Maths

Let $l = \frac{2 b^{2}}{a} = \frac{2 (1)}{3}$ be a twice differentiable function such that $\Rightarrow 9 (e^{2} + l)$ $= 9 (\frac{10}{9} + \frac{2}{3})$ , for all $= 10 + 6$ . If $= 16$ , then the value of $= \frac{10!}{5! 3! 2!} = 2520$ is

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Contributor-Level 10

$f : R \to R$

$(s i n ? x c o s ? y) (f (2 x + 2 y) - f (2 x - 2 y)) = (c o s ? x$ Put $s i n ? y) (f (2 x + 2 y) + f (2 x - 2 y))$

$x, y \in R$ Put $f^{'} (0) = \frac{1}{2}$

$24 f^{''} (\frac{5 π}{3})$

$(s i n ? x c o s ? y) (f (2 x + 2 y) - f (2 x - 2 y)) = (c o s ? x s i n ? y)$

$(f (2 x + 2 y) + f (2 x - 2 y))$

$f (2 x + 2 y) (s i n ? (x - y)) = f (2 x - 2 y) s i n ? (x + y)$

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New answer posted

a year ago

Let A be a matrix of order $3 * 3$ and $| A | = 5$ . If $| 2 a d j (3 A a d j (2 A)) | = 2^{α} \cdot 3^{β} \cdot 5^{γ} α, β, γ \in N$ then $α + β + γ$ is equal to

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alok kumar singh

Contributor-Level 10

: $? 2$ adj $(3 A$ adj(2A))|
${= 2}^{3} . ? 3 A$ adj(2A)| $|^{2}$

${= 2}^{3} \cdot {(3^{3})}^{2} \cdot | A |^{2} \cdot | a d j (2 A) |^{2}$ intersect the line $= 2^{3} \cdot 3^{6} \cdot | A |^{2} \cdot {(| 2 A |^{2})}^{2}$ at the point

$= 2^{3} \cdot 3^{6} \cdot | A$

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New answer posted a year ago

Let one focus of the hyperbola $= 22$ be at ( $H : \frac{x^{2}}{a^{2}} - \frac{y^{2}}{b^{2}} = 1$ ) and the corresponding directrix be $\sqrt[]{10}, 0$ . If e and $x = \frac{9}{\sqrt[]{10}}$ respectively are the eccentricity and the length of the latus rectum of H, then $l$ is equal to

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Contributor-Level 10

Using the standard equations of a hyperbola:

$9 (e^{2} + l)$ and directrix $f o c u s a e = \sqrt[]{10}$

By multiplying both focus and directrix, we get
$\frac{a}{e} = \frac{9}{\sqrt[]{10}}$ and $\Rightarrow a^{2} = 9$
Now $e = \frac{\sqrt[]{10}}{3}$
$(a e)^{2} = a^{2} + b^{2}$

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New answer posted

a year ago

The largest $n \in N$ such that $3^{n}$ divides 50 ! is

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Contributor-Level 10

We can convert 50! In terms of prime factor: $2^{α} \cdot 3^{β} \cdot 5^{γ}$ & using the greatest integer function.

n $= [\frac{50}{3}] + [\frac{50}{3^{2}}] + [\frac{50}{3^{3}}] + [\frac{50}{3^{4}}]$
$= 16 + 5 + 1 + 0$ The maximum value of n is 22.

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New answer posted

a year ago

Let a be an integer such that $\underset{x \to 7}{l i m} \frac{18 - [1 - x]}{[x - 3 a]}$ $\underset{}{} \frac{}{}$ exists, where [t] is greatest integer $\leq$ t. Then a is equal to:

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Contributor-Level 9

$\underset{x \to 7}{l i m} \frac{18 - [1 - x]}{[x - 3 a]}$

exist & $a \in I .$

$= \underset{x \to 7}{l i m} \frac{17 - [- x]}{[x] - 3 a}$

exist

RHL = $\underset{x \to 7^{+}}{l i m} \frac{17 - [- x]}{[x] - 3 a} = \frac{25}{7 - 3 a} [a \neq \frac{7}{3}]$

