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2 months ago

Class 12th Maths

Let f : R ® R be a function defined by f(x) $\Rightarrow {(2 (1 - \frac{x^{25}}{2}) (2 + x^{25}))}^{\frac{1}{50}} .$ If the function g(x) = $f (f (f (x))) + f (f (x)),$ then the greatest integer less than or equal to g(1) is…….

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Raj Pandey

Contributor-Level 9

g (1) =?

$= f (f (f (1))) + f (f (1))$

$f (1) = {(2 (1 - \frac{1}{2}) (2 + 1))}^{\frac{1}{50}} = 3^{\frac{1}{50}}$

$3^{\frac{1}{50}} + 1 3^{\frac{1}{50}} > 1^{\frac{1}{50}}$

3 > 1

$3^{\frac{1}{50}} > 2$ $2 > 3^{\frac{1}{50}} > 1$

$3 < 2^{50}$ $[3^{\frac{1}{50}} + 1] = 1 + 1 = 2$

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New answer posted

2 months ago

Class 12th Maths

Let q be the angle between the vectors $\vec{a} a n d \vec{b},$ where $| \vec{a} | = 4, | \vec{b} | = 3$ and $θ \in (\frac{π}{4}, \frac{π}{3})$ . Then ${| (\vec{a} - \vec{b}) * (\vec{a} + \vec{b}) |}^{2} + 4 {(\vec{a} . \vec{b})}^{2}$ is equal to…………

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Raj Pandey

Contributor-Level 9

${| 2 (\vec{a} * \vec{b}) |}^{2} + 4 {(\vec{a} . \vec{b})}^{2}$

= $4 {| \vec{a} |}^{2} {| \vec{b} |}^{2}$

=4 * 16 * 9 = 576

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2 months ago

Class 12th Ncert Solutions Maths class 12th

Let a circle C in complex plane pass through the points z₁ = 3 + 4i, z₂ = 4 + 3i and z₃ = 5i. If $z (\neq z_{1})$ is a point on C such that the line through z and z₁ is perpendicular to the line through z₂ and z₃, then arg(z) is equal to:

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Raj Pandey

Contributor-Level 9

$\frac{2}{- 4} = \frac{- 1}{2}$

y – 4 = 2 (x – 3)

y = 2x – 2

x² + (2x – 2)² = 25

$5 x^{2} - 8 x - 21 = 0$

$z (\frac{- 7}{5}, \frac{- 24}{5})$

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2 months ago

Class 12th Differential Equations

If the solution curve y = y(x) of the differential equation y²dx + (x² – xy + y²)dy = 0, which passes through the point (1, 1) and intersects the line y = $\sqrt[]{3} x$ at the point $(α, \sqrt[]{3} α),$ then value of $l o g_{e} (\sqrt[]{3} α)$ is equal to:

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Raj Pandey

Contributor-Level 9

$y^{2} d x + (x^{2} - x y + y^{2}) d y = 0$

$\frac{d x}{d y} + \frac{x^{2} - x y + y^{2}}{y^{2}} = 0 \Rightarrow \frac{d x}{d y} + {(\frac{x}{y})}^{2} - (\frac{x}{y}) + 1 = 0$

Put x = vy $\Rightarrow v + y \frac{d v}{d y} + v^{2} - v + 1 = 0$

$\Rightarrow \frac{y d v}{d y} + v^{2} + 1 = 0$

$\Rightarrow \frac{π}{6} + l n | y | = \frac{π}{4}$

$l n | y | = \frac{π}{12}$

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2 months ago

Class 12th Maths

Let Q be the mirror image of the point P(1, 0, 1) with respect to the plane S : x + y + z = 5. If a line L passing through (1, -1, -1) parallel to the line PQ meets the plane S at R, then QR² is equal to:

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Raj Pandey

Contributor-Level 9

$l i m P Q : \frac{x - 1}{1} = \frac{y - 0}{1} = \frac{3 - 1}{1} = λ$

$M (1 + λ, λ, 1 + λ)$

x + y + z = 5

$\Rightarrow λ = 1$

M (2, 1, 2)

Q (3, 2, 3)

$L : \frac{x - 1}{1} = \frac{y + 1}{1} = \frac{z + 1}{1} = t$

R (3, 1, 1), QR² = 1 + 4 = 5

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New answer posted

2 months ago

Class 12th Differential Equations

Let y = y(x) be the solution of the differential equation (x + 1)y' – y = e^3x (x + 1)², with y(0) = $\frac{1}{3} .$ Then, the point x = $- \frac{4}{3}$ for the curve y = y(x) is:

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Raj Pandey

Contributor-Level 9

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

$(x + 1) d y - y d x = e^{3 x} {(x + 1)}^{2} d x$

$\frac{(x + 1) d y - y d x}{{(x + 1)}^{2}} = e^{3 x} d x$

$\Rightarrow d (\frac{y}{x + 1}) = e^{3 x} d x \Rightarrow \frac{y}{x + 1} = \frac{e^{3 x}}{3} + c$

