Tags Class 12th

Class 12th

Get insights from 11.9k 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

11.9k

Questions

Discussions

Active Users

Followers

New answer posted

11 months ago

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

Let $P (3,3)$ be a point on the hyperbola, $\frac{x^{2}}{a^{2}} - \frac{y^{2}}{b^{2}} = 1$ . If the normal to it at $P$ intersects the $x$ -axis at $(9,0)$ and $e$ is its eccentricity, then the ordered pair $(a^{2}, e^{2})$ is equal to:

0 Follower 5 Views

alok kumar singh

Contributor-Level 10

Since $(3,3)$ lies on $\frac{x^{2}}{a^{2}} - \frac{y^{2}}{b^{2}} = 1$

$\frac{9}{a^{2}} - \frac{9}{b^{2}} = 1$

Now, normal at $(3,3)$ is $y - 3 = - \frac{a^{2}}{b^{2}} (x - 3)$ ,

which passes through $(9,0) \Rightarrow b^{2} = 2 a^{2}$

So, $e^{2} = 1 + \frac{b^{2}}{a^{2}} = 3$

Also, $a^{2} = \frac{9}{2}$

(From (i) and (ii)

Thus, $(a^{2}, e^{2}) = (\frac{9}{2}, 3)$

0 0

New answer posted

11 months ago

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

A triangle $A B C$ lying in the first quadrant has two vertices as $A (1,2)$ and $B (3,1)$ . If $∠ B A C = 90^{\circ}$ , and $a r (△ A B C) = 5 \sqrt[]{5}$ sq. units, then the abscissa of the vertex $C$ is:

0 Follower 25 Views

alok kumar singh

Contributor-Level 10

$(\frac{K - 2}{h - 1}) (\frac{1 - 2}{3 - 1}) = - 1 \Rightarrow K = 2 h$

$? [? A B C] = 5 \sqrt[]{5}$

$\Rightarrow \frac{1}{2} (\sqrt[]{5}) \sqrt[]{(h - 1)^{2} + (K - 2)^{2}} = 5 \sqrt[]{5}$

0 0

New answer posted

11 months ago

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

Let $u = \frac{2 z + i}{z - k i}, z = x + i y$ and $k > 0$ . If the curve represented by $R e (u) + I m (u) = 1$ intersects the $y$ -axis at the points $P$ and $Q$ where $P Q = 5$ , then the value of $k$ is:

0 Follower 5 Views

alok kumar singh

Contributor-Level 10

$u = \frac{2 z + i}{z - k i}$

$= \frac{2 x^{2} + (2 y + 1) (y - k)}{x^{2} + (y - k)^{2}} + i \frac{(x (2 y + 1) - 2 x (y - k))}{x^{2} + (y - k)^{2}}$

Since $R e ? (u) + I m ? (u) = 1$

$\Rightarrow 2 x^{2} + (2 y + 1) (y - k) + x (2 y + 1) - 2 x (y - k) = x^{2} + (y - k)^{2}$

$\begin{matrix} P (0, y_{1}) \\ Q (0, y_{2}) \end{matrix}? \Rightarrow y^{2} + y - k - k^{2} = 0 ?\begin{matrix} y_{1} + y_{2} = - 1 \\ y_{1} y_{2} = - k - k^{2} \end{matrix}$

$? P Q = 5$

$\Rightarrow |y_{1} - y_{2}| = 5 \Rightarrow k^{2} + k - 6 = 0$

$\Rightarrow k = - 3,2$

So, $k = 2 (k > 0)$

0 0

New answer posted

11 months ago

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

Given the following two statements:

$(S_{1}) : (q \lor p) \to (p \leftrightarrow \sim q)$ is a tautology.

$(S_{2}) : \sim q \land (\sim p \leftrightarrow q)$ is a fallacy. Then:

0 Follower 3 Views

alok kumar singh

Contributor-Level 10

Let TV (r) denotes truth value of a statement $r$ .

$Now, if T V (p) = T V (q) = T$

$\Rightarrow T V (S_{1}) = F$

Also, if $T V (p) = T$ and $T V (q) = F$

$\Rightarrow T V (S_{2}) = T$

0 0

New question posted

11 months ago

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

Given the following two statements:

$(S_{1}) : (q \lor p) \to (p \leftrightarrow \sim q)$ is a tautology.

$(S_{2}) : \sim q \land (\sim p \leftrightarrow q)$ is a fallacy. Then:

0 Follower 1 View

New answer posted

11 months ago

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

If $1 + (1 - 2^{2} \cdot 1) + (1 - 4^{2} \cdot 3) + (1 - 6^{2} \cdot 5) + \dots \dots . + (1 - 20^{2} \cdot 19) = α - 220 β$ , then an ordered pair $(α, β)$ is equal to:

0 Follower 8 Views

alok kumar singh

Contributor-Level 10

$1 + (1 - 2^{2} . 1) + (1 - 4^{2} . 3) + \dots \dots . + (1 - 20^{2} . 19)$

$= α - 220 β$

$= 11 - (2^{2} \cdot 1 + 4^{2} \cdot 3 + \dots . . + 20^{2} \cdot 19)$

$= 11 - 2^{2} \cdot \sum_{r = 1}^{10} ? r^{2} (2 r - 1) = 11 - 4 (\frac{110^{2}}{2} - 35 * 11)$

