Maths

Le the mirror image of the point (a, b, c), with respect to the plane 3x – 4y + 12z + 19 = 0 be $(a - 6, β, γ)$ . If a + b + c = 5, then $7 β - 9 γ$ is equal to……….s

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Contributor-Level 10

$\frac{x - a}{3} = \frac{y - b}{- 4} = \frac{z - c}{12} = \frac{- 2 (3 a - 4 b + 12 c + 19)}{3^{2} + {(- 4)}^{2} + 1 2^{2}}$

$\frac{x - a}{3} = \frac{y - b}{- 4} = \frac{z - c}{12} = \frac{- 6 a + 8 b - 24 c - 38}{169}$

$(x, y, z) \equiv (a - 6, β, γ)$

$\frac{(a - b) - a}{3} = \frac{β - b}{- 4} = \frac{γ - c}{12} = \frac{- 6 a + 8 b - 24 c - 38}{169}$

$\frac{β - b}{- 4} = - 2$

$\Rightarrow β = 8 + b$

$\Rightarrow 3 a - 4 b + 12 c = 150$ ….(i)

a + b + c = 5

$\Rightarrow 3 a + 3 b + 3 c = 15$ ….(ii)

Applying (i) – (ii), we get :

= 56 + 216 + 7b – 9c = 56 + 216 – 135 = 137

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New answer posted

a year ago

A circle of radius 2 unit passes through the vertex and the focus of the parabola y² = 2x and touches the parabola y = 2x and touches the parabola $y = {(x - \frac{1}{4})}^{2} + α,$ where a > 0. Then (4a - 8)² is equal to………….

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Contributor-Level 10

Let the equation of circle be

$x (x - \frac{1}{2}) + y^{2} + λ y = 0$

$\Rightarrow x^{2} + y^{2} - \frac{1}{2} x + λ y = 0$

Radius = $\sqrt[]{\frac{1}{16} + \frac{λ^{2}}{4}} = 2$

$\Rightarrow λ^{2} = \frac{63}{4} \Rightarrow {(x - \frac{1}{4})}^{2} + {(y + \frac{λ}{2})}^{2} = 4$

$?$ This circle and parabola

$y - α = {(x - \frac{1}{4})}^{2}$ touch each other, so

$α = - \frac{λ}{2} + 2 \Rightarrow α - 2 = - \frac{λ}{2} \Rightarrow {(α - 2)}^{2} = \frac{λ^{2}}{4} = \frac{63}{16}$

${(4 α - 8)}^{2} = 63$

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New answer posted

a year ago

A rectangle R with end points of one of its sides as (1, 2) and (3, 6) is inscribed in a circle. If the equation of a diameter of the circle is 2x – y + 4 = 0, then the area of R is………….

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Contributor-Level 10

As slope of line joining (1, 2) and (3, 6) is 2 given diameter is parallel to side

$? a = \sqrt[]{{(3 ? 1)}^{2} + {(6 ? 2)}^{2}} = \sqrt[]{20} ? ? a n d$

$a b = 2 \sqrt[]{5} . \frac{8}{\sqrt[]{5}} = 16$

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New answer posted

a year ago

If the sum of the first ten terms of the series

$\frac{1}{25} + \frac{2}{65} + \frac{3}{325} + \frac{4}{1025} + \frac{5}{2501} + . . . . . . i s \frac{m}{n},$ where m and n co-prime numbers, then m + n is equal to…….

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Contributor-Level 10

$T_{r} = \frac{r}{{(2 r^{2})}^{2} + 1}$

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

$= \frac{1}{4} \frac{4 r}{(2 r^{2} + 2 r + 1) (2 r^{2} - 2 r + 1)}$

$S_{10} = \frac{1}{4} \sum_{r = 1}^{10} (\frac{1}{(2 r^{2} - 2 r + 1)} - \frac{1}{(2 r^{2} + 2 r + 1)})$

$\Rightarrow S_{10} = \frac{1}{4} . \frac{220}{221} = \frac{55}{221} = \frac{m}{n}$

$∴ m + m = 276$

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New answer posted

a year ago

Let

A₁ = ${(x, y) : | x | \leq y^{2}, | x | + 2 y \leq 8} a n d$

A₂ = ${(x, y) : | x | + | y | \leq k} :$ If 27 (Area A₁) = 5 (Area A₂), then k is equal to:

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Contributor-Level 10

Required area (above x-axis)

$A_{1} = 2 \int_{0}^{4} (\frac{8 - x}{2} - \sqrt[]{x}) d x$

$= 2 (16 - \frac{16}{4} - \frac{8}{3 / 2}) = \frac{40}{3}$

and $A_{2} = 4 (\frac{1}{2} . k^{2}) = 2 k^{2}$

$∴ 27 . \frac{40}{3} = 5 . (2 k^{2})$

-> k = 6

for above x-axis.

