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4 months ago

Maths Class 11th

Let 3, 6, 9, 12,…. upto 78 terms and 5, 9, 13, 17,…. upto 59 terms be two series. Then, the sum of the terms common to both the series is equal to

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Payal Gupta

Contributor-Level 10

First common term to both AP's is 9

t₇₈ of $(3, 6, 9, . . . . . .) = 78 * 3 = 234$

t₅₉ of $(5, 9, 13, . . . . . . . .) = 5 + (5 - 1) 4 = 237$

n^th common term $\leq 234$

9 + (n – 1) 12 $\leq$ 234

n < $\frac{237}{12} \Rightarrow n = 19$

Now sum of 19 terms with a = 9, d = 12

$= \frac{19}{2} (2.9 + (19 - 1) \cdot 12) = 2223$

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New answer posted

4 months ago

Maths Class 12th

Let y = y(x), x > 1, be the solution of the differential equation

$(x - 1) \frac{d y}{d x} + 2 x y = \frac{1}{x - 1}, w i t h y (2) = \frac{1 + e^{4}}{2 e^{4}} .$ If y(3) = $\frac{e^{α} + 1}{β e^{α}}$ , then the value of α+β is equal to

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Payal Gupta

Contributor-Level 10

$\frac{d y}{d x} + \frac{2 x}{x - 1} \cdot y = \frac{1}{{(x - 1)}^{2}}$

IF $= e^{\int \frac{2 x}{x - 1} d x}$

= $e^{2 x} \cdot {(x - 1)}^{2}$

$y \cdot e^{2 x} {(x - 1)}^{2} = {e^{2 x} \frac{{(x - 1)}^{2}}{{(x - 1)}^{2}} d x + C$

y = $\frac{e^{2 x}}{2 {(x - 1)}^{2}} + \frac{C}{{(x - 1)}^{2}}$

y (2) = $\frac{1 + e^{4}}{2 e^{4}}, \Rightarrow C = \frac{1}{2}$

y (3) = $\frac{e^{α} + 1}{β e^{α}} = \frac{e^{6} + 1}{8 e^{6}}$

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New answer posted

4 months ago

Maths Class 12th

Let $\vec{a} = \hat{i} - 2 \hat{j} + 3 \hat{k}, \vec{b} = \hat{i} + \hat{j} + \hat{k} a n d \vec{c}$ be a vector such that $\vec{a} + (\vec{b} * \vec{c}) = \vec{0} a n d \vec{b} . \vec{c} = 5 .$ Then, the value of $3 (\vec{c} . \vec{a})$ is equal to

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Payal Gupta

Contributor-Level 10

Data contradiction.

$\vec{a} * (\vec{b} * \vec{c}) = (\vec{a} \cdot \vec{c}) \vec{b} - (\vec{a} \cdot \vec{b}) \vec{c}$

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New answer posted

4 months ago

Maths Maths Binomial Theorem

Let $n \geq 5$ be an integer. If 9ⁿ – 8n – 1 = 64α and 6ⁿ – 5n – 1 = 25β, then α - β is equal to

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Payal Gupta

Contributor-Level 10

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New answer posted

4 months ago

Maths Class 11th

Negation of the Boolean statement $(p \lor q) \Rightarrow ((\sim r) \lor p)$ is equivalent to

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Payal Gupta

Contributor-Level 10

pva ⇒ $(\sim r v p)$

$\sim (p \lor q) \lor (\sim r v p)$

its negation as asked in question

$(p \lor q) \land (\sim p \land r)$

= $(p \land \sim p \land r) \lor (q \land r \land \sim p)$

= $(p \land r \land \sim p) [a s p \land \sim p i s f a l s e]$

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New answer posted

4 months ago

Maths Class 11th

From the base of a pole of height 20 meter, the angle of elevation of the top of tower is 60°. The pole subtends an angle 30° at the top of the tower. Then the height of the tower is

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Payal Gupta

Contributor-Level 10

Let base = b

$t a n 60 ° = \frac{h}{b}$

$t a n 30 ° = \frac{h ? 20}{b}$

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New question posted

4 months ago

Maths Class 11th

From the base of a pole of height 20 meter, the angle of elevation of the top of tower is 60°. The pole subtends an angle 30° at the top of the tower. Then the height of the tower is

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New answer posted

4 months ago

Maths Maths Statistics

The number of values of $a \in N$ such that the variance of 3, 7, 12, a, 43 a is a natural number is

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Payal Gupta

Contributor-Level 10

$M e a n = \frac{3 + 12 + 7 + a + (43 - a)}{5} = 13$

Variance = $\frac{3^{2} + 1 2^{2} + 7^{2} + a^{2} + {(43 - a)}^{2}}{5} - {(13)}^{2}$

$\Rightarrow \frac{2 a^{2} - a + 1}{5} \to N a t u r a l n u m b e r$

Let 2a² – a + 1 = 5x

D = 1 – 4 (2) (1 – 5n)

= 40n – 7, which is not $4 λ o r 4 λ + 1 f r o m .$

As each square form is $4 λ o r 4 λ + 1$

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New answer posted

4 months ago

Maths Relations and Functions

The probability that a relation R from ${x, y} t o {x, y}$ is both symmetric and transitive, is equal to

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Payal Gupta

Contributor-Level 10

Total number of possible relation = $2^{n^{2}} = 2^{4} = 16$

Favourable relations = $?, {(x, x)}, {(y, y)}$

${(x, x), (y, y)}$

${(x, x), (y, y), (x, y), (y, x)}$

Probability = $\frac{5}{16}$

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New answer posted

4 months ago

Maths Class 12th

Let A, B, C be three points whose position vectors respectively are

$\vec{a} = \hat{i} + 4 \hat{j} + 3 \hat{k}$

$\vec{b} = 2 \hat{i} + α \hat{j} + 4 \hat{k}, α \in R$

$\vec{c} = 3 \hat{i} - 2 \hat{j} + 5 \hat{k}$

If is the smallest positive integer for which $\vec{a}, \overset{\leftarrow}{b}, \vec{c}$ are noncollinear, then the length of the median, in $Δ A B C,$ through A is

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Payal Gupta

Contributor-Level 10

Mid point of BC is $\frac{1}{2} (5 \hat{i} + (α - 2) \hat{j} + 9 \hat{k})$

$\bar{A B} = \hat{i} + (α - 4) \hat{j} + \hat{k}$

$\bar{A C} = \hat{i} + (- 2 - α) \hat{j} + \hat{k}$

For = 1, $\bar{A B}$ and $\bar{A C}$ will be collinear. So for non collinearity

= 2

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