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4 months ago

Maths Maths NCERT Exemplar Solutions Class 12th Chapter Five

Let $α$ and $β$ be roots of $x^{2} - 3 x + p = 0$ and $γ$ and $δ$ be the roots of $x^{2} - 6 x + q = 0$ . If $α, β$ , $γ, δ$ , form a geometric progression. Then ratio $(2 q + p) : (2 q - p)$ is:

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alok kumar singh

Contributor-Level 10

$x^{2} - 3 x + p = 0$

$α, β, γ, δ$ in G.P.

$α + α r = 3$

$x^{2} - 6 x + q = 0$

$α r^{2} + α r^{3} = 6$

$(2) \div (1) \Rightarrow r^{2} = 2$

So, $\frac{2 q + p}{2 q - p} = \frac{2 r^{5} + r}{2 r^{5} - r} = \frac{2 r^{4} + 1}{2 r^{4} - 1} = \frac{9}{7}$

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4 months ago

Maths Limits and Derivatives

Let the lines y + 2x = $\sqrt[]{11} + 7 \sqrt[]{7}$ and $2 y + x = 2 \sqrt[]{11} + 6 \sqrt[]{7}$ be normal to a circle C : ${(x - h)}^{2} + {(y - k)}^{2} = r^{2} .$ If the line $\sqrt[]{11} y - 3 x = \frac{5 \sqrt[]{77}}{3} + 11$ is tangent to the circle C, then the value of ${(5 h - 8 k)}^{2} + 5 r^{2}$ is equal to

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Payal Gupta

Contributor-Level 10

$y + 2 x = \sqrt[]{11} + 7 \sqrt[]{7}$ ……… (i)

$2 y + x = 2 \sqrt[]{11} + 6 \sqrt[]{7}$ ……… (ii)

$\Rightarrow x + y = \sqrt[]{11} + \frac{13}{3} \sqrt[]{7}$ ……… (iii)

Centre of the circle given by solving (i) & (ii)

$a s (\frac{8 \sqrt[]{7}}{3}, \sqrt[]{11} + \frac{5 \sqrt[]{7}}{3})$

Again $\sqrt[]{11} y - 3 x = \frac{5 \sqrt[]{77}}{3}$ is tangent to the circle.

$∴ r = | \frac{\sqrt[]{11} (\sqrt[]{11} + \frac{5 \sqrt[]{7}}{3}) - 8 \sqrt[]{7} - \frac{5 \sqrt[]{77}}{3}}{\sqrt[]{11 + 9}} | = | \frac{11 - 8 \sqrt[]{7}}{\sqrt[]{20}} |$

$∴ {(5 h - 8 k)}^{2} + 5 r^{2} = 816$

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4 months ago

Maths Linear Programming

The number of elements in the set ${z = a + i b \in C : a, b \in Z a n d 1 < | z - 3 + 2 i | < 4}$ is

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Payal Gupta

Contributor-Level 10

$1 < | z - 3 + 2 i | < 4$

$z = a + i b a, b \in I$

⇒ 1 < (a 3)² + (b + 2)² < 16

in equivalent to 1 < ² + β² < 16 $α = a - 3 \in l$

$(\pm 2, 0) (\pm 3, 0), (0, \pm 2) (0, \pm 3)$ $β = b + 2 \in l$

Total 40 such points are possible

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4 months ago

Maths Probability

The number of positive integers k such that the constant term in the binomial expansion of ${(2 x^{3} + \frac{3}{x^{k}})}^{12}, x \neq 0 i s 2^{8} . l, w h e r e l$ is an odd integer, is……….

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Payal Gupta

Contributor-Level 10

${(2 x^{3} + \frac{3}{x^{k}})}^{12}$

gen term =

$=^{12} C_{r} 2^{12 - r} . 3^{r} . x^{36 - 3 r - r k}$

For constant term

36 – 3r – rk = 0

$\Rightarrow k = \frac{36 - 3 r}{r}$

for r = 1, 2, 4

¹²C_r2^12-r>2⁸

Possible values of k = 3, 1

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4 months ago

Maths Straight Lines

Let A = ${1, a_{1}, a_{2} . . . . a_{18}, 77}$ be a set of integers with 1 < a₁ < a₂ < < a₁₈ < 77. Let the set A + A = ${x + y : x, y \in A}$ contains exactly 39 elements. Then, the value of a₁ + a₂ +…. + a₁₈ is equal to

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Payal Gupta

Contributor-Level 10

1 < a₁ < a₂ ……18 < 77

77 = 1 + (20 – 1) . d If n numbers are in A.P.

