Straight Lines

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New answer posted•a year 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:

40 Views|Posted a year ago

Asked by Vishal Baghel

Answered by

Vishal Baghel | Contributor-Level 10

a year ago

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•a year 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

69 Views|Posted a year ago

Asked by Payal Gupta

Answered by

Payal Gupta | Contributor-Level 10

a year ago

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•a year 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 _________.

17 Views|Posted a year ago

Asked by Alok Kumar Singh

Answered by

Alok Kumar Singh | Contributor-Level 10

a year ago

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

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

6 Views|Posted a year ago

Asked by Alok Kumar Singh

Answered by

Alok Kumar Singh | Contributor-Level 10

a year ago

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

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

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

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

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New answer posted•a year 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:

2 Views|Posted a year ago

Asked by Alok Kumar Singh

Answered by

Alok Kumar Singh | Contributor-Level 10

a year ago

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

Graph

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

&nbs

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New answer posted•a year 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

2 Views|Posted a year ago

Asked by Payal Gupta

Answered by

Payal Gupta | Contributor-Level 10

a year ago

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

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

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

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

Clearly $α = \frac{1}{64}$

Hence 16 + a^-1= 80

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New answer posted•a year ago

Straight Lines Class 11th

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

8 Views|Posted a year ago

Asked by Vishal Baghel

Answered by

Vishal Baghel | Contributor-Level 10

a year ago

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•a year 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:

3 Views|Posted a year ago

Asked by Vishal Baghel

Answered by

Vishal Baghel | Contributor-Level 10

a year ago

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

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

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

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

$x = - 1, 1$

x = 1 local maxima. Then maximum value is

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

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New answer posted•a year ago

Straight Lines Class 11th

Let S be the set of all the natural numbers, for which the line $\frac{x}{a} + \frac{y}{b} = 2$  is a tangent to the curve ${(\frac{x}{a})}^{n} + {(\frac{y}{b})}^{n} = 2$  at the point (a, b), ab  Then

10 Views|Posted a year ago

Asked by Payal Gupta

Answered by

Payal Gupta | Contributor-Level 10

a year ago

${(\frac{x}{a})}^{n} + {(\frac{y}{b})}^{n} = 2$

$\Rightarrow \frac{n}{a} {(\frac{x}{a})}^{n - 1} + \frac{n}{b} {(\frac{y}{b})}^{n - 1} \frac{d y}{d x} = 0$

$\Rightarrow \frac{d y}{d x} = - \frac{b}{a} {(\frac{b x}{a y})}^{n - 1}$

$\frac{d y}{d x_{(a, b)}} = - \frac{b}{a}$

So line always touches the given curve.

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New answer posted•a year ago

Straight Lines Maths NCERT Exemplar Solutions Class 12th Chapter One

The distance between the two points A and A’ which lie on y = 2 such that both the line segments AB and A’B (where B is the point (2, 3)) subtend angle $\frac{π}{4}$ at the origin, is equal to:

23 Views|Posted a year ago

Asked by Alok Kumar Singh

Answered by

Alok Kumar Singh | Contributor-Level 10

a year ago

Let A, A’ be (, 2) AB and A’B subtends $\frac{π}{4}$ angle at (0, 0) slope of OA = $\frac{2}{α}$

slope of OB = $\frac{3}{2}$

$t a n \frac{π}{4} = | \frac{\frac{2}{α} - \frac{3}{2}}{1 + \frac{2}{α} \cdot \frac{3}{2}} |$

$\Rightarrow 1 + \frac{3}{α} = \pm (\frac{4 - 3 α}{2 α})$

$\Rightarrow \frac{α + 3}{α} = \pm (\frac{4 - 3 α}{2 α}) \Rightarrow α = 10, - \frac{2}{5}$

now distance between A’A, (10, 2) & $(\frac{- 2}{5}, 2) i s \frac{52}{5}$

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