\right)(2 x+3 y-6)=0\). If the point \((-2, \lambda)\) lies inside and \((\mu, 1)\) lies outside the triangle, then
(D) : Three lines of triangle are given by \(\left(x^{2}-y^{2}\right)(2 x+3 y-6)=0\) \(\Rightarrow \quad(x-y)(x+y)(2 x+3 y-6)=0\) \(\therefore\) The three lines of triangle are \(x-y=0, x+y=0\) and \(\quad 2 x+3 y-6=0\) From given lines of triangle, the required \(\triangle \mathrm{OAB}\) is formed. \(\because(-2, \lambda)\) lines inside the triangle. \(\therefore 2(-2)+3(\lambda)-6\lt 0\) and \(-2+\lambda>0\) \(\Rightarrow \quad-4+3 \lambda-6\lt 0\) and \(\lambda>2\) \(\Rightarrow \quad 3 \lambda\lt 10\) and \(\lambda>2\) \(\Rightarrow \quad \lambda\lt \frac{10}{3}\) and \(\lambda>2\) \(\therefore \quad \lambda \in\left(2, \frac{10}{3}\right)\) Now, \((\mu, 1)\) lies outside the triangle. To find value of \(\mu\), we find the interval \([\mathrm{M}, \mathrm{N}]\) for values of \(\mathrm{x}\). \(\mathrm{x}+1 \geq 0 \text { and } \mathrm{x}-1 \leq 0\) \(\Rightarrow \mathrm{x} \in[-1,1]\) \(\because(\mu, 1)\) lies outside the triangle. \(\therefore \mu \in(-\infty,-1) \cup(1, \infty)\) \(\text { or } \mu \in R-[-1,1]\)
CG PET-2015
Application of the Integrals
87027
Let \(q\) be the maximum integral value of \(p\) in \([0\), 10] for which the roots of the equation \(x^{2}-p x+\frac{5}{4} p=0\) are rational. Then the area of the region \(\left\{(x, y): 0 \leq y \leq(x-q)^{2}, 0 \leq x \leq q\right)\) is
1 164
2 243
3 \(\frac{125}{3}\)
4 25
Explanation:
(B) : Maximum integral value of \(\mathrm{p}=\mathrm{q}\) Roots of equation \(=x^{2}-p x+\frac{5}{4} p=0\) The graphs of curve are as shown in the figure. \(\Rightarrow \quad \mathrm{x}^{2}-\mathrm{px}+\frac{5}{4} \mathrm{p}=0\) \(\mathrm{a}=1, \quad \mathrm{~b}=-\mathrm{p}, \quad \mathrm{c}=\frac{5}{4} \mathrm{p}\) \(D=b^{2}-4 a c=(-p)^{2}-4 \times 1 \times \frac{5}{4} p\) \(=\mathrm{p}^{2}-5 \mathrm{p}=\mathrm{p}(\mathrm{p}-5)\) \(q=9\) Area of shared region \(=\int_{0}^{9}(x-9)^{2} d x\) \(=\left[\frac{(x-9)^{3}}{3}\right]_{0}^{9}=\frac{(+9)^{3}}{3}=243 \text { sq. units }\)
Shift-II
Application of the Integrals
87028
The area of the region given by \(\{(x, y): x y \leq 8\), \(\left.1 \leq \mathrm{y} \leq \mathrm{x}^{2}\right\}\) is
87029
If an equilateral triangle is inscribed in the circle \(x^{2}+y^{2}=a^{2}\), the length of its each side is
1 \(\sqrt{2} \mathrm{a}\)
2 \(\sqrt{3} \mathrm{a}\)
3 \(\frac{\sqrt{3}}{2} \mathrm{a}\)
4 \(\frac{1}{\sqrt{3}} \mathrm{a}\)
Explanation:
(B) : Given, equation of the circle is \(x^{2}+y^{2}=a^{2}\) Where, \(\text { radius }=\mathrm{a}\) Let \(\mathrm{ABC}\) be the inscribed equilateral triangle and length of its each sides be \(l\). Area of \(\triangle \mathrm{ABC}=\frac{\sqrt{3}}{4} l^{2}\) Also, \(\mathrm{OD}=\sqrt{\mathrm{a}^{2}-\frac{l^{2}}{4}}\) [ from \(\Delta \mathrm{ODB}]\) \(\therefore\) Area of \(\triangle \mathrm{OBC}=\frac{1}{2} \times l \times \sqrt{\mathrm{a}^{2}-\frac{l}{4}}\) Now, Area of \(\triangle \mathrm{ABC}=3 \times\) Area of \(\triangle \mathrm{OBC}\) \(\Rightarrow \frac{\sqrt{3}}{4} l^{2}=3 \times \frac{l}{2} \sqrt{\mathrm{a}^{2}-\frac{l^{2}}{4}} \Rightarrow \frac{\sqrt{3}}{2} l=3 \frac{\sqrt{4 \mathrm{a}^{2}-l^{2}}}{2}\) \(\Rightarrow \sqrt{3} l=3 \sqrt{4 \mathrm{a}^{2}-l^{2}} \quad \Rightarrow 3 l^{2}=9\left(4 \mathrm{a}^{2}-l^{2}\right)\) \(\Rightarrow l^{2}=12 \mathrm{a}^{2}-3 l^{2} \quad \Rightarrow 4 l^{2}=12 \mathrm{a}^{2}\) \(\Rightarrow l^{2}=3 \mathrm{a} \quad \Rightarrow l=\sqrt{3} \mathrm{a}\)
