Maths

Contributor-Level 10

$\sum_{n = 1}^{21} \frac{3}{(4 n - 1) (4 n + 3)}$

$= \frac{3}{4} \sum_{n = 1}^{21} \frac{(4 n + 3) - (4 n - 1)}{(4 n - 1) (4 n + 3)}$

$= \frac{3}{4} \frac{4 n + 3 - 2}{3 (4 n + 3)} = \frac{n}{4 n + 3}$

for n = 21

$S_{21} = \frac{21}{84 + 3} = \frac{21}{87} = \frac{7}{29}$

0 0

New answer posted

a year ago

The remainder when (11)¹⁰¹¹ + (1011)¹¹ is divided by 9 is

0 Follower 9 Views

Maths Inverse Trigonometric Functions

Contributor-Level 10

$11 \equiv 2 m o d (9)$

${(11)}^{1011} \equiv 2^{1011} m o d (9)$

Again 23 $\equiv$ 1 mod (9)

$\Rightarrow {(2^{3})}^{337} \equiv {(- 1)}^{337} m o d (9)$

$∴ {(11)}^{1011} + {(1011)}^{11} \equiv 8 m o d (9)$

$∴$ Remainder = 8

0 0

New answer posted

a year ago

The number of bijective functions f : {1, 3, 5, 7,…., 99} {2, 4, 6, 8,…., 100}, such that $f (3) \geq f (9) \geq f (15) \geq f (21) \geq . . . . \geq f (99),$ is

0 Follower 16 Views

Contributor-Level 10

f : {1, 3, 5, 7, ….,99} {2, 4, 6, 8, …….100}

$f (3) \geq f (9) \geq f (15) . . . . . . \geq f (99)$

$3 + (n - 1) 6 = 99 \Rightarrow n = 17$

cases f (3) > f (9) > f (15) ……. > f (99)

$∴$ from the set {2, 4, 6, …., 100}

17 distinct numbers can be selected in 50C17 ways again remaining {1, 5, 7, 11, ….} can map in 33! ways

$∴$ total number of such required functions

$=^{50} C_{17} * 33!$

$= \frac{50!}{33! 17!} * 33! =^{50} P_{33}$

0 0

New answer posted

a year ago

Maths Maths Determinants

The number of real values of $λ$ , such that the system of linear equations

2x – 3y + 5z = 9

x + 3y – z = 18

3x – y $(λ^{2} - | λ |) z = 16$ has no solutions, is

0 Follower 34 Views

Contributor-Level 10

System of equation can be written as

$(\begin{matrix} 2 & - 3 & 5 \\ 1 & 3 & - 1 \\ 3 & - 1 & λ^{2} - | λ | \end{matrix}) (\begin{array}{l} x \\ y \\ z \end{array}) = (\begin{array}{l} 9 \\ - 18 \\ 16 \end{array})$

for no solution

$\frac{9}{10} (λ^{2} - | λ | + 3) - 7 = 0$

$\Rightarrow 9 λ^{2} - 9 | λ | - 43 = 0$

$\Rightarrow | λ | = \frac{9 + \sqrt[]{81 + 36 * 43}}{18} > 0$

$∴$ only two possible value of $λ$ are there.

0 0

New answer posted

a year ago

For $z \in C$ if the minimum value of $(| z - 3 \sqrt[]{2} | + | z - p \sqrt[]{2} i |) i s 5 \sqrt[]{2}$ , then a value of p is

0 Follower 3 Views

Contributor-Level 10

$| z - 3 \sqrt[]{2} | + | z - p \sqrt[]{2} i |$ is minimum for z, $3 \sqrt[]{2} & p \sqrt[]{2} i$ are collinear.

$\Leftrightarrow {(3 \sqrt[]{2})}^{2} + {(p \sqrt[]{2})}^{2} = {(5 \sqrt[]{2})}^{2}$

$\Rightarrow 18 + 2 p^{2} = 50$

$2 p^{2} = 32$

$p = \pm 4$

0 0

New answer posted

a year ago

Let $S = {(x, y) \in N * N : 9 {(x - 3)}^{2} + 16 {(y - 4)}^{2} \leq 144}$ and $T = {(x, y) \in R * R : {(x - 7)}^{2} + {(y - 4)}^{2} \leq 36} .$

Then $n (S \cap T)$ is equal to………….

