Ncert Solutions Maths class 12th

Get insights from 2.5k questions on Ncert Solutions Maths class 12th, answered by students, alumni, and experts. You may also ask and answer any question you like about Ncert Solutions Maths class 12th

2.5k
Questions
0
Discussions
0
Active Users
65
Followers

Have a question about Ncert Solutions Maths class 12th? Ask our experts

Ask Your Question

Keep it short & simple. Type complete word. Avoid abusive language.

New answer posted•a year ago

Ncert Solutions Maths class 12th Class 12th

Let θ 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 ${| (\vec{a} - \vec{b}) \times (\vec{a} + \vec{b}) |}^{2} + 4 {(\vec{a} . \vec{b})}^{2}$  is equal to…………

4 Views|Posted a year ago

Asked by Vishal Baghel

Answered by

Vishal Baghel | Contributor-Level 10

a year ago

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

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

= 4 × 16 × 9 = 576

Upvote

New answer posted•a year ago

Ncert Solutions Maths class 12th Class 12th

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:

2 Views|Posted a year ago

Asked by Vishal Baghel

Answered by

Vishal Baghel | Contributor-Level 10

a year ago

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

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

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

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

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

$l n | y | = \frac{?}{12}$

Upvote

New answer posted•a year ago

Ncert Solutions Maths class 12th Class 12th

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:

1 Views|Posted a year ago

Asked by Vishal Baghel

Answered by

Vishal Baghel | Contributor-Level 10

a year ago

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

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

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

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

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

$l n | y | = \frac{?}{12}$

Upvote

New answer posted•a year ago

Ncert Solutions Maths class 12th Class 12th

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:

4 Views|Posted a year ago

Asked by Vishal Baghel

Answered by

Vishal Baghel | Contributor-Level 10

a year ago

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

$M (1 + λ, λ, 1 + λ)$

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

Upvote

New answer posted•a year ago

Ncert Solutions Maths class 12th Class 12th

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:

11 Views|Posted a year ago

Asked by Vishal Baghel

Answered by

Vishal Baghel | Contributor-Level 10

a year ago

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

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

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

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

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

Upvote

New answer posted•a year ago

Ncert Solutions Maths class 12th Class 12th

$\vec{a} = a_{1} \hat{i} + a_{2} \hat{j} + a_{3} \hat{k} a_{i} > 0, i = 1, 2, 3$ 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  with  and x-axis are coplanar, then projection of a vector $\vec{b}$  on $3 \hat{i} + 4 \hat{j}$  is equal to:

2 Views|Posted a year ago

Asked by Vishal Baghel

Answered by

Vishal Baghel | Contributor-Level 10

a year ago

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

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

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

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

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

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

Upvote

New answer posted•a year ago

Ncert Solutions Maths class 12th Class 12th

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?

7 Views|Posted a year ago

Asked by Vishal Baghel

Answered by

Vishal Baghel | Contributor-Level 10

a year ago

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

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)

$\Rightarrow a < b (a s g ↓)$

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

Upvote

New answer posted•a year ago

Ncert Solutions Maths class 12th Class 12th

Let E₁ and E₂ be two events such that conditional probabilities

^{$P (E_{1} | E_{2}) = \frac{1}{2}, P (E_{2} | E_{1}) = \frac{3}{4} a n d P (E_{1} \cap E_{2}) = \frac{1}{8} .$}Then:

4 Views|Posted a year ago

Asked by Vishal Baghel

Answered by

Vishal Baghel | Contributor-Level 10

a year ago

$P (E_{1} / E_{2}) = \frac{1}{2}, \Rightarrow \frac{P (E_{1} \cap E_{2})}{P (E_{2})} = \frac{1}{2}$ f

$P (E_{2} / E_{1}) = \frac{3}{4}$ $P (E_{1}) = \frac{1}{6}$

$P (E_{1} \cap E_{2}) = \frac{1}{8}$ $P (E_{2}) = \frac{1}{4}$

$P (E_{1} \cup E_{2}) = \frac{1}{6} + \frac{1}{4} - \frac{1}{8} = \frac{4 + 6 - 3}{24} = \frac{7}{24}$

