Ncert Solutions Maths class 12th

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New answer posted•a year ago

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

If the value of a third-order determinant is 12, then the value of the determinant formed by replacing each element by its cofactor will be 144.

23 Views|Posted a year ago

Asked by Vishal Baghel

Answered by

Vishal Baghel | Contributor-Level 10

a year ago

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

Sol:

$\begin{array}{l} T r u e . \\ Since | A | = 12 \\ I f A i s a s q u a r e m a t r i x o f o r d e r n \\ t h e n | A d j A | = {| A |}^{n - 1} \\ ∴ | A d j A | = {| A |}^{3 - 1} = {| A |}^{2} = {(12)}^{2} = 144 [n = 3] \end{array}$

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New answer posted•a year ago

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

If $A$ and $B$ are matrices of order 3 and $? A ? = 5$ , $? B ? = 3$ , then $? 3 A B ? = 27 \times 5 \times 3 = 405 .$

11 Views|Posted a year ago

Asked by Vishal Baghel

Answered by

Vishal Baghel | Contributor-Level 10

a year ago

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

Sol:

$\begin{array}{l} T r u e . \\ | 3 A B | = 3^{3} | A B | = 27 | A | | B | = 27 \times 5 \times 3 [? | K A | = K^{n} | A |] \end{array}$

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New answer posted•a year ago

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

Kindly consider the following

$f (x) = {\begin{matrix} \frac{2^{x + 2} - 16}{4^{x} - 16}, & i f x \neq 2 \\ k, & i f x = 2 \end{matrix}$ at $x = 2$

5 Views|Posted a year ago

Asked by Alok Kumar Singh

Answered by

Alok Kumar Singh | Contributor-Level 10

a year ago

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

Sol:

$\begin{array}{l} f (x) = \frac{2^{x + 2} - 16}{4^{x} - 16} = \frac{2^{2} . 2^{x} - 16}{{(2^{x})}^{2} - {(4)}^{2}} = \frac{4 (2^{x} - 4)}{(2^{x} - 4) (2^{x} + 4)} \\ = \frac{4}{(2^{x} + 4)} \\ \underset{x \to 2^{-}}{l i m} f (x) = \underset{h \to 0}{l i m} \frac{4}{(2^{2 - h} + 4)} = \frac{4}{2^{2} + 4} = \frac{4}{4 + 4} = \frac{4}{8} = \frac{1}{2} \\ \underset{x \to 2}{l i m} f (x) = k \\ A s t h e f u n c t i o n i s c o n t i n u o u s a t x = 2 \\ ∴ \underset{x \to 2^{-}}{l i m} f (x) = \underset{x \to 2}{l i m} f (x) \\ ∴ k = \frac{1}{2} \\ H e n c e, t h e v a l u e o f k i s \frac{1}{2} . \end{array}$

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New answer posted•a year ago

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

$∣ A^{- 1} ∣ \neq ∣ A ∣^{- 1}$ , where $A$ is a non-singular matrix.

7 Views|Posted a year ago

Asked by Vishal Baghel

Answered by

Vishal Baghel | Contributor-Level 10

a year ago

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

Sol:

$\begin{array}{l} F a l s e . \\ Since ? ? | A^{? 1} | = {| A |}^{? 1} f o r ? ? a ? ? n o n ? singular ? ? m a t r i x . \end{array}$

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New answer posted•a year ago

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

${(a A)}^{- 1} = \frac{1}{a} A^{- 1}$ , where $a$ is any real number and $A$ is a square matrix.

5 Views|Posted a year ago

Asked by Vishal Baghel

Answered by

Vishal Baghel | Contributor-Level 10

a year ago

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

Sol:

$\begin{array}{l} I f a n o n - singular s q u a r e m a t i x, t h e n f o r a n y n o n - z e r o s c a l a r' a', a A i s i n v e r t i b l e . \\ ∴ (a A) . (\frac{1}{a} A^{- 1}) = a . \frac{1}{a} . A . A^{- 1} = I \\ S o, (a A) i s i n v e r s e o f (\frac{1}{a} A^{- 1}) \\ \Rightarrow {(a A)}^{- 1} = \frac{1}{a} A^{- 1} i s t r u e . \end{array}$

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New answer posted•a year ago

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

Find the value of $k$ in each of the Exercises 11 to 14 so that the function $f$ is continuous at the indicated point:

$f (x) = {\begin{matrix} 3 x - 8, & i f x \leq 5 \\ 2 k, & i f x > 5 \end{matrix}$ at $x = 5$

10 Views|Posted a year ago

Asked by Alok Kumar Singh

Answered by

Alok Kumar Singh | Contributor-Level 10

a year ago

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

Sol:

$\begin{array}{l} \underset{x \to 5}{l i m} f (x) = 3 x - 8 = \underset{h \to 0}{l i m} 3 (5 - h) - 8 = 15 - 8 = 7 \\ \underset{x \to 5^{+}}{l i m} f (x) = 2 k \\ A s t h e f u n c t i o n i s c o n t i n u o u s a t x = 5 \\ ∴ \underset{x \to 1^{-}}{l i m} f (x) = \underset{x \to 5^{+}}{l i m} f (x) \\ ∴ 7 = 2 k \