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8 months ago

Maths Class 11th

Let $S = {(\begin{matrix} - 1 & a \\ 0 & b \end{matrix}) a, b \in {1, 2, 3, . . . 100}};$ and let Tn = ${A \in S : A^{n (n + 1)} = I} .$ Then the number of elements in $\cap_{n = 1}^{100} T_{n} i s . . . . . . . . . . . .$

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Payal Gupta

Contributor-Level 10

$S = {(\begin{matrix} 1 & a \\ 0 & b \end{matrix}), a, b \in 1, 2, 3, . . . . . 100$

$∴ A = (\begin{matrix} - 1 & a \\ 0 & b \end{matrix})$ then even power of A as A = $(\begin{matrix} 1 & 0 \\ 0 & 1 \end{matrix}) .$

If b = 1 & $a \in {1, 2, 3 . . . . . 100}$ and n (n + 1) is always even

$∴ T_{1}, T_{2}, T_{3}, . . . . . . . ., T_{n}$ are all 1 for b = 1 and each value of a.

$∴ \cap_{n = 1}^{1000} T_{n} = 100$

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New answer posted

8 months ago

Maths Class 12th

Let S = ${z \in C : | z - 3 | \leq 1 a n d z (4 + 3 i) \leq 24} .$ If $α + i β$ is the point in S which in closest to 4i, then 25 $(α + β)$ is equal to

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Payal Gupta

Contributor-Level 10

Circle $| z - 3 | \leq 1 \Rightarrow {(x - 3)}^{2} + y^{2} \leq 1$

and line $z (4 + 3 i) + \bar{z} (4 - 3 i) \leq 24$

$\Rightarrow 4 x - 3 y \leq 12 ∴ s l o p e = t a n θ = \frac{4}{3}$

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New answer posted

8 months ago

Maths Class 12th

Let $λ$ be the largest value of $λ$ for which the function $f_{λ} (x) = 4 λ x^{3} - 36 λ x^{2} + 36 x + 48$ is increasing for all $x \in R .$ Then $f_{λ} . (1) + f_{λ} . (- 1)$ is equal to

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Payal Gupta

Contributor-Level 10

$? f_{λ} (x) = 4 λ x^{3} - 36 λ x^{2} + 36 x + 48$

$f_{λ} (x) = 12 (λ x^{2} - 6 λ x + 3)$

For increasing $f_{λ} (x) \geq 0$

$f_{λ}^{*} (x) = \frac{4}{3} x^{3} - 12 x^{2} + 36 x + 48 ∴ f_{λ}^{*} (1) + f_{λ}^{*} (- 1) = 73 \frac{1}{2} - 1 \frac{1}{2} = 72$

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New answer posted

8 months ago

Maths Class 12th

The slope of normal at any point (x,y), X > 0, y > 0 on the curve y = y(x) is given by $\frac{x^{2}}{x y - x^{2} y^{2} - 1} .$ If the curve passes through the point (1, 1), then e.y (e) is equal to

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Payal Gupta

Contributor-Level 10

$? - \frac{d x}{d y} = \frac{x^{2}}{x y - x^{2} y^{2} - 1}$

$∴ \frac{d y}{d x} = \frac{x y - x^{2} y^{2} - 1}{x^{2}}$

$L e t x y = v \Rightarrow x \frac{d y}{d x} + y = \frac{d v}{d x}$

Put x = 1, y = 1 $\Rightarrow t a n^{- 1} = c \Rightarrow c = \frac{π}{4}$

$∴ t a n^{- 1} (x y) = l n x = \frac{π}{4}$

$e (y (e)) = t a n (1 + \frac{π}{4}) = \frac{t a n 1 + 1}{1 - t a n 1}$

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New answer posted

8 months ago

Maths Class 11th

Consider the following statements:

A : Rishi is a judge.

B : Rishi is honest.

C : Rishi is not arrogant.

