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

7 months ago

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

The value of the determinant $[\begin{matrix} x & x + y & x + 2 y \\ x + 2 y & x & x + y \\ x + y & x + 2 y & x \end{matrix}]$

(A) $9 x^{2} (x + y)$

(B) $9 y^{2} (x + y)$

(C) $3 y^{2} (x + y)$

(D) $7 x^{2} (x + y)$

0 Follower 4 Views

Vishal Baghel

Contributor-Level 10

This is a Objective Type Questions as classified in NCERT Exemplar

Sol:

$\begin{array}{l} L e t Δ = | \begin{matrix} x & x + y & x + 2 y \\ x + 2 y & x & x + y \\ x + y & x + 2 y & x \end{matrix} | \\ C_{1} \to C_{1} + C_{2} + C_{3} \\ = | \begin{matrix} 3 x + 3 y & x + y & x + 2 y \\ 3 x + 3 y & x & x + y \\ 3 x + 3 y & x + 2 y & x \end{matrix} | \\ = (3 x + 3 y) | \begin{matrix} 1 & x + y & x + 2 y \\ 1 & x & x + y \\ 1 & x + 2 y & x \end{matrix} | [T a k i n g (3 x + 3 y) c o m m o n f r o m C_{1}] \\ R_{1} \to R_{1} - R_{2}, R_{2} \to R_{2} - R_{3} \\ \Rightarrow = (3 x + 3 y) | \begin{matrix} 0 & y & y \\ 0 & - 2 y & y \\ 1 & x + 2 y & x \end{matrix} | \\ E x p a n d i n g a l o n g C_{1} \\ \Rightarrow = 3 (x + y) [1 | \begin{matrix} y & y \\ - 2 y & y \end{matrix} |] \\ \Rightarrow = 3 (x + y) (y^{2} + 2 y^{2}) \Rightarrow 3 (x + y) (3 y^{2}) \Rightarrow 9 y^{2} (x + y) \\ H e n c e, t h e c o r r e c t o p t i o n i s (b) . \end{array}$

0 0

New answer posted

7 months ago

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

If $x, y, z$ are all different from zero and $[\begin{matrix} 1 + x & 1 & 1 \\ 1 & 1 + y & 1 \\ 1 & 1 & 1 + z \end{matrix}] = 0,$ then value of

x^–1 + y^–1 + z^–1 is

(A) $x y z$

(B) $x^{- 1} y^{- 1} z^{- 1}$

(C) $- x - y - z$

(D) $- 1$

0 Follower 2 Views

Vishal Baghel

Contributor-Level 10

This is a Objective Type Questions as classified in NCERT Exemplar

Sol:

$\begin{array}{l} G i v e n t h a t | \begin{matrix} 1 + x & 1 & 1 \\ 1 & 1 + y & 1 \\ 1 & 1 & 1 + z \end{matrix} | = 0 \\ T a k i n g x, y a n d z c o m m o n f r o m R_{1}, R_{2} a n d R_{3} r e s p e c t i v e l y . \\ \Rightarrow x y z | \begin{matrix} \frac{1}{x} + 1 & \frac{1}{x} & \frac{1}{x} \\ \frac{1}{y} & \frac{1}{y} + 1 & \frac{1}{y} \\ \frac{1}{z} & \frac{1}{z} & \frac{1}{z} + 1 \end{matrix} | = 0 \\ R_{1} \to R_{1} + R_{2} + R_{3} \\ \Rightarrow x y z | \begin{matrix} \frac{1}{x} + \frac{1}{y} + \frac{1}{z} + 1 & \frac{1}{x} + \frac{1}{y} + \frac{1}{z} + 1 & \frac{1}{x} + \frac{1}{y} + \frac{1}{z} + 1 \\ \frac{1}{y} & \frac{1}{y} + 1 & \frac{1}{y} \\ \frac{1}{z} & \frac{1}{z} & \frac{1}{z} + 1 \end{matrix} | = 0 \\ T a k i n g \frac{1}{x} + \frac{1}{y} + \frac{1}{z} + 1 c o m m o n f r o m R_{1} \\ \Rightarrow x y z (\frac{1}{x} + \frac{1}{y} + \frac{1}{z} + 1) | \begin{matrix} 1 & 1 & 1 \\ \frac{1}{y} & \frac{1}{y} + 1 & \frac{1}{y} \\ \frac{1}{z} & \frac{1}{z} & \frac{1}{z} + 1 \end{matrix} | = 0 \\ C_{1} \to C_{1} - C_{2}, C_{2} \to C_{2} - C_{3} \\ \Rightarrow x y z (\frac{1}{x} + \frac{1}{y} + \frac{1}{z} + 1) | \begin{matrix} 0 & 0 & 1 \\ - 1 & 1 & \frac{1}{y} \\ 0 & - 1 & \frac{1}{z} + 1 \end{matrix} | = 0 \\ E x p a n d i n g a l o n g R_{1} \\ \Rightarrow x y z (\frac{1}{x} + \frac{1}{y} + \frac{1}{z} + 1) [1 | \begin{matrix} - 1 & 1 \\ 0 & - 1 \end{matrix} |] = 0 \\ \Rightarrow x y z (\frac{1}{x} + \frac{1}{y} + \frac{1}{z} + 1) (1) = 0 \\ \Rightarrow \frac{1}{x} + \frac{1}{y} + \frac{1}{z} + 1 = 0 a n d x y z \neq 0 (x \neq y \neq z \neq 0) \\ ∴ x^{- 1} + y^{- 1} + z^{- 1} = - 1 \\ H e n c e, t h e c o r r e c t o p t i o n i s (d) . \end{array}$

0 0

New answer posted

7 months ago

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

There are certain material developed in laboratories which have a negative refractive index figure. A ray incident from air (Medium 1) into such a medium (Medium 2) shall follow a path given by

0 Follower 3 Views

Payal Gupta

Contributor-Level 10

This is a multiple choice answer as classified in NCERT Exemplar

(a) The negative refractive index metamaterials are those in which incident ray from air (Medium 1) to them refract or bend differently to that of positive refractive index medium.

0 0

New answer posted

7 months ago

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

If $A$ and $B$ are invertible matrices, then which of the following is not correct?

0 Follower 7 Views

Vishal Baghel

Contributor-Level 10

This is a Objective Type Questions as classified in NCERT Exemplar

Sol:

$\begin{array}{l} I f A a n d B a r e t w o i n v e r t i b l e m a t r i c e s t h e n \\ (a) a d j A = | A | . A^{- 1} i s c o r r e c t \\ (b) d e t {(A)}^{- 1} = {[d e t (A)]}^{- 1} = \frac{1}{d e t (A)} i s c o r r e c t \\ (c) A l s o, {(A B)}^{- 1} = B^{- 1} A^{- 1} i s c o r r e c t \\ (d) {(A + B)}^{- 1} = \frac{1}{| A + B |} . a d j (A + B) \\ ∴ {(A + B)}^{- 1} \neq B^{- 1} + A^{- 1} \\ H e n c e, t h e c o r r e c t o p t i o n i s (d) . \end{array}$

0 0

New answer posted

7 months ago

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

If $A = [\begin{matrix} 2 & λ & - 3 \\ 0 & 2 & 5 \\ 1 & 1 & 3 \end{matrix}]$ , then $A^{- 1}$ exists if

(A) $\begin{matrix} λ \end{matrix}$ $= 2$

(B) $\begin{matrix} λ \end{matrix}$ $\neq 2$

(C) $\begin{matrix} λ \end{matrix}$ $\neq - 2$

(D) None of these

0 Follower 6 Views

Vishal Baghel

Contributor-Level 10

This is a Objective Type Questions as classified in NCERT Exemplar

Sol:

$\begin{array}{l} W e h a v e, \\ A = [\begin{matrix} 2 & λ & - 3 \\ 0 & 2 & 5 \\ 1 & 1 & 3 \end{matrix}] \Rightarrow | A | = | \begin{matrix} 2 & λ & - 3 \\ 0 & 2 & 5 \\ 1 & 1 & 3 \end{matrix} | \\ E x p a n d i n g a l o n g R_{1} = 2 | \begin{matrix} 2 & 5 \\ 1 & 3 \end{matrix} | - λ | \begin{matrix} 0 & 5 \\ 1 & 3 \end{matrix} | - 3 | \begin{matrix} 0 & 2 \\ 1 & 1 \end{matrix} | \\ = 2 (6 - 5) - λ (0 - 5) - 3 (0 - 2) \\ = 2 + 5 λ + 6 = 8 + 5 λ \\ I f A^{- 1} e x i s t s t h e n | A | \neq 0 \\ ∴ 8 + 5 λ \neq 0 S o λ \neq \frac{- 8}{5} \\ H e n c e, t h e c o r r e c t o p t i o n i s (d) . \end{array}$

0 0

New answer posted

7 months ago

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

A far is moving with a constant speed of 60 km h^-1on a straight road. Looking at the rear view mirror, the driver finds that the car following him is at a distance of 100 m and is approaching with a speed of 5 km h^-1.
In order to keep track of the car in the rear, the driver begins to glance alternatively at the rear and side mirror of his car after every 2 s till the other car overtakes. If the two cars were maintaining their speeds, which of the following statement (s) is/are correct?

(a) The speed of the car in the rear is 65 km h^-1

(b) In the side mirror, the car in the-rear would appear to approach with a speed of 5 km h-1 to

...more

0 Follower 12 Views

Payal Gupta

Contributor-Level 10

This is a multiple choice answer as classified in NCERT Exemplar

(d) The speed of the image of the car would appear to increase as the distance between the cars decreases.

0 0

New answer posted

7 months ago

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

If $f (x) = [\begin{matrix} 0 & x - a & x - b \\ x + a & 0 & x - c \\ x + b & x + c & 0 \end{matrix}]$ , then

(A) $f (a) = 0$

(B) $f (b) = 0$

(C) $f (0) = 0$

(D) $f (1) = 0$

0 Follower 2 Views

Vishal Baghel

Contributor-Level 10

This is a Objective Type Questions as classified in NCERT Exemplar

Sol:

$\begin{array}{l} G i v e n t h a t f (x) = | \begin{matrix} 0 & x - a & x - b \\ x + a & 0 & x - c \\ x + b & x + c & 0 \end{matrix} | \\ f (a) = | \begin{matrix} 0 & 0 & a - b \\ 2 a & 0 & a - c \\ a + b & a + c & 0 \end{matrix} | \\ E x p a n d i n g a l o n g R_{1} = (a - b) | \begin{matrix} 2 a & 0 \\ a + b & a + c \end{matrix} | \\ = (a - b) [2 a (a + c)] = (a - b) . 2 a . (a + c) \neq 0 \\ f (b) = | \begin{matrix} 0 & b - a & 0 \\ b + a & 0 & b - c \\ 2 b & b + c & 0 \end{matrix} | \\ E x p a n d i n g a l o n g R_{1} = - (b - a) | \begin{matrix} b + a & b - c \\ 2 b & 0 \end{matrix} | \\ = - (b - a) [(- 2 b) (b - c)] = 2 b (b - a) (b - c) \neq 0 \\ f (0) = | \begin{matrix} 0 & - a & - b \\ a & 0 & - c \\ b & c & 0 \end{matrix} | \\ E x p a n d i n g a l o n g R_{1} = a | \begin{matrix} a & - c \\ b & 0 \end{matrix} | - b | \begin{matrix} a & 0 \\ b & c \end{matrix} | \\ = a (b c) - b (a c) = a b c - a b c = 0 \\ H e n c e, t h e c o r r e c t o p t i o n i s (c) . \end{array}$

0 0

New answer posted

7 months ago

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

$f (x) = {\begin{matrix} \frac{1 - c o s 2 x}{x^{2}}, & i f x \neq 0 \\ 5, & i f x = 0 \end{matrix}$ at $x = 2$ , $x = 0$

0 Follower 2 Views

alok kumar singh

Contributor-Level 10

This is a Short Answer Type Questions as classified in NCERT Exemplar

Sol:

$\begin{array}{l} \underset{x \to 0^{-}}{l i m} f (x) = \frac{1 - c o s 2 x}{x^{2}} = \underset{h \to 0}{l i m} \frac{1 - c o s 2 (0 - h)}{{(0 - h)}^{2}} \\ = \underset{h \to 0}{l i m} \frac{1 - c o s (- 2 h)}{h^{2}} = \underset{h \to 0}{l i m} \frac{1 - c o s (2 h)}{h^{2}} \\ = \underset{h \to 0}{l i m} \frac{2 s i n^{2} h}{h^{2}} [? 1 - c o s θ = 2 s i n^{2} \frac{θ}{2}] \\ = \underset{h \to 0}{l i m} \frac{2 s i n h}{h} . \frac{s i n h}{h} = 2.1.1 = 2 [\underset{x \to 0}{l i m} \frac{s i n x}{x} = 1] \\ \underset{x \to 0^{+}}{l i m} f (x) = \frac{1 - c o s 2 x}{x^{2}} \\ = \underset{h \to 0}{l i m} \frac{1 - c o s 2 (0 + h)}{{(0 + h)}^{2}} = \underset{h \to 0}{l i m} \frac{1 - c o s (2 h)}{h^{2}} \\ = \underset{h \to 0}{l i m} \frac{2 s i n^{2} h}{h^{2}} = \underset{h \to 0}{l i m} \frac{2 s i n h}{h} . \frac{s i n h}{h} = 2.1.1 = 2 \\ \underset{x \to 0}{l i m} f (x) = 5 \\ A s \underset{x \to 0^{-}}{l i m} f (x) = \underset{x \to 0^{+}}{l i m} f (x) \neq \underset{x \to 0}{l i m} f (x) \\ H e n c e, f (x) i s d i s c o n t i n u o u s a t x = 0 . \end{array}$

0 0

New answer posted

7 months ago

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

The optical density of turpentine is higher than that of water while its mass density is lower. Figure shows a layer of turpentine floating over water in a container. For which one of the four rays incident on turpentine in figure, the path shown is correct?

(a) 1

(b) 2

(c) 3

(d) 4

0 Follower 5 Views

Payal Gupta

Contributor-Level 10

This is a multiple choice answer as classified in NCERT Exemplar

Here, light ray goes from (optically) rarer medium air to optically denser terpentine, then it bends towards the normal i.e., i>r whereas when it goes from to optically denser medium terpentine to rarer medium water. then it bends away the normal i.e., i

0 0

New answer posted

7 months ago

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

The maximum value of = $[\begin{matrix} 1 & 1 & 1 \\ 1 & 1 + s i n θ & 1 \\ 1 + c o s θ & 1 & 1 \end{matrix}]$ is ( $θ$ is real number)

(A) $\frac{1}{2}$

(B) √3/2

(C) √2

(D) $\frac{2√3}{4}$

0 Follower 3 Views

Vishal Baghel

Contributor-Level 10

This is a Objective Type Questions as classified in NCERT Exemplar

Sol:

$\begin{array}{l} G i v e n t h a t Δ = | \begin{matrix} 1 & 1 & 1 \\ 1 & 1 + s i n θ & 1 \\ 1 + c o s θ & 1 & 1 \end{matrix} | \\ C_{1} \to C_{1} - C_{2}, C_{2} \to C_{2} - C_{3} \\ \Rightarrow = | \begin{matrix} 0 & 0 & 1 \\ - s i n θ & s i n θ & 1 \\ c o s θ & 0 & 1 \end{matrix} | \\ E x p a n d i n g a l o n g R_{1} \\ \Rightarrow = 1 | \begin{matrix} - s i n θ & s i n θ \\ c o s θ & 0 \end{matrix} | = - s i n θ c o s θ \\ \Rightarrow = - \frac{1}{2} . 2 s i n θ c o s θ = - \frac{1}{2} s i n 2 θ \\ b u t maximum value of sin2 θ =1 \Rightarrow | - \frac{1}{2} . 1 | = \frac{1}{2} \\ H e n c e, t h e c o r r e c t o p t i o n i s (a) . \end{array}$

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
684k Reviews
1800k 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