76 = 19 * d ⇒ d = 4

⇒ a₁ = 5

$\begin{array}{l} ∴ a_{1} + a_{2} + . . . . + a_{18} \\ = \frac{18}{2} [2 * 5 + 17 * 4] = 702 \end{array}$

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4 months ago

Maths Conic Sections

Let $l$ be a line which is normal to the curve y = 2x² + x + 2 at a point P on the curve. If the point Q(6, 4) lies on the line $l$ and O is origin, then the area of the triangle OPQ is equal to

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Payal Gupta

Contributor-Level 10

$y = 2 x^{2} + x + 2 . . . . (i)$

$\frac{d y}{d x} = 4 x + 1$

Slope of normal

$- \frac{d x}{d y} = - \frac{1}{4 x + 1}$

Equation of PQ y - β = $- \frac{1}{4 α + 1} (x - α)$

It passes (6, 4)

$\Rightarrow (4 - β) (4 α + 1) = - (6 - α)$

$4 α^{3} + 3 α^{2} - 3 α - 3 = 0$

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4 months ago

Maths Differential Equations

A ray of light passing through the point P(2, 3) reflects on the x-axis at point A and the reflected ray passes through the point Q(5, 4). Let R be the point that divides the line segment AQ internally into the ration 2 : 1. Let the co-ordinates of the foot of the perpendicular M from R on the bisector of the angle PAQ be (a, b). Then the value of 7a + 3b is equal to……….

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Payal Gupta

Contributor-Level 10

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New answer posted

4 months ago

Maths Maths NCERT Exemplar Solutions Class 12th Chapter Eight

If $\vec{a} = 2 \hat{i} + \hat{j} + 3 \hat{k}, \vec{b} = 3 \hat{i} + 3 \hat{j} + \hat{k} a n d \vec{c} = c_{1} \hat{i} + c_{2} \hat{j} + c_{3} \hat{k}$ are coplanar vectors and $\vec{a} . \vec{c} = 5, \vec{b} ⊥ \vec{c}, t h e n 122 (c_{1} + c_{2} + c_{3})$ is equal to……….

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Payal Gupta

Contributor-Level 10

$\vec{a} = 2 \hat{i} + \hat{j} + 3 \hat{k}$

$\begin{array}{l} \vec{b} = 3 \hat{i} + 3 \hat{j} + \hat{k} \\ \vec{c} = c_{1} \hat{i} + c_{2} \hat{j} + c_{3} \hat{k} \end{array}$

$C o p l n a n a r \Rightarrow | \begin{matrix} 2 & 1 & 3 \\ 3 & 3 & 1 \\ c_{1} & c_{2} & c_{3} \end{matrix} | = 0$

$\begin{array}{l} \Rightarrow - 8 c_{1} + 7 c_{2} + 12 c_{3} = 0 . . . . . . . . (i) \\ \vec{a} . \vec{c} = 5 \Rightarrow 2 c_{1} + c_{2} + 3 c_{3} = 5 . . . . . . . . (i i) \\ \vec{b} . \vec{c} = 0 \Rightarrow 3 c_{1} + 3 c_{2} + c_{3} = 0 . . . . . . . . (i i i) \end{array}$

Solving (i), (ii), (iii)

$C_{1} = \frac{10}{122}, c_{2} = \frac{- 85}{122}, c_{3} = \frac{225}{122}$

$∴ 122 (c_{1} + c_{2} + c_{3}) = 150$

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4 months ago

Maths Maths NCERT Exemplar Solutions Class 12th Chapter Six

The mean and standard deviation of 15 observations are found to be 8 and 3 respectively. On rechecking it was found that, in the observations, 20 was misread as 5. Then, the correct variance is equal to…………….

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Payal Gupta

Contributor-Level 10

$M e a n \frac{\sum_{i = 1}^{15} x_{i}}{15} = 8$

$\Rightarrow \sum_{i = 1}^{15} x_{i} = 120 . . . . . . . (i)$

S.D = 3

$\Rightarrow \sum_{i = 1}^{15} {x_{i}}^{2} = 15 (9 + 64) = 15 * 73 . . . . . . . . . . (i i)$

Given 20 has misread as 5

$∴$ in new case

$\sum_{q = 1}^{15} x_{i^{2}} = 15 * 73 - 25 + 400 = 1470$

mean in new case

$\bar{x} = \frac{120 - 5 + 20}{15}$

$∴ v$ ariance in new case

$σ_{n e w}^{2} = \frac{1}{15} (1470) - 81 = 17$

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4 months ago

Maths Maths NCERT Exemplar Solutions Class 12th Chapter Five

The number of real solutions of the equation e^4x + 4e^3x – 58e^2x + 4e^x + 1 = 0, is……………

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Payal Gupta

Contributor-Level 10

$e^{4 x} + 4 e^{3 x} - 58 e^{2 x} + 4 e^{x} + 1 = 0$

$(e^{2 x} + \frac{1}{e^{2 x}}) + 4 (e^{x} + \frac{1}{e^{x}}) - 58 = 0$

$\Rightarrow {(e^{x} + \frac{1}{e^{x}} + 2)}^{2} = 64$

$e^{x}$ $=$ $\frac{6 \pm 32}{2}$ $=$ $3$ $\pm$ $2$

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