$= \frac{e^{3 x}}{3} (3 x + 4)$

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New answer posted

2 months ago

Class 12th Vector Algebra

$\vec{a} = a_{1} \hat{i} + a_{2} \hat{j} + a_{3} \hat{k} a_{i} > 0, i = 1, 2, 3$ be a vector which makes equal angles with the coordinate axes OX, OY and OZ. Also, let the projection of $\vec{a}$ on the vector $3 \hat{i} + 4 \hat{j} b e 7 .$ Let $\vec{b}$ be a vector obtained by rotating $\vec{a}$ with $90 ° . I f \vec{a}, \vec{b}$ and x-axis are coplanar, then projection of a vector $\vec{b}$ on $3 \hat{i} + 4 \hat{j}$ is equal to:

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Raj Pandey

Contributor-Level 9

$\vec{a} = (t, t, t)$

$\frac{3 t + 4 t}{5} = 7 \Rightarrow t = 5$

$\vec{a} = (5, 5, 5)$

$\vec{b} = l \vec{a} + m \hat{i}$

$= (5 l + m, 5 l, 5 l)$

$\frac{\vec{b} . (3, 4, 0)}{5} = \frac{\pm \frac{5}{\sqrt[]{2}} (- 6 + 4 + 0)}{5} = \pm (\sqrt[]{2})$

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New answer posted

2 months ago

Class 12th Maths Continuity and Differentiability

Let f : R® R and g : R ® R be two functions defined by f(x) = log_e (x² + 1) – e^-x + 1 and $g (x) = \frac{1 - 2 e^{2 x}}{e^{x}} .$ Then, for which of the following range of a, the inequality $f (g (\frac{{(α - 1)}^{2}}{3})) > f (g (α - \frac{5}{3}))$ holds?

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Raj Pandey

Contributor-Level 9

$f (x) = l n (x^{2} + 1) - e^{- x} + 1, g (x) = e^{- x} - 2 e^{x}$

for $f (g (a)) > f (g (b)) g' (x) = \bar{e^{x}} - 2 e^{x} < 0 \forall x \in R$

$f' (x) = \frac{2 x}{x^{2} + 1} + e^{- x}$

${\begin{cases} < 1 f o r n \geq 0, g (x) ↓ \\ \geq 1 f o r n < 0 \end{cases}$

g(a) > g(b)

$\Rightarrow a < b (a s g ↓)$

$\begin{array}{l} \Rightarrow α^{2} - 5 α + 6 < 0 \\ (α - 3) (α - 2) < 0 \end{array}$

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New answer posted

2 months ago

Class 12th Differential Equations

Let g : $(0, \infty) \to R$ be a differentiable function such that $\int (\frac{x (c o s x - s i n x)}{e^{x} + 1} + \frac{g (x) (e^{x} + 1 - x e^{x})}{{(e^{x} + 1)}^{2}})$ dx = $\frac{x g (x)}{e^{x} + 1} + c,$ for all x > 0, where c is an arbitrary constant. Then:

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Raj Pandey

Contributor-Level 9

$\frac{x (c o s x - s i n x)}{e^{x} + 1} + \frac{g (x) (e^{x} + 1 - x e^{x})}{{(e^{x} + 1)}^{2}}$

$= \sqrt[]{2} s i n (x + \frac{π}{4}) + c, x > 0$

$g' (x) = \sqrt[]{2} c o s (x + \frac{π}{4}), x > 0$

g + g' = 2cos x + c, x > 0

g – g' = 2 sin x + c, x > 0

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New answer posted

2 months ago

Class 12th Maths Matrices

Let $A = [\begin{matrix} 0 & - 2 \\ 2 & 0 \end{matrix}]$ . If M and N are two matrices given by $M = \sum_{k = 1}^{10} A^{2 k} a n d N = \sum_{k = 1}^{10} A^{2 k - 1}$ then MN² is:

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Raj Pandey

Contributor-Level 9

$A = [\begin{matrix} 0 & - 2 \\ 2 & 0 \end{matrix}]$

$A^{2} = [\begin{matrix} 0 & - 2 \\ 2 & 0 \end{matrix}] [\begin{matrix} 0 & - 2 \\ 2 & 0 \end{matrix}]$

$= [\begin{matrix} - 4 & 0 \\ 0 & - 4 \end{matrix}] = - 4 [\begin{matrix} 1 & 0 \\ 0 & 1 \end{matrix}] = - 4 l$

$N^{2} = \frac{{(2^{20} - 1)}^{2}}{25} l$

$M N^{2} = \frac{4}{125} {(2^{20} - 1)}^{3} l$

$= [\begin{matrix} λ & 0 & 0 \\ 0 & λ & 0 \\ 0 & 0 & λ \end{matrix}], λ \neq k$

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