$= 11 - 220 (103)$

$\Rightarrow α = 11, β = 103$

0 0

New answer posted

11 months ago

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

Two vertical poles $A B = 15 m$ and $C D = 10 m$ are standing apart on a horizontal ground with points $A$ and $C$ on the ground. If $P$ is the point of intersection of $B C$ and $A D$ , then the height of $P$ (in $m$ ) above the line $A C$ is:

0 Follower 4 Views

alok kumar singh

Contributor-Level 10

$t a n ? θ = \frac{10}{x} = \frac{h}{x_{2}} \Rightarrow x_{2} = \frac{h x}{10}$

$t a n ? ? = \frac{15}{x} = \frac{h}{x} \Rightarrow x_{1} = \frac{h x}{15}$

Now, $x_{1} + x_{2} = x = \frac{h x}{15} + \frac{h x}{10}$

$\Rightarrow 1 = \frac{h}{10} + \frac{h}{15} \Rightarrow h = 6$

0 0

New answer posted

11 months ago

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

If $(a + \sqrt[]{2} b c o s x) (a - \sqrt[]{2} b c o s y) = a^{2} - b^{2}$ , where $a > b > 0$ , then $\frac{d x}{d y}$ at $(\frac{π}{4}, \frac{π}{4})$ is:

0 Follower 5 Views

alok kumar singh

Contributor-Level 10

$(a + \sqrt[]{2} b c o s ? x) (a - \sqrt[]{2} b c o s ? y) = a^{2} - b^{2}$

$\Rightarrow a^{2} - \sqrt[]{2} a b c o s ? y + \sqrt[]{2} a b c o s ? x - 2 b^{2} c o s ? x c o s ? y = a^{2} - b^{2}$

Differentiating both sides:

$0 - \sqrt[]{2} a b (- s i n ? y \frac{d y}{d x}) + \sqrt[]{2} a b (- s i n ? x) - 2 b^{2} [c o s ? x (- s i n ? y \frac{d y}{d x}) + c o s ? y (- s i n ? x)] = 0$

At $(\frac{π}{4}, \frac{π}{4})$ :

$\begin{array}{r} a b \frac{d y}{d x} - a b - 2 b^{2} (- \frac{1}{2} \frac{d y}{d x} - \frac{1}{2}) = 0 \\ \Rightarrow \frac{d x}{d y} = \frac{a b + b^{2}}{a b - b^{2}} = \frac{a + b}{a - b}; a, b > 0 \end{array}$

0 0

New answer posted

11 months ago

Class 12th Vector Algebra

Let $x_{0}$ be the point of local maxima of $f (x) = \vec{a} \cdot (\vec{b} * \vec{c})$ where $\vec{a} = x \hat{i} - 2 \hat{j} + 3 \hat{k}, \vec{b} = - 2 \hat{i} + x \hat{j} - \hat{k}$ and $\vec{c} = 7 \hat{i} - 2 \hat{j} + x \hat{k}$ . Then the value of $\vec{a} \cdot \vec{b} + \vec{b} \cdot \vec{c} + \vec{c} \cdot \vec{a}$ at $x = x_{0}$ is

0 Follower 7 Views

alok kumar singh

Contributor-Level 10

$f (x) = \overset{?}{a} \cdot (\overset{?}{b} * \overset{?}{c}) = |\begin{matrix} x & - 2 & 3 \\ - 2 & x & - 1 \\ 7 & - 2 & x \end{matrix}|$

$= x^{3} - 27 x + 26$

$f^{'} (x) = 3 x^{2} - 27 = 0 \Rightarrow x = \pm 3$ and $f^{''} (- 3) < 0$

$\Rightarrow$ local maxima at $x = x_{0} = - 3$

Thus, $\overset{?}{a} = - 3 \hat{i} - 2 \hat{j} + 3 \hat{k}, \overset{?}{b} = 2 \hat{i} - 3 \hat{j} - \hat{k}$ , and $\overset{?}{c} = 7 \hat{i} - 2 \hat{j} - 3 \hat{k}$

$\Rightarrow \overset{?}{a} \cdot \overset{?}{b} + \overset{?}{b} \cdot \overset{?}{c} + \overset{?}{c} \cdot \overset{?}{a} = 9 - 5 - 26 = - 22$

0 0

New answer posted

11 months ago

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

The mean and variance of 8 observations are 10 and 13.5, respectively. If 6 of these observations are $5,7, 10,12,14,15$ , then the absolute difference of the remaining two observations is:

0 Follower 2 Views

alok kumar singh

Contributor-Level 10

$\overline{x} = 10$

$\Rightarrow \overset{?}{x} = \frac{63 + a + b}{8} = 10$

$\Rightarrow a + b = 17$

Since, variance is independent of origin.

So, we subtract 10 from each observation.

So, $σ^{2} = 13.5 = \frac{79 + (a - 10)^{2} + (b - 10)^{2}}{8}$

$\Rightarrow a^{2} + b^{2} - 20 (a + b) = - 171$

$\Rightarrow a^{2} + b^{2} = 169$

From (1) and (2) ; $a = 12$ and $b = 5$

0 0

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

On Shiksha, get access to

66k Colleges
1.2k Exams
703k 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