0 Follower 16 Views

Contributor-Level 10

$\frac{{(x - 3)}^{2}}{16} + \frac{{(y - 4)}^{2}}{9} \leq 1, x, y \in N, {(x - 7)}^{2} + {(y - 4)}^{2} \geq 36, x, y \in R$

Total number of common point = 1 + 5 + 7 + 5 + 5 + 3 + 1 = 27

0 0

New answer posted

a year ago

Let $\vec{a} a n d \vec{b}$ be two vectors such that $| \vec{a} + \vec{b} | = {| \vec{a} |}^{2} + 2 {| \vec{b} |}^{2}, \vec{a} . \vec{b} = 3 a n d {| \vec{a} * \vec{b} |}^{2} = 75 .$ Then ${| \overset{\leftrightarrow}{a} |}^{2}$ is equal to…………

0 Follower 3 Views

Contributor-Level 10

Given : ${| \vec{a} + \vec{b} |}^{2} = {| \vec{a} |}^{2} + 2 {| \vec{b} |}^{2} & \vec{a} \cdot \vec{b} = 3$

$| \vec{a} | | \vec{b} | c o s θ = 3$

$| \vec{a} | c o s θ = \frac{3}{\sqrt[]{6}} = \sqrt[]{\frac{9}{6}} = \sqrt[]{\frac{3}{2}}$

${| \vec{a} |}^{2} {| \vec{b} |}^{2} s i n^{2} θ = 75$

$| \vec{a} | s i n θ = \sqrt[]{\frac{75}{6}} = \frac{5}{\sqrt[]{2}}$

$| \vec{a} | c o s θ = \sqrt[]{\frac{3}{2}}$

${| \vec{a} |}^{2} = \frac{25}{2} + \frac{3}{2} = \frac{28}{2} = 14$

0 0

New answer posted

a year ago

Maths Class 11th

Let AB be a chord of length 12 of the circle ${(x - 2)}^{2} + {(y + 1)}^{2} = \frac{169}{4} .$

If tangents drawn to the circle at points A and B intersect at the point P, then five times the distance of point P form chord AB is equal to……………

0 Follower 4 Views

Contributor-Level 10

Let C be the centre and M be the mid point of AB

$Δ A P C : s i n θ = \frac{13 / 2}{p c} = \frac{5}{13} \Rightarrow p C = \frac{169}{10}$

$Δ A M C : c o s θ = \frac{6}{13 / 2} = \frac{12}{13}$

PC = $\frac{169}{10}, M C = \frac{13}{2} s i n θ = \frac{13}{2} \cdot \frac{5}{13}$

PM = PC – MC = $\frac{169}{10} - \frac{5}{2} = \frac{144}{10}$

5PM = 72

0 0

New answer posted

a year ago

If the tangent to the curve $y = x^{3} - x^{2} + x$ at the point (a, b) is also tangent to the curve y = 5x² + 2x – 25 at the point (2, -1), then $| 2 a + 9 b |$ is equal to………….

0 Follower 9 Views

Contributor-Level 10

y = 5x² + 2x – 25

P(2, -1)

T_(p) : T = 0

->y – 1 = 10x(2) + 2(x + 2) – 50

$\Rightarrow y = 22 x - 45$ is also tangent to y = x³ – x² + x at point (a, b)

For y = x³ -x² + x

$\frac{d y}{d x} = 3 x^{2} - 2 x + 1 > 0$

y = 22x $- \frac{1939}{27}$ which is not tangent to the curve.

$3 a^{2} - 2 a + 1 = 22$ (slope of tangent)

$\Rightarrow 3 a^{2} - 2 a - 21 = 0 \to a = \frac{2 \pm \sqrt[]{4 + 252}}{6} = \frac{2 \pm 16}{6} = 3, \frac{- 7}{3}$

b = 27 – 9 + 3 = 21

tangent : y – 21 = 22(x – 3)

->y = 22x – 45

a = 3, b = 21

2a + 9b = 6 + 189 = 195

Also, $a^{3} - a^{2} + a = b$

For a = $\frac{- 7}{3}$

b = $\frac{- 343}{27} - \frac{49}{9} - \frac{7}{3}$

= $\frac{- 343 - 147 - 63}{27} = \frac{- 553}{27}$

0 0

New answer posted

a year ago

If [t] denotes the greatest integer $\leq t,$ then the number of points, at which the function $f (x) = 4 | 2 x + 3 | + 9 [x + \frac{1}{2}] - 12 [x + 20]$ is not differentiable in the open interval (-20, 20), is……….

0 Follower 6 Views