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2 months ago

Straight Lines Class 11th

Let A(a, -2), B(a, 6) and $C (\frac{α}{4}, - 2)$ be vertices of a $Δ A B C .$ If $(5, \frac{α}{4})$ is the circumcentre of $Δ A B C,$ then which of the following is NOT correct about $Δ A B C ?$

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alok kumar singh

Contributor-Level 10

Circumcentre (D) $\equiv (5, \frac{α}{4})$

${(5 - α)}^{2} + {(\frac{α}{4} + 2)}^{2} = {(5 - α)}^{2} + {(\frac{α}{4} - 6)}^{2} . . . . . . . . . . . . . . . (i)$

${(5 - \frac{α}{4})}^{2} + {(\frac{α}{4} + 2)}^{2} . . . . . . . . . . . . . . . . . . . . (i i)$

(i) -> $\frac{α}{4} + 2 = \pm (\frac{α}{4} - 6)$

(ii) -> 9 + 16 = 9 + 16

$\oplus \to x$

$(-) \to \frac{α}{2} = 4 \Rightarrow α = 8$

ar $(A B C) = 24$

2S = 24

R = 5, r = $\frac{Δ}{s} = \frac{24}{12} = 2$

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2 months ago

Straight Lines Class 11th

Let m₁, m₂ be the slopes of two adjacent sides of a square of side a such that a² + 11a + $3 (m_{1}^{2} + m_{2}^{2}) = 220 .$ If one vertex of the square is $(10 (c o s α - s i n α), 10 (s i n α + c o s α)),$ where $α \in (0, \frac{π}{2})$ and the equation of one diagonal is $(c o s α - s i n α) x + (s i n α + c o s α) y = 10,$ then $72 (s i n^{4} α + c o s^{4} α) + a^{2} - 3 a + 13$ is equal to:

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alok kumar singh

Contributor-Level 10

$m_{1} m_{2} = - 1,$ for square a,b,c,d let

$A (10 (c o s α - s i n α), 10 (s i n α + c o s α))$

Diagonal : (cos a - sina)x + (sina + cosa)y = 10

BD (diagonal)

Dist. Of BD from A is

$\frac{| 10 {(c o s α - s i n α)}^{2} + 10 {(s i n α + c o s α)}^{2} - 10 |}{\sqrt[]{2}} = \frac{a}{\sqrt[]{2}}$

$\frac{10}{\sqrt[]{2}} = \frac{a}{\sqrt[]{2}} \Rightarrow a = 10$

Also, a² + 11a + 3 $(m_{1}^{2} + m_{2}^{2}) = 220$

-> 210 + 3 $(c m_{1}^{2} + m_{2}^{2}) = 220$

$m_{1}^{2} + m_{2}^{2} = \frac{10}{3}$

Also, m₁ m₂ = -1

->m² + $\frac{1}{m^{2}} = \frac{10}{3}$

or $- \sqrt[]{3}, \frac{1}{\sqrt[]{3}}$

m = $\sqrt[]{3}, \frac{- 1}{\sqrt[]{3}}$

$m^{4} - \frac{10}{3} m^{2} + 1 = 0 \Rightarrow m^{2} = \frac{\frac{10}{3} \pm \sqrt[]{\frac{100}{9} - 4}}{2} - \frac{\frac{10}{3} \pm \frac{8}{3}}{2} = 3, \frac{1}{3}$

m = $\pm \sqrt[]{3}, \pm \frac{1}{\sqrt[]{3}}$

Diagonal AC:

$(s i n α + c o s α) x - (c o s α - s i n α) y$

=10 cos2a - 10cos2a = 0

Slope of AC = $\frac{s i n α + c o s α}{c o s α - s i n α} = \frac{t a n α + 1}{1 - t a n α} = t a n (α + \frac{π}{4}) α = 30 °$

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$m_{1} m_{2} = - 1,$ for square a,b,c,d let

$A (10 (c o s α - s i n α), 10 (s i n α + c o s α))$

Diagonal : (cos a - sina)x + (sina + cosa)y = 10

BD (diagonal)

Dist. Of BD from A is

$\frac{| 10 {(c o s α - s i n α)}^{2} + 10 {(s i n α + c o s α)}^{2} - 10 |}{\sqrt[]{2}} = \frac{a}{\sqrt[]{2}}$