$L H L = \underset{x \to 7^{-}}{l i m} \frac{17 - [- x]}{[x] - 3 a} = \frac{24}{6 - 3 a} [a \neq 2]$

LHL = RHL

$\Rightarrow \frac{25}{7 - 3 a} = \frac{8}{2 - a}$

$∴ a = - 6$

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New answer posted

a year ago

Let $\frac{d y}{d x} = \frac{a x - b y + a}{b x + c y + a},$ $\frac{}{} \frac{}{}$ where a, b, c, are constants, represent a circle passing through the point (2, 5). Then the shortest distance of the point (11, 6) from this circle is:

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Contributor-Level 9

$\frac{d y}{d x} = \frac{a x - b y + a}{b x + c y + a}$

$= b x d y + c y d y + a d y = a x d x - b y d x + a d x$

$= \frac{c y^{2}}{2} + a y - \frac{a x^{2}}{2} - a x + b x y = k$

$a x^{2} + a y^{2} + 2 a x - 2 a y = k$

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

Short distance of (11,6)

= $\sqrt[]{1 2^{2} + 5^{2}} - 5$ $^{}^{}$

= 13 – 5

= 8

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New answer posted

a year ago

If $^{}$ $x = \sum_{n = 0}^{\infty} a^{n},$ $y = \sum_{n = 0}^{\infty} b^{n}, z = \sum_{n = 0}^{\infty} c^{n},$ $^{}^{}$ where a, b, c are in A.P. and $| a | < 1, | b | < 1, | c | < 1, a b c \neq 0,$ then:

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Contributor-Level 9

$x = \sum_{n = 0}^{\infty} a^{n} = \frac{1}{1 - a} y = \sum_{n = 0}^{\infty} b^{n} = \frac{1}{1 - b} \sum_{n = 0}^{\infty} c^{n} = \frac{1}{1 - c}$

Now,

a, b, c AP

1 – a, 1 – b, 1 – c AP

$\frac{1}{1 - a}, \frac{1}{1 - b}, \frac{1}{1 - c} \to H P$

x, y, z HP

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New answer posted

a year ago

Maths Maths Matrices

Le the system of linear equations x + 2y + z = 2, x + 3y – z = , x + y + 2z = be inconsistent. Then is equal to:

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Contributor-Level 9

x + 2y + z = 2

$α x + 3 y - z = α$

$- α x + y + 2 z = - α$

$Δ = | \begin{matrix} 1 & 2 & 1 \\ α & 3 & - 1 \\ - α & 1 & 2 \end{matrix} | = 1 (6 + 1) - 2 (2 α - α) + 1 (α + 3 α) = 7 + 2 a$

$α = - \frac{7}{2}$

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New answer posted

a year ago

The area of the polygon, whose vertices are the non-real roots of the equation $\bar{z} = i z^{2} i s :$ $^{}$

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Maths Quantitative Aptitude Prep Tips for MBA

Contributor-Level 9

$\bar{z} = i z^{2}$

Let z = x + iy

x – iy = I (x2 – y2 + 2xiy)

Case-I

x = 0

y2 = y

y = 0, 1

Case – II

$y = - \frac{1}{2}$

$\Rightarrow x^{2} - \frac{1}{4} = \frac{1}{2} \Rightarrow x = \pm \frac{\sqrt[]{3}}{2}$

Area of polygon

$= \frac{1}{2} | \begin{matrix} 0 & 1 & 1 \\ \frac{\sqrt[]{3}}{2} & \frac{- 1}{2} & 1 \\ \frac{- \sqrt[]{3}}{2} & \frac{- 1}{2} & 1 \end{matrix} | = \frac{1}{2} | - \sqrt[]{3} - \frac{\sqrt[]{3}}{2} | = \frac{3 \sqrt[]{3}}{4}$

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New answer posted

a year ago

There are three containers of equal capacity. The ratio of milk and water in three containers are 2 : 3, 2 : 5 and 3 : 7. If mixture of all three containers put into a large container, then what will be the ratio of milk and water in the large container?

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