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Maths Sets

The number of elements in the set S = ${x \in R : 2 c o s (\frac{x^{2} + x}{6}) = 4^{x} + 4^{- x}} i s :$

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alok kumar singh

Contributor-Level 10

2cos $(\frac{x^{2} + x}{6}) = 4^{x} + 4^{- x}$

$- 2 \leq L H S \leq 2 L H S = 2 & R H S = 2 \Rightarrow x = 0 o n l y \to t h e n L H S = 2 a l s o$

RHS $\geq$ 2

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New answer posted

a year ago

Maths Class 12th

If the minimum value of f(x) = $\frac{5 x^{2}}{2} + \frac{α}{x^{5}}, x > 0,$ $\frac{^{}}{} \frac{}{^{}}$ is 14, then the value of is equal to:

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Vishal Baghel

Contributor-Level 10

$\frac{5 x^{2}}{2} + \frac{α}{2 x^{5}} + \frac{α}{2 x^{5}} \geq 7 {(\frac{α^{2}}{2^{7}})}^{\frac{1}{7}}$

$\frac{7 . {(α)}^{2 / 7}}{2} = 14$

${(α^{2})}^{\frac{1}{7}} = 2^{2} \Rightarrow α = {(2^{2})}^{\frac{7}{2}} = 2^{7}$

=128

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New answer posted

a year ago

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

...more

$m_{1} m_{2} = - 1,$ 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

a year ago

Maths Class 12th

The foot of the perpendicular from a point on the circle x² + y² = 1, z = 0 to the plane 2x + 3y + z = 6 lies on which one of the following curves?

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Vishal Baghel

Contributor-Level 10

$\frac{h - c o s θ}{2} = \frac{k - s i n θ}{3} = \frac{w - 0}{1}$

$= \frac{- 1 (2 c o s θ + 3 s i n θ - 6)}{14} \Rightarrow h = \frac{c o s - 2 (2 c o s θ + 3 s i n θ - 6)}{14}$

$k = \frac{5 s i n θ - 6 c o s θ + 18}{14}$

${(5 h + 6 k - 12)}^{2} + 4 {(3 h + 5 k - 9)}^{2} = 1$

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New answer posted

a year ago

Maths Class 12th

Let $S_{1} = {z_{1} \in C : | z_{1} - 3 | = \frac{1}{2}}$ and $S_{2} = {z_{2} \in C : | z_{2} - | z_{2} + 1 | | = | z_{2} + | z_{2} - 1 | |} .$ Then, for $z_{1} \in S_{1}$ and $s_{2} \in S_{2},$ the least value of $| z_{2} - z_{1} | i s :$

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Vishal Baghel

Contributor-Level 10

$S_{1} = {z, \in c : ⌊ z_{1} - 3 ⌋ = \frac{1}{2}} a n d S_{2} = {z_{2} \in c : | z_{2} - | z_{2} + 1 | | = | z_{2} + | z_{2} - 1 | |}$

Then, for $z_{1} \in S_{1}$ and $z_{2} \in S_{2},$ the least value of |z₂ – z₁|

${| z_{2} + | z_{2} - 1 | |}^{2} = {| z_{2} - | z_{2} + 1 | |}^{2}$

$\Rightarrow (z_{2} + {\bar{z}}_{2}) (| z_{2} - 1 | + | z_{2} + 1 | - 2) = 0$

$∴ z_{2} + {\bar{z}}_{2} = 0 o r | z_{2} - 1 | + | z_{2} + 1 | - 2 = 0$

$∴$ z₂ lies on imaginary axis or on real axis with in [-1, 1]

also $| z_{1} - 3 | = \frac{1}{2}$ lie on circle having centre 3 and radius $\frac{1}{2}$

Clearly $| z_{1} - z_{2} | m i n = \frac{5}{2} - 1 = \frac{3}{2}$

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New answer posted

a year ago

Maths Matrices

Let the matrix $A = [\begin{matrix} 0 & 1 & 0 \\ 0 & 0 & 1 \\ 1 & 0 & 0 \end{matrix}]$ and the matrix B₀ = A²⁹ + 2A⁹⁸. If B_n = Adj(B_n=1) for all $n \geq 1,$ then det

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Vishal Baghel

Contributor-Level 10

$A^{2} = [\begin{matrix} 0 & 1 & 0 \\ 0 & 0 & 1 \\ 1 & 0 & 0 \end{matrix}] [\begin{matrix} 0 & 1 & 0 \\ 0 & 0 & 1 \\ 1 & 0 & 0 \end{matrix}]$

$= [\begin{matrix} 0 & 0 & 1 \\ 1 & 0 & 0 \\ 0 & 1 & 0 \end{matrix}]$

$a \leftrightarrow$ R2

$= - [\begin{matrix} 1 & 0 & 0 \\ 0 & 0 & 1 \\ 0 & 1 & 0 \end{matrix}]$

$R_{2} \leftrightarrow R_{3}$

$= [\begin{matrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{matrix}]$ =1

$\begin{array}{l} B_{0} = A^{49} + 2 A^{98} = A + 2 I \\ B_{n} = A d j (B_{n} - 1) \\ B_{4} = A d j (A d j (A d j (A d j B_{0}))) \end{array}$