$\frac{10}{\sqrt[]{2}} = \frac{a}{\sqrt[]{2}} \Rightarrow a = 10$

Also, a² + 11a + 3 $(m_{1}^{2} + m_{2}^{2}) = 220$

-> 210 + 3 $(c m_{1}^{2} + m_{2}^{2}) = 220$

$m_{1}^{2} + m_{2}^{2} = \frac{10}{3}$

Also, m₁ m₂ = -1

->m² + $\frac{1}{m^{2}} = \frac{10}{3}$

or $- \sqrt[]{3}, \frac{1}{\sqrt[]{3}}$

m = $\sqrt[]{3}, \frac{- 1}{\sqrt[]{3}}$

$m^{4} - \frac{10}{3} m^{2} + 1 = 0 \Rightarrow m^{2} = \frac{\frac{10}{3} \pm \sqrt[]{\frac{100}{9} - 4}}{2} - \frac{\frac{10}{3} \pm \frac{8}{3}}{2} = 3, \frac{1}{3}$

m = $\pm \sqrt[]{3}, \pm \frac{1}{\sqrt[]{3}}$

Diagonal AC:

$(s i n α + c o s α) x - (c o s α - s i n α) y$

=10 cos2a - 10cos2a = 0

Slope of AC = $\frac{s i n α + c o s α}{c o s α - s i n α} = \frac{t a n α + 1}{1 - t a n α} = t a n (α + \frac{π}{4}) α = 30 °$

FIGURE

? = $72 (\frac{1}{16} + \frac{9}{16}) + 100 - 30 + 13 = \frac{720}{16} + 83 = 128$

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New answer posted

2 months ago

Straight Lines Class 11th

For $t \in (0, 2 π),$ if ABC is an equilateral triangle with vertices $A (s i n t, - c o s t)$ , $B (c o s t, s i n t)$ and C(a, b) such that its orthocenter lies on a circle with centre $(1, \frac{1}{3})$ $\frac{}{}$ , then $(a^{2} - b^{2})$ $^{}^{}$ is equal to:

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Vishal Baghel

Contributor-Level 10

S $\equiv s i n t, C \equiv c o s t$

Let orthocenter be (h, k)

Since it if an equilateral triangle hence orthocenter coincides with centroid.

$∴ a + s + c = 3 h, b + s - c = 3 k$

$\frac{a}{3} = 1, \frac{b}{3} = \frac{1}{3} \Rightarrow a = 3, b = 1$

$∴ a^{2} - b^{2} - 8$

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New answer posted

2 months ago

Straight Lines Class 11th

Let the point P(α,β) be at a unit distance from each of the two lines L₁ : 3x – 4y + 12 = 0, and L₂ : 8x + 6y + 11 = 0. If P lies below L₁ and above L₂, then 100(α + β) is equal to

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Payal Gupta

Contributor-Level 10

L₁ : 3x – 4y + 12 = 0

L₂ : 8x + 6y + 11 = 0

$(α, β)$ lies on that angle which contain origin

$∴$ Equation of angle bisector of that angle which contain origin is

$\frac{3 x - 4 y + 12}{5} = \frac{8 x + 6 y + 11}{10}$

$\Rightarrow 2 x + 14 y - 13 = 0$

$(α, β)$ lies on it

$\Rightarrow 2 α + 14 β - 13 = 0$ …… (i)

$\Rightarrow 3 α - 4 β + 7 = 0$ ……. (ii)

Solving (i) & (ii)

$α = - \frac{23}{25} & β = \frac{53}{50}$

$∴ α + β = \frac{7}{50}$

$\Rightarrow 100 (α + β) = 14$

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New answer posted

2 months ago

Straight Lines Class 11th

Let $A (\frac{3}{\sqrt[]{a}}, \sqrt[]{a}), a > 0$ , be a fixed point in the xy-plane. The image of A in y-axis be B and the image of B in x-axis be C. If D (3 cos q, a sin q) is a point in the fourth quadrant such that the maximum area of $Δ A C D$ is 12 square units, then a is equal to _________.