The negation of the statement “if Rishi is a judge and he is not arrogant, then he is honest” is

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Payal Gupta

Contributor-Level 10

$?$ given statement is

$(A \land C) \to B$ then its negation is ${(A \land C) \to B}$

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New answer posted

8 months ago

Maths Class 12th

If y = tan^-1 (sec x³ tanx³), $\frac{π}{2} < x^{3} < \frac{3 π}{2}, t h e n$

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Payal Gupta

Contributor-Level 10

$L e t x^{3} = θ \Rightarrow \frac{θ}{2} \in (\frac{π}{4}, \frac{3 π}{4})$

$∴ y = t a n^{- 1} (s e c θ - t a n θ)$

$t a n^{- 1} (\frac{1 - s i n θ}{c o s θ})$

$\Rightarrow \frac{d y}{d x} = \frac{- 3 x^{2}}{2} \Rightarrow \frac{d^{2} y}{d x^{2}} = - 3 x$

$∴ x^{2} \frac{d^{2} y}{d x^{2}} - 6 y + \frac{3 π}{2} = 0 \Rightarrow x^{2} y^{11} - 6 y + \frac{3 π}{2} = 0$

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New answer posted

8 months ago

Maths Class 12th

Let $\hat{a}$ and $\hat{b}$ be two unit vectors such that $| (\hat{a} + \hat{b}) + 2 (\hat{a} * \hat{b}) |$ = 2. If $θ \in (0, π)$ is the angle between $\hat{a} a n d \hat{b}$ , then among the statements:

$(S 1) : 2 | \hat{a} * \hat{b} | = | \hat{a} - \hat{b} |$

(S1) : The projection of $\hat{a} o n (\hat{a} + \hat{b}) i s \frac{1}{2}$

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Payal Gupta

Contributor-Level 10

$| \vec{a} + \vec{b} + 2 (\vec{a} * \vec{b}) | = 2, θ \in (0, π)$

squaring on both sides, we get = $\frac{2 π}{3}$

where θ is angle between $\hat{a} a n d \hat{b} .$

$2 | \hat{a} * \hat{b} | = \sqrt[]{3} = | \hat{a} - \hat{b} |, S_{1}$ is correct.

and projection of $\hat{a} o n \hat{a} + \hat{b} = | \frac{\hat{a} . (\hat{a} + \hat{b})}{| \hat{a} + \hat{b} |} | = \frac{1}{2}$

So (S₂) is correct.

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New answer posted

8 months ago

Maths Class 12th

Let the points on the plane P be equidistance from the points (–4, 2, 1) and (2, –2, 3). Then the acute angle between the plane P and the plane 2x + y + 3z = 1 is

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Payal Gupta

Contributor-Level 10

Let P (x, y, z) be any point on plane P₁ then

${(x + 4)}^{2} + {(y - 2)}^{2} + {(z - 1)}^{2} = {(x - 2)}^{2} + {(y + 2)}^{2} + {(z - 3)}^{2}$

$∴ c o s θ = \frac{| 6 - 2 + 3 |}{14} \Rightarrow θ = \frac{π}{3}$

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New answer posted

8 months ago

Maths Class 12th

If the shortest distance between the lines $\frac{x - 1}{2} = \frac{y - 2}{3} = \frac{z - 3}{λ}$ and $\frac{x - 2}{1} = \frac{y - 4}{4} = \frac{z - 5}{5} i s \frac{1}{\sqrt[]{3}},$ then the sum of all possible values of $λ$ is

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Payal Gupta

Contributor-Level 10

S. D. = $\frac{1}{\sqrt[]{3}}$

$b_{1} * b_{2} = | \begin{matrix} \hat{i} & \hat{j} & \hat{k} \\ 2 & 3 & λ \\ 1 & 4 & 5 \end{matrix} | = \hat{i} (15 - 4 λ) + \hat{j} (λ - 10) + \hat{k} (5)$

$\frac{| (- \hat{i} - 2 \hat{j} - 2 \hat{k}) . {(15 - 4 λ) \hat{i} + (λ - 10) \hat{j} + 5 \hat{k}} |}{\sqrt[]{{(15 - 4 λ)}^{2} + {(λ - 10)}^{2} + 25}} = \frac{1}{\sqrt[]{3}}$

$\Rightarrow λ = 16$

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New answer posted

8 months ago

Maths Class 11th

The number of solutions of the equation $c o s (x + \frac{π}{3}) c o s (\frac{π}{3} - x) = \frac{1}{4} c o s^{2} 2 x, x \in [- 3 π, 3 π] i s :$

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Payal Gupta

Contributor-Level 10

$c o s (x + \frac{π}{3}) c o s (\frac{π}{3} - x) = \frac{1}{4} c o s^{2} 2 x$

$x = - 3 π, - 2 π, - π, 0, π, 2 π, 3 π$

$∴$ total number of solution = 7.

0 0

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