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New answer posted

a year ago

If the coefficient of x¹⁰ in the binomial expansion of ${(\frac{\sqrt[]{x}}{5^{\frac{1}{4}}} + \frac{\sqrt[]{5}}{x^{\frac{1}{3}}})}^{60}$ is $5^{k} . l,$ where $l, k \in N a n d l$ is co-prime to 5, then k is equal to……….

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Contributor-Level 10

$T_{r + 1} =^{60} C_{r} {(x^{\frac{1}{2}})}^{60 - r} {(x^{- \frac{1}{3}})}^{r} {(5^{\frac{- 1}{4}})}^{60 - r} {(5^{\frac{1}{2}})}^{r}$

for

$x^{10} \frac{60 - r}{2} - \frac{r}{3} = 10$

$\Rightarrow 180 - 3 r - 2 r = 60$

->r = 24

k = 3 + exponent of 5 in

= $3 + ([\frac{60}{5}] + [\frac{60}{5^{2}}] - [\frac{24}{5}] - [\frac{24}{5^{2}}] - [\frac{36}{5}] - [\frac{36}{5^{2}}])$

= 3 + (12 + 2 – 4 – 0 – 7 – 1)

= 3 + 2 = 5

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New answer posted

a year ago

The number of ways, 16 identical cubes, of which 11 are blue and rest are red, can be placed in a row so that between any two red cubes there should be at least 2 blue cubes, is…………

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Contributor-Level 10

First we arrange 5 red cubes in a row and assume number of blue cubes between them

$H e r e, x_{1} + x_{2} + x_{3} + x_{4} + x_{5} + x_{6} = 11$

and $x_{2}, x_{3}, x_{4}, x_{5} \geq 2$

so $x_{1} + x_{2} + x_{3} + x_{4} + x_{5} + x_{6} = 3$

No. of solutions = ⁸C₅ = 56

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New answer posted

a year ago

The positive value of the determination of the matrix A, whose $A d j (A d j (A)) = (\begin{matrix} 14 & 28 & - 14 \\ - 14 & 14 & 28 \\ 28 & - 14 & 14 \end{matrix}), i s$ ………………

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Contributor-Level 10

$| a d j (a d j (A)) | = {| A |}^{2^{2}} = {| A |}^{4}$

$∴ {| A |}^{4} = | \begin{matrix} 14 & 28 & - 14 \\ - 14 & 14 & 28 \\ 25 & - 14 & 14 \end{matrix} |$

$= {(14)}^{3} | \begin{matrix} 1 & 2 & - 1 \\ - 1 & 1 & 2 \\ 2 & - 1 & 1 \end{matrix} |$

$= {(14)}^{3} (3 - 2 (- 5) - 1 (- 1))$

${| A |}^{4} = {(14)}^{4} \Rightarrow | A | = 14$

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New answer posted

a year ago

If the sum of all the roots of the equation $e^{2 x} - 11 e^{x} - 45 e^{- x} + \frac{81}{2} = 0$ is log_e p, then p is equal to………….

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Contributor-Level 10

Let e^x = t then equation reduces to

$t^{2} - 11 t - \frac{45}{t} + \frac{81}{2} = 0$

$\Rightarrow 2 t^{3} - 22 t^{2} + 81 t - 45 = 0$ ……. (i)

If roots of

$e^{2 x} - 11 e^{x} - 45 e^{- x} + \frac{81}{2} = 0$

$\Rightarrow α_{1} + α_{2} + α_{3} = l n 45 \Rightarrow p = 45$

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New answer posted

a year ago

Let f : R -> R be a function defined by f(x) $= \frac{2 e^{2 x}}{e^{2 x} + e} .$

Then $f (\frac{1}{100}) + f (\frac{2}{100}) + f (\frac{3}{100}) + . . . . + f (\frac{99}{100})$ is equal to…………….

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