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Ncert Solutions Maths class 12th Physics NCERT Exemplar Solutions Class 12th Chapter Five

Kindly consider the following

$f (x) = {\begin{matrix} | x | c o s \frac{1}{x}, & i f x \neq 0 \\ 0, & i f x = 0 \end{matrix}$

4 Views|Posted a year ago

Asked by Alok Kumar Singh

S

Answered by

Alok Kumar Singh | Contributor-Level 10

a year ago

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

Sol:

$\begin{array}{l} \underset{x \to 0^{-}}{l i m} f (x) = | x | c o s \frac{1}{x} \\ = \underset{h \to 0}{l i m} | 0 - h | c o s \frac{1}{(0 - h)} = \underset{h \to 0}{l i m} h . c o s \frac{1}{h} = 0 [? c o s \frac{1}{x} o s c i l l a t e b e t w e e n - 1 a n d 1] \\ \underset{x \to 0^{+}}{l i m} f (x) = | x | c o s \frac{1}{x} = \underset{h \to 0}{l i m} | 0 + h | c o s \frac{1}{(0 + h)} = \underset{h \to 0}{l i m} h . c o s \frac{1}{h} = 0 \\ \underset{x \to 0}{l i m} f (x) = 0 \\ ∴ \underset{x \to 0^{-}}{l i m} f (x) = \underset{x \to 0^{+}}{l i m} f (x) = \underset{x \to 0}{l i m} f (x) = 0 \\ H e n c e, f (x) i s c o n t i n u o u s a t x = 0 . \end{array}$

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

If $A$ is a matrix of order $3 \times 3$ , then $? 3 A ? =$ ___.

5 Views|Posted a year ago

Asked by Vishal Baghel

S

Answered by

Vishal Baghel | Contributor-Level 10

a year ago

This is a Fill in the blanks as classified in NCERT Exemplar

Sol:

$\begin{array}{l} W e k n o w t h a t f o r a m a t r i x o f o r d e r 3 \times 3, \\ | K A | = K^{3} | A | \\ ∴ | 3 A | = 3^{3} | A | = 27 | A | \end{array}$

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Ncert Solutions Maths class 12th Physics NCERT Exemplar Solutions Class 12th Chapter Five

Kindly consider the following
$f (x) = {\begin{matrix} \frac{| x - 4 |}{2 (x - 4)}, & x \neq 4 \\ 0, & x = 4 \end{matrix}$ at $x = 2$ , $x = 4$

7 Views|Posted a year ago

Asked by Alok Kumar Singh

S

Answered by

Alok Kumar Singh | Contributor-Level 10

a year ago

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

Sol:

$\begin{array}{l} \underset{x \to 4^{-}}{l i m} f (x) = \frac{| x - 4 |}{2 (x - 4)} [\begin{array}{l} f o r x < 4, | x - 4 | = - (x - 4) \\ f o r x > 4, | x - 4 | = (x - 4) \end{array}] \\ = \underset{h \to 0}{l i m} \frac{- [4 - h - 4]}{2 [4 - h - 4]} = \underset{h \to 0}{l i m} \frac{h}{- 2 h} = - \frac{1}{2} \\ \underset{x \to 4^{+}}{l i m} f (x) = \frac{| x - 4 |}{2 (x - 4)} = \underset{h \to 0}{l i m} \frac{[4 + h - 4]}{2 [4 + h - 4]} = \underset{h \to 0}{l i m} \frac{h}{2 h} = \frac{1}{2} \\ \underset{x \to 4}{l i m} f (x) = 0 \\ ∴ \underset{x \to 4^{-}}{l i m} f (x) \neq \underset{x \to 4^{+}}{l i m} f (x) \neq \underset{x \to 4}{l i m} f (x) \\ H e n c e, f (x) i s d i s c o n t i n u o u s a t x = 4 . \end{array}$

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

There are two values of $a$ which make the determinant $= [\begin{matrix} 1 & - 2 & 5 \\ 2 & a & - 1 \\ 0 & 4 & 2 a \end{matrix}] = 86,$ then the sum of these numbers is

(A) 4

(B) 5

(C) -4

(D) 9

9 Views|Posted a year ago

Asked by Vishal Baghel

S

Answered by

Vishal Baghel | Contributor-Level 10

a year ago

This is a Objective Type Questions as classified in NCERT Exemplar

Sol:

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Ncert Solutions Maths 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) $7 x^{2} (x + y)$

4 Views|Posted a year ago

Asked by Vishal Baghel

S

Answered by

Vishal Baghel | Contributor-Level 10

a year ago

This is a Objective Type Questions as classified in NCERT Exemplar

Sol:

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

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Ncert Solutions Maths 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) $- 1$

12 Views|Posted a year ago

Asked by Vishal Baghel

S

Answered by

Vishal Baghel | Contributor-Level 10

a year ago

This is a Objective Type Questions as classified in NCERT Exemplar

Sol:

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

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Ncert Solutions Maths 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?

14 Views|Posted a year ago

Asked by Vishal Baghel

S

Answered by

Vishal Baghel | Contributor-Level 10

a year ago

This is a Objective Type Questions as classified in NCERT Exemplar

Sol:

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

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Ncert Solutions Maths 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

11 Views|Posted a year ago

Asked by Vishal Baghel

S

Answered by

Vishal Baghel | Contributor-Level 10

a year ago

This is a Objective Type Questions as classified in NCERT Exemplar

Sol:

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

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Ncert Solutions Maths 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) $f (1) = 0$

4 Views|Posted a year ago

Asked by Vishal Baghel

S

Answered by

Vishal Baghel | Contributor-Level 10

a year ago

This is a Objective Type Questions as classified in NCERT Exemplar

Sol:

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

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Ncert Solutions Maths 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 $x = 2$ , $x = 0$

2 Views|Posted a year ago

Asked by Alok Kumar Singh

S

Answered by

Alok Kumar Singh | Contributor-Level 10

a year ago

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

Sol:

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

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Kindly consider the following

$f (x) = {\begin{matrix} | x | c o s \frac{1}{x}, & i f x \neq 0 \\ 0, & i f x = 0 \end{matrix}$

If $A$ is a matrix of order $3 \times 3$ , then $? 3 A ? =$ ___.

Kindly consider the following
$f (x) = {\begin{matrix} \frac{| x - 4 |}{2 (x - 4)}, & x \neq 4 \\ 0, & x = 4 \end{matrix}$ at $x = 2$ , $x = 4$

There are two values of $a$ which make the determinant $= [\begin{matrix} 1 & - 2 & 5 \\ 2 & a & - 1 \\ 0 & 4 & 2 a \end{matrix}] = 86,$ then the sum of these numbers is

(A) 4

(B) 5

(C) -4

(D) 9

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

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) $- 1$

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

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

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) $f (1) = 0$

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