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alok kumar singh

Contributor-Level 10

$B (- \frac{3}{\sqrt[]{a}}, \sqrt[]{a}) a n d c (- \frac{3}{\sqrt[]{a}}, - \sqrt[]{a})$

$∴ A r e a o f Δ A C D = \frac{1}{2} | \begin{matrix} \frac{3}{\sqrt[]{a}} & \sqrt[]{a} & 1 \\ \frac{- 3}{\sqrt[]{a}} & - \sqrt[]{a} & 1 \\ 3 c o s θ & a s i n θ & 1 \end{matrix} |$ = 12

$\Rightarrow Δ = 3 \sqrt[]{a} | c o s θ + s i n θ | = 12$

$∴ Δ m a x = 3 \sqrt[]{a} . \sqrt[]{2} = 12 \Rightarrow a = 8$

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New answer posted

2 months ago

Straight Lines Class 11th

Let for n = 1, 2,……, 50, S_n be the sum of the infinite geometric progression whose first term is n² and whose common ratio is Then the value of $\frac{1}{26} + \sum_{n = 1}^{50} (S_{n} + \frac{2}{n + 1} - n - 1)$ is equal to………….

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alok kumar singh

Contributor-Level 10

$S_{n} = \frac{n^{2}}{1 - \frac{1}{{(n + 1)}^{2}}} = \frac{n {(n + 1)}^{2}}{(n + 2)} = {(n + 1)}^{2} - 2 (n + 1) + 2 - \frac{2}{(n + 2)}$

$\frac{1}{26} + \sum_{n = 1}^{50} (S_{n} + \frac{2}{(n + 1)} - (n + 1)) = \frac{1}{26} + \sum_{n = 1}^{50} {(n + 1)}^{2} - 3 (n + 1) + 2 + 2 (\frac{1}{(n + 1)} - \frac{1}{(n + 2)})$

$= \frac{1}{26} + 45525 - 3 * 1325 + 2 * 50 + 2 (\frac{1}{2} - \frac{1}{52})$

$= \frac{1}{26} + 41550 + 100 + 1 - \frac{1}{26} = 41651$

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New answer posted

2 months ago

Straight Lines Class 11th

The area of the bounded region enclosed by the curve y = 3 $| x - \frac{1}{2} | - | x + 1 |$ - and the x-axis is:

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alok kumar singh

Contributor-Level 10

$y = 3 ? | x ? \frac{1}{2} | = | x + 1 |$

Graph

Area = $(\frac{1}{2} * \frac{3}{2} * \frac{3}{4}) * 2 + \frac{3}{2} * \frac{3}{2} = \frac{3}{2} * \frac{3}{2} * \frac{3}{2} = \frac{27}{8}$

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New answer posted

2 months ago

Straight Lines Class 11th

If $s i n^{2} (1 0^{0}) s i n (2 0^{0}) s i n (4 0^{0}) s i n (5 0^{0}) s i n (7 0^{0}) = α - \frac{1}{16} s i n (1 0^{0}),$ then 16 + 1 is equal to

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Payal Gupta

Contributor-Level 10

$(s i n 10 ° . s i n 50 ° . s i n 70 °) . (s i n 10 ° . s i n 20 ° . s i n 40 °)$

$= (\frac{1}{4} s i n 30 °) . [\frac{1}{2} s i n 10 ° (c o s 20 ° - c o s 60 °)]$

$= \frac{1}{32} [s i n 30 ° - s i n 10 ° - s i n 10 °]$

$\frac{1}{64} - \frac{1}{16} s i n 10 °$

Clearly $α = \frac{1}{64}$

Hence 16 + a^-1= 80

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New answer posted

2 months ago

Straight Lines Class 11th

The total number of 3-digit numbers, whose greatest common divisor with 36 is 2, is……….

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Vishal Baghel

Contributor-Level 10

36 = 2 * 2 * 3 * 3

Number should be odd multiple of 2 and does not having factor 3 and 9

Odd multiple of 2 are

102, 106, 110, 114….998 (225 no.)

No. of multiplies of 3 are

102, 114, 126 ….990 (75 no.)

Which are also included multiple of 9

Hence,

Required = 225 – 75 = 150

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New answer posted

2 months ago

Straight Lines Class 11th

If y = y(x) is the solution of the differential equation $\frac{}{}^{}$ then the local maximum value of the function z(x) = x²y(x) – e^x, x is:

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Vishal Baghel

Contributor-Level 10

$x \frac{d y}{d x} + 2 y = x e^{x}$ $\frac{}{}^{}$

$\frac{d y}{d x} + \frac{2 y}{x} = e^{x}$

$\frac{d z}{d x} = e^{x} . 2 (x - 1) + e^{x} {(x - 1)}^{2} = 0$

$e^{x} (x - 1) (2 + x - 1) = 0$

$x = - 1, 1$

x = 1 local maxima. Then maximum value is

$z (- 1) = \frac{4}{e} - e$

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