Maths

Let A = $([\begin{matrix} [x + 1] & [x + 2] & [x + 2] \\ [x] & [x + 3] & [x + 3] \\ [x] & [x + 2] & [x + 4] \end{matrix}])$ where [t] denotes the greatest integer less than or equal to t. If det(A) = 192, then the set of values of x is the interval.

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Contributor-Level 10

$| A | = | \begin{matrix} [x + 1] & [x + 2] & [x + 3] \\ [x] & [x + 3] & [x + 3] \\ [x] & [x + 2] & [x + 4] \end{matrix} | = | \begin{matrix} [x] + 1 & [x] + 2 & [x] + 3 \\ [x] & [x] + 3 & [x] + 3 \\ [x] & [x] + 2 & [x] + 4 \end{matrix} |$

$R_{1} \to R_{1} - R_{3} & R_{2} \to R_{2} - R_{3}$

$| \begin{matrix} 1 & 0 & - 1 \\ 0 & 1 & - 1 \\ [x] & [x] + 2 & [x] + 4 \end{matrix} | = 192$

$\Rightarrow [x] = 62$

$\Rightarrow x \in [62, 63)$

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3 months ago

If $y^{1 / 4} + y^{- 1 / 4} = 2 x a n d (x^{2} - 1) \frac{d^{2} y}{d x^{2}} + α x \frac{d y}{d x} + β y = 0, t h e n | α - β |$ is equal to_________

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Contributor-Level 10

$y^{1 / 4} + \frac{1}{y^{1 / 4}} = 2 x$

${(y^{1 / 4})}^{2} - 2 x y^{1 / 4} + 1 = 0$

=> ${(y^{1 / 4})}^{2} - 2 x y^{1 / 4} + 1 = 0$

$\Rightarrow \frac{d y}{d x} = \frac{4 y}{\sqrt[]{x^{2} - 1}} . . . . . . . . . . (i)$

$\Rightarrow (x^{2} - 1) \frac{d^{2} y}{d x^{2}} + \frac{x d y}{d x} - 16 y = 0 f r o m (i)$

=>α = 1, β = -16

|α - β| = 17

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3 months ago

Maths Class 11th

The set of all values of k > -1, for which the equation ${(3 x^{2} + 4 x + 3)}^{2} - (k + 1) (3 x^{2} + 4 x + 3) (3 x^{2} + 4 x + 2) + k {(3 x^{2} + 4 x + 2)}^{2} = 0$ has real roots is:

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Maths Differential Equations

Contributor-Level 10

${(3 x^{2} + 4 x + 3)}^{2} - (k + 1) (3 x^{2} + 4 x + 3) (3 x^{2} + 4 x + 2) + k {(3 x^{2} + 4 x + 2)}^{2} = 0$

Let $3 x^{2} + 4 x + 3 = a & 3 x^{2} + 4 x + 2 = b \Rightarrow a - 1$

So, (i) becomes a² – (k + 1)ab + kb² = 0

(a – kb) (a – b) = 0 Þ a = kb or a = b ® not possible

->3x² + 4x + 3 = k (3x² + 4x + 2)

For real roots $D \geq 0$

$16 {(k - 1)}^{2} - 12 (k - 1) (2 k - 3) \geq 0$

So, $k \in (1, \frac{5}{2}]$

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3 months ago

If $\int \frac{d x}{{(x^{2} + x + 1)}^{2}} = a t a n^{- 1} (\frac{2 x + 1}{\sqrt[]{3}}) + b (\frac{2 x + 1}{x^{2} + x + 1}) + C, x > 0$ where C is the constant of integration, then the value of 9 $(\sqrt[]{3} a + b)$ is equal to

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Contributor-Level 10

Let $l = \int \frac{d x}{{(x^{2} + x + 1)}^{2}} . . . . . . . . . . . . . (i)$

Now $\int \frac{d x}{x^{2} + x + 1} = \int (\frac{1}{x^{2} + x + 1}) d x$

by integration by parts

=> $\int \frac{d x}{{(x^{2} + x + 1)}^{2}} = \frac{4}{3 \sqrt[]{2}} t a n^{- 1} \frac{2 x + 1}{\sqrt[]{3}} + \frac{1}{3} \frac{(2 x + 1)}{(x^{2} + x + 1)} + c$