$= {| B_{0} |}^{{(n - 1)}^{4}} = {| B_{0} |}^{16}$

$B_{0} = [\begin{matrix} 0 & 1 & 0 \\ 0 & 0 & 1 \\ 1 & 0 & 0 \end{matrix}] + [\begin{matrix} 2 & 0 & 0 \\ 0 & 2 & 0 \\ 0 & 0 & 2 \end{matrix}] = [\begin{matrix} 2 & 1 & 0 \\ 0 & 2 & 1 \\ 1 & 0 & 2 \end{matrix}]$

$= 2 (4 - 0) - 1 (0 - 1) = 9$

$B_{4} {(9)}^{16} = 3^{32}$

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New answer posted

a year ago

Maths Class 12th

Let y = y(x) be the solution curve of the differential equation $\frac{d y}{d x} + (\frac{2 x^{2} + 11 x + 13}{x^{3} + 6 x^{2} + 11 x + 6}) y = \frac{(x + 3)}{x + 1}, x > - 1,$ which passes through the point (0, 1). Then y(1) is equal to:

0 Follower 3 Views

alok kumar singh

Contributor-Level 10

$\frac{d y}{d x} + (\frac{2 x^{2} + 11 x + 13}{x^{3} + 6 x^{2} + 11 x + 6}) y = \frac{x + 3}{x + 1}, x > - 1$

IF = $e^{\int p d x} = \frac{{(x + 1)}^{2} (x + 2)}{x + 3}$

$\int P d x = \int \frac{2 x^{2} + 11 x + 13}{x^{3} + 6 x^{2} + 11 x + 6} d n = \int (\frac{2}{x + 1} + \frac{1}{x + 2} - \frac{1}{x + 3}) d x$

$= l n ({(x + 1)}^{2} \cdot (x + 2) / (x + 3))$

$\frac{2 x^{2} + 11 x + 13}{(x + 1) (x + 2) (x + 3)} = \frac{A}{x + 1} + \frac{B}{x + 2} + \frac{C}{x + 3}$

$2 x^{2} + 11 x + 13 = A (x + 2) (x + 3) + B (x + 1) (x + 3) + C (x + 1) (x + 2)$

x = -1

->4 = 2A Þ A = 2

x = -2

-> -1 = -B Þ B = 1

x₂ – 3 Þ -2 = 2c

c = -1

$y \cdot \frac{{(x + 1)}^{2} (x + 2)}{x + 3} = \int \frac{x + 3}{x + 1} \cdot \frac{{(x + 1)}^{2} (x + 2)}{x + 3} d x$

= $\int (x + 1) (x + 2) d x$

= $\frac{x^{3}}{3} + \frac{3 x^{2}}{2} + 2 x + c$

$(0, 1) \Rightarrow 1 \cdot \frac{2}{3} = c$

x = 1 $y \cdot (3) = \frac{1}{3} + \frac{3}{2} + 2 + \frac{2}{3} = \frac{3}{2} + 3 = \frac{9}{2}$

y = $\frac{3}{2}$

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New answer posted

a year ago

Maths Complex Numbers and Quadratic Equations

The remainder when 7²⁰²² + 3²⁰²² is divided by 5 is:

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Vishal Baghel

Contributor-Level 10

$7^{2022} + 3^{2022}$

$= {(49)}^{1011} + {(9)}^{1011}$

$= {(50 - 1)}^{1011} + {(10 - 1)}^{1011}$

$= 5 λ - 1 + 5 k - 1$

= 5m – 2

Remainder = 5 – 2 = 3

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New answer posted

a year ago

Maths Class 11th

Let C be the centre of the circle $x^{2} + y^{2} - x + 2 y = \frac{11}{4}$ $^{}^{} \frac{}{}$ and P be a point on the circle. A line passes through the point C, makes an angle of $\frac{π}{4}$ $\frac{}{}$ with the line CP and intersects the circle at the points Q and R. Then the area of the triangle PQR (in unit2) is:

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Vishal Baghel

Contributor-Level 10

$x^{2} + y^{2} - x + 2 y = \frac{11}{4}$

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

or $Δ P Q R P R = Q R s i n 2 \geq \frac{1}{3}$

$= 4.6 s i n \frac{π}{8}$

$a s Δ P Q R = \frac{1}{2} P R * P Q$

$= 4 s i n \frac{π}{4} = \frac{4}{\sqrt[]{2}} = 2 \sqrt[]{2}$

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New answer posted

a year ago

Maths Class 11th

If the tangents drawn at the points P and Q on the parabola y² = 2x – 3 intersect at the point R(0, 1), then the orthocenter of the triangle PQR is:

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Vishal Baghel

Contributor-Level 10

$y^{2} = 2 x - 3$ ……. (i)

Equation of chord of contact

PQ : r = O

(y * 1) = (x + 0) – 3

y = x – 3 ……… (ii)

From (i) and (ii)

y = 1 or 3

$M P Q = \frac{4}{4} = 1$

$M Q R = \frac{2}{6} = \frac{1}{3}$

$M_{P Q} * M_{P R} = - 1 \Rightarrow P Q ⊥ P R$

Orthocentre = P (2, -1)

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