Upvote

New answer posted•a year ago

Ncert Solutions Maths class 12th Class 12th

If the shortest distance between the lines $\vec{r} = (- \hat{i} + 3 \hat{k}) + λ (\hat{i} - a \hat{j}) a n d \vec{r} = (- \hat{j} + 2 \hat{k}) + μ (\hat{i} - \hat{j} + \hat{k})$ is $\sqrt[]{\frac{2}{3}}$ , then the integral value of ‘a’ is equal to __________.

6 Views|Posted a year ago

Asked by Alok Kumar Singh

Answered by

Alok Kumar Singh | Contributor-Level 10

a year ago

${\vec{b}}_{1} \times {\vec{b}}_{2} = | \begin{matrix} \hat{i} & \hat{j} & \hat{k} \\ 1 & - a & 0 \\ 1 & - 1 & 1 \end{matrix} |$ = $- a \hat{i} - \hat{j} + (a - 1) \hat{k}$

${\vec{a}}_{1} - {\vec{a}}_{2} = - \hat{i} + \hat{j} + \hat{k}$

Shortest distance = $| \frac{({\vec{a}}_{1} - {\vec{a}}_{2}) . ({\vec{b}}_{1} \times {\vec{b}}_{2})}{| {\vec{b}}_{1} \times {\vec{b}}_{2} |} |$

$= \frac{2 (a - 1)}{\sqrt[]{a^{2} + 1 + {(a + 1)}^{2}}} = \sqrt[]{\frac{2}{3}} = \frac{4 {(a - 1)}^{2}}{a^{2} + 1 + {(a - 1)}^{2}} = \frac{2}{3}$

$12 {(a - 1)}^{2} = 2 (a^{2} + 1) + 2 {(a - 1)}^{2}$

$10 {(a - 1)}^{2} = 2 (a^{2} + 1)$

$5 a^{2} - 10 a + 5 = a^{2} + 1 \Rightarrow 4 a^{2} - 10 a + 4 = 0$

$2 a^{2} - 5 a + 2 = 0$

$(2 a - 1) (a - 2) = 0$

$∴ a = \frac{1}{2}, 2$

Upvote

New answer posted•a year ago

Ncert Solutions Maths class 12th Class 12th

Let S be the region bonded by the curves y = x³ and y² = x. The curve y = 2|x| divides S into two regions of areas R₁, and R₂ _________.

If max {R₁, R₂} = R₂, then $\frac{R_{2}}{R_{1}}$ is equal to _________.

9 Views|Posted a year ago

Asked by Alok Kumar Singh

Answered by

Alok Kumar Singh | Contributor-Level 10

a year ago

$R_{2} = \int_{0}^{1} (x - x^{3}) d x = {(\frac{x^{2}}{2} - \frac{x^{4}}{4})}_{0}^{1} = \frac{1}{2} - \frac{1}{4} = \frac{1}{4}$

R_{1} = \int_{0}^{\frac{1}{4}} (\sqrt[]{x} - 2 x) d x = {(\frac{2 x^{3 / 2}}{3} - x^{2})}_{0}^{\frac{1}{4}} = \frac{1}{12} - \frac{1}{16}

= \frac{4 - 3}{48} = \frac{1}{48}

R_{1} + R_{2} = {\int_{0}^{1} (\sqrt[]{x} - x^{3}) d x = (\frac{2 x^{3 / 2}}{3} - \frac{x^{4}}{4})}_{0}^{1} = \frac{2}{3} - \frac{1}{4} = \frac{5}{12}

\frac{R_{1} + R_{2}}{R_{1}} = 1 + \frac{R_{2}}{R_{1}} = \frac{\frac{5}{12}}{\frac{1}{48}} = 20

∴ \frac{R_{2}}{R_{1}} = 19

Upvote

Ncert Solutions Maths class 12th

Questions

Discussions

Active Users

Followers