Tags Class 12th

Class 12th

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

Follow Ask Question

12k

Questions

Discussions

Active Users

Followers

New answer posted

a year ago

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

Let S be the set of all a $\in$ R for which the angel between the vectors $\vec{u} = a (l o g_{e} b) \hat{i} - 6 \hat{j} + 3 \hat{k}$ and $\vec{v} = (l o g_{e} b) \hat{i} + 2 \hat{j} + 2 a (l o g_{e} b) \hat{k}, (b > 1)$ is acute. Then S is equal to:

0 Follower 47 Views

Vishal Baghel

Contributor-Level 10

For angle to the acute $\vec{U} . \vec{V} > 0$

$\Rightarrow a {(l o g_{e} b)}^{2} - 12 + 6 a (l o g_{e} b) > 0$

$\forall b > 1$

Let log_eb = t > 0 as b > 1

Y = at² + 6at – 12 & y > 0 $\forall$ t > 0

$\Rightarrow a \in ?$

0 0

New answer posted

a year ago

Class 12th Maths NCERT Exemplar Solutions Class 12th Chapter Eleven

A plane P is parallel to two lines whose direction ratios are 2, 1 3 and 1, 2, 2 and it contains the point (2, 2, 2). Let P intersect the co-ordinate axes at the points A, B, C making the intercepts $α, β, γ .$ If V is the volume of the tetrahedron OABC, where O is the origin, and p = $α+$
$β$ + $γ$ , then the ordered pair (V, p) is equal to:

0 Follower 3 Views

Vishal Baghel

Contributor-Level 10

Normal of plane P : = $| \begin{matrix} \hat{i} & \hat{j} & \hat{k} \\ - 2 & 1 & - 3 \\ - 1 & 2 & - 2 \end{matrix} | = 4 \hat{i} - j - 3 \hat{k}$

Equation of plane P which passes through (2, 2) is 4x – y – 3z – 12 = 0

Now, A (3, 0, 0), B (0, 12 0), C (0, 4)

$\begin{array}{l} \Rightarrow α = 3, β - 12, γ = - 4 \\ \Rightarrow P = α + β + γ = - 13 \end{array}$

Now, volume of tetrahedron OABC

$V = | \frac{1}{6} \vec{O A} . (\vec{O B} * \vec{O C}) | = 24$

(V, P) = (24, 13)

0 0

New answer posted

a year ago

Class 12th Maths NCERT Exemplar Solutions Class 12th Chapter Eleven

Let the lines $\frac{x - 1}{λ} = \frac{y - 2}{1} = \frac{z - 3}{2} a n d \frac{x + 26}{- 2} = \frac{y + 18}{3} = \frac{z + 28}{λ}$ be coplanar and P be the plane containing these two lines. Then which of the following points does NOT lie on P?

0 Follower 8 Views

Vishal Baghel

Contributor-Level 10

Given, $L_{1} : \frac{x - 1}{λ} = \frac{y - 2}{1} = \frac{z - 3}{2}$

and $L_{2} : \frac{x + 26}{- 2} = \frac{y + 18}{3} = \frac{z + 28}{λ}$

are coplanar

$\Rightarrow | \begin{matrix} 27 & 20 & 31 \\ λ & 1 & 2 \\ 2 & 3 & λ \end{matrix} | = 0$

$λ = 3$

Now, normal of plane P, which contains L1 and L2

equation of required plane P : 3x + 13y – 11z + 4 = 0

(0, 4, 5) does not lie on plane P.

0 0

New answer posted

a year ago

Class 12th Maths NCERT Exemplar Solutions Class 12th Chapter Eleven

Let the hyperbola H : $\frac{x^{2}}{a^{2}} - \frac{y^{2}}{b^{2}} = 1$ pass through the point $(2 \sqrt[]{2}, - 2 \sqrt[]{2}) .$ A parabola is drawn whose focus is same as the focus of H with positive abscissa and the directrix of the parabola passes through the other focus of H. If the length of the latus rectum of the parabola is e times the length of the latus rectum of H, where e is the eccentricity of H, then which of the following points lies on the parabola?

0 Follower 16 Views

Vishal Baghel

Contributor-Level 10

$H : \frac{x^{2}}{a^{2}} - \frac{y^{2}}{b^{2}} = 1$

Foci : S (ae, 0), S' (ae, 0)

Focus of parabola is (ae, 0)

Now, semi latus rectum of parabola = |SS'| = 2ae

$G i v e n, 4 a e = e (\frac{2 b^{2}}{a})$

B² = 2a²……… (i)

$G i v e n, (2 \sqrt[]{2}, - 2 \sqrt[]{2})$ lies on H

$\Rightarrow \frac{1}{a^{2}} - \frac{1}{b^{2}} = \frac{1}{8} . . . . . . . . (i i)$

From (i) and (ii)

$a^{2} = 4, b^{2} = 8$

$? b^{2} = a^{2} (e^{2} - 1)$

$∴ e = \sqrt[]{3}$

$\Rightarrow$ equation of parabola is y² $= 8 \sqrt[]{3} x$

0 0

New answer posted

a year ago

Class 12th Maths NCERT Exemplar Solutions Class 12th Chapter Eleven

Let the tangents at two points A and B on the circle x² + y² – 4x + 3 = 0 meet at origin O(0, 0). Then the area of the triangle OAB is:

0 Follower 21 Views

Vishal Baghel

Contributor-Level 10

C : (x – 2)² + y² = 1

Equation of chord AB : 2x = 3

$O A = O B = \sqrt[]{3}$

$A M = \frac{\sqrt[]{3}}{2}$

$A r e a o f Δ O A B = \frac{1}{2} (2 A M) (O M)$

$= \frac{3 \sqrt[]{3}}{4} s q . u n i t$

0 0

New answer posted

a year ago

Class 12th Maths NCERT Exemplar Solutions Class 12th Chapter Nine

The differential equation of the family circles passing through the points (0, 2) and (0, 2) is:

0 Follower 5 Views

Vishal Baghel

Contributor-Level 10

Equation of circle passing through (0, 2) and (0, 2) is

$x^{2} + (y^{2} - 4) + λ x = 0, (λ \in R)$

Divided by x we get

$\frac{x^{2} + (y^{2} - 4)}{x} + λ = 0$

Differentiating w.r.t. x

$\frac{x [2 x + 2 y . \frac{d y}{d x}] - [x^{2} + y^{2} - 4] . 1}{x^{2}} = 0$

$\Rightarrow 2 x y . \frac{d y}{d x} + (x^{2} - y^{2} + 4) = 0$

0 0

New answer posted

a year ago

Class 12th Maths NCERT Exemplar Solutions Class 12th Chapter Nine

Let y = y(x) be the solution curve of the differential equation $\frac{d y}{d x} + \frac{1}{x^{2} - 1} y = {(\frac{x - 1}{x + 1})}^{1 / 2}, x > 1$ passing through the point $(2, \sqrt[]{\frac{1}{3}}) .$ Then $\sqrt[]{7} y (8)$ is equal to:

0 Follower 4 Views

Vishal Baghel

Contributor-Level 10

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

$\frac{d y}{d x} + p y = Q$

$I . F = e^{\int P d x} = {(\frac{x - 1}{x +})}^{\frac{1}{2}}$

$x - 2 l o g_{e} | x + 1 | + C$

Curves passes through $(2, \frac{1}{\sqrt[]{3}})$

$\Rightarrow C = 2 l o g_{e} 3 - \frac{5}{3}$

$a t x = 8, \sqrt[]{7} y (8) = 19 - 6 l o g_{e} 3$

0 0

New answer posted

a year ago

Class 12th Maths NCERT Exemplar Solutions Class 12th Chapter Eight

The area enclosed by the curves y = $l o g_{e} (x + e^{2}), x = l o g_{e} (\frac{2}{y}) a n d x = l o g_{e} 2,$ above the line y = 1 is:

0 Follower 26 Views

Vishal Baghel

Contributor-Level 10

Required area is

= $\int_{e - e^{2}}^{0} l n (x + e^{2}) - 1 d x + \int_{0}^{l n 2} 2 e^{- x} - 1 d x$

$= 1 + e - l n 2$

0 0

New answer posted

a year ago

Class 12th Maths NCERT Exemplar Solutions Class 12th Chapter Seven

Let I_n (x) = $\int_{0}^{x} \frac{1}{{(t^{2} + 5)}^{n}} d t, n = 1, 2, 3, . . . .$ Then:

0 Follower 9 Views

Vishal Baghel

Contributor-Level 10

$l_{n} (x) = \int_{0}^{x} \frac{d t}{{(t^{2} + s)}^{n}}$

Applying integral by parts

$l_{n} (x) = {[\frac{t}{{(t^{2} + 5)}^{n}}]}_{0}^{x} - \int_{0}^{x} n {(t^{2} + 5)}^{- n - 1} . 2 t^{2}$

$10 n l_{n + 1} (x) + (1 - 2 n) l_{n} (x) = \frac{x}{{(x^{2} + 5)}^{n}}$

Put n = 5

0 0

New answer posted

a year ago

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

Let x(t) = $2 \sqrt[]{2} c o s t \sqrt[]{s i n 2 t} a n d y (t) = 2 \sqrt[]{2} s i n t \sqrt[]{s i n 2 t}, t \in (0, \frac{π}{2}) .$ Then $\frac{1 + {(\frac{d y}{d x})}^{2}}{\frac{d^{2} y}{d x^{2}}}$ at $t = \frac{π}{4}$ is equal to:

0 Follower 5 Views

Vishal Baghel

Contributor-Level 10

$x = 2 \sqrt[]{2} c o s t \sqrt[]{s i n 2 t}$

$\frac{d x}{d t} = \frac{2 \sqrt[]{2} c o s 3 t}{\sqrt[]{s i n 2 t}}$

$y (t) = 2 \sqrt[]{2} s i n t \sqrt[]{s i n 2 t}$

$\frac{d y}{d x} = \frac{2 \sqrt[]{2} s i n 3 t}{\sqrt[]{s i n 2 t}}$

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

0 0

Get authentic answers from experts, students and alumni that you won't find anywhere else

On Shiksha, get access to

67k Colleges
1.2k Exams
718k Reviews
1850k Answers

Community GuidelinesUser Points System

QuestionsDiscussionsTags

Need guidance on career and education? Ask our experts

Your Question

Class 12th

Questions

Discussions

Active Users

Followers

All

Discussions

Unanswered Questions

Share Your College Life Experience