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Maths NCERT Exemplar Solutions Class 12th Chapter Ten

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2 months ago

Maths NCERT Exemplar Solutions Class 12th Chapter Ten 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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Maths NCERT Exemplar Solutions Class 12th Chapter Ten 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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Maths NCERT Exemplar Solutions Class 12th Chapter Ten 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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Maths NCERT Exemplar Solutions Class 12th Chapter Ten 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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2 months ago

Maths NCERT Exemplar Solutions Class 12th Chapter Ten 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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Maths NCERT Exemplar Solutions Class 12th Chapter Ten 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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2 months ago

Maths NCERT Exemplar Solutions Class 12th Chapter Ten Maths NCERT Exemplar Solutions Class 12th Chapter Five

Let R₁ and R₂ be relations on the set {1,2,…., 50} such that

R₁ = ${(p, p^{n}) : p i s a p r i m e a n d n \geq 0 i s a n i n t ? e g e r} a n d$

R₂ = ${(p, p^{n}) : p i s a p r i m e a n d n = 0 o r 1} .$

Then, the number of elements in R₁ – R₂ is…………

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Payal Gupta

Contributor-Level 10

A = {1, 2, 3, ….50}

R₁ = (2, 1), (2, 2), (2, 4)…. (2, 32)

(3, 1) (3, 3) (3, 9) (3, 27)

(13, 1) (13, 13) etc.

R₂ = { (2, 1), (2, 2), (3, 1), (3, 3), (5, 1), (5, 5), (7, 1), (7, 7), (1, 1), (11, 11)…}

$∴ R_{1} - R_{2}$ contains only 9 elements.

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2 months ago

Maths NCERT Exemplar Solutions Class 12th Chapter Ten Probability

The probability, that in a randomly selected 3-digit number at least two digits are odd, is

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Payal Gupta

Contributor-Level 10

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Maths NCERT Exemplar Solutions Class 12th Chapter Ten Three Dimensional Geometry

Let the plane $P : \vec{r} . \vec{a} = d$ contains the line of intersection of two planes $\vec{r} . (\hat{i} + 3 \hat{j} - \hat{k}) = 6$ and $\vec{r} . (6 \hat{i} + 5 \hat{j} - \hat{k}) = 7 .$ If the plane P passes through the point $(2, 3, \frac{1}{2})$ , then the value of $\frac{{| 13 \vec{a} |}^{2}}{d^{2}}$ is equal to

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Payal Gupta

Contributor-Level 10

$P : (x + 3 y - z - 6) = λ (- 6 x + 5 y - z - 7) = 0$

Passes $(2, 3, \frac{1}{2})$

$\Rightarrow (2 + 9 - \frac{1}{2} - 6) = λ (- 12 + 15 - \frac{1}{2} - 7) = 0$

$\Rightarrow λ = 1$

$\frac{{| 13 \vec{a} |}^{2}}{d^{2}} = \frac{{(13)}^{2} (93)}{{(13)}^{2}} = 93$

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Maths NCERT Exemplar Solutions Class 12th Chapter Ten Three Dimensional Geometry

The acute angle between the planes P₁ and P₂, when P₁ and P₂ are the planes passing through the intersection of the planes 5x + 8y + 13z – 29 = 0 and 8x – 7y + z – 20 = 0 and the points (2, 1, 3) and (0, 1, 2), respectively, is

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Payal Gupta

Contributor-Level 10

$(5 x + 8 y + 13 z - 29) + λ (8 x - 7 y + z - 20) = 0$

P₁ passing (2, 1, 3)

(10 + 8 + 39 – 29) + $λ (16 - 7 + 3 - 20) = 0$

$28 - 8 λ = 0 λ = 7 / 2$

$\Rightarrow$ 2X – Y + Z – 6 = 0 ….(i)

For P₂ passes (0, 1, 2)

$(8 + 26 - 29) + λ (- 7 + 2 - 20) = 0$

$5 - 25 λ = 0 \Rightarrow λ = \frac{1}{5}$

$P_{2} : x + y + 2 z - 5 = 0 . . . . . . . . (i i)$

Acute angle between the planes

$c o s θ = \frac{2 - 1 + 2}{\sqrt[]{4 + 1 + 1} \sqrt[]{1 + 1 + 4}} = \frac{1}{2}$

$\Rightarrow θ = \frac{π}{3}$

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