$\Rightarrow a = \frac{4}{3 \sqrt[]{3}}, b = \frac{1}{3}$

Now, $9 (\sqrt[]{3} a + b) = 9 (\frac{4}{3} + \frac{1}{3}) = 15$

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3 months ago

The equation of the plane passing through the line of intersection of the planes $\vec{r} . (\hat{i} + \hat{j} + \hat{k}) = 1$ and $\vec{r} . (2 \hat{i} + 3 \hat{j} - \hat{k}) + 4 = 0$ and parallel to the x –axis is

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Contributor-Level 10

Equation of the plane will be ${\vec{r} . (\hat{i} + \hat{j} + \hat{k}) - 1} + λ {\vec{r} . (2 \hat{i} + 3 \hat{j} - \hat{k}) + 4} = 0$

$\vec{r} . {(1 + 2 λ) i + (1 + 3 λ) \hat{j} + (1 - λ) \hat{k}} + (4 λ - 1) = 0$

-> $(1 + 2 λ) x + (1 + 3 λ) y + (1 - λ) z + (4 λ - 1) = 0 . . . . . . . . . (i)$

(i) is parallel to x-axis so its d.r.s will be (1, 0, 0)

-> $1 + 2 λ = 0 s o λ = \frac{- 1}{2}$

Hence required equation will be

$\vec{r} {- \frac{1}{2} \hat{j} . + \frac{3}{2} \hat{k}} + (- 3) = 0$

$\vec{r} . (\hat{j} - 3 \hat{k}) + 6 = 0$

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3 months ago

Maths Statistics

Let n be an odd natural number such that the variance of 1, 2, 3, 4, …, n is 14. Then n is equal to ________

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Contributor-Level 10

Variance = $\frac{\sum {(x_{i} - \bar{x})}^{2}}{n} = \frac{\sum ({x_{i}}^{2} + {\bar{x}}^{2} - 2 \bar{x} x i)}{n}$

$= \frac{\sum {x_{i}}^{2} + {\bar{x}}^{2} \sum 1 - 2 \bar{x} \sum x_{i}}{n}$

$= \frac{\frac{n (n + 1) (2 n + 1)}{6} + {(\frac{n (n + 1)}{2 n})}^{2} . n - \frac{2 (n + 1)}{2 n} . \frac{n (n + 1)}{2}}{n}$ $= \frac{n^{2} - 1}{12}$

Now, $\frac{n^{2} - 1}{12} = 14 s o n = 13$

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3 months ago

A differential equation representing the family of parabolas with axis parallel to y – axis and whose length of latus rectum is the distance of the point (2, -3) form the line 3x + 4y = 5, is given by:

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Contributor-Level 10

$L . R . = \frac{| 3 * 2 + 4 x - 3 - 5 |}{\sqrt[]{3^{2} + 4^{2}}} = \frac{11}{5}$

Equation of family of parabolas

${(x - h)}^{2} = \frac{11}{5} (y - k)$

Differentiate 2 (x – h) = $\frac{11}{5} \frac{d y}{d x}$

Again differentiate 2 = $\frac{11}{5} \frac{d^{2} y}{d x^{2}}$

$11 \frac{d^{2} y}{d x^{2}} = 10$

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3 months ago

If the system of linear equations

2x + y – z = 3

x – y – z = a

3x + 3y + bz = 3

has infinitely many solution, then α + β - αβ is equal to__________

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Contributor-Level 10

Given 2x + y – z = 3 . (i)

x – y – z = α . (ii)

3x + 3y + βz = 3 . (iii)

(i) x 2 – (ii) – (iii) – (1 + β) z = 3 - α

For infinite solution 1 + β = 0 = 3 - α

=> α = 3, β = -1

So, α + β - αβ = 5

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3 months ago

Maths Class 11th

The area of the region bounded by the parabola (y – 2)² = (x – 1), the tangent to it at the point whose ordinate is 3 and the x – axis is:

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Contributor-Level 10

(y – 2)² = (x – 1)

2 (y – 2) $\frac{d y}{d x} = 1$

$\Rightarrow \frac{d y}{d x} (2, 3) = \frac{1}{2 (3 - 2)} = \frac{1}{2}$

Equation of tangent at P (2, 3):

$y - 3 = \frac{1}{2} (x - 2)$

2y – 6 = x – 2

x – 2y + 4 = 0

Q (-4, 0)

Required area = $\int_{0}^{3} ({(y - 2)}^{2} + 1 - (2 y - 4)) d y = 9$

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3 months ago

Maths Class 11th

A number is called a palindrome if it reads the same backward as well as forward. For example 285582 is a six digit palindrome. The number of six digit palindromes, which are divisible by 55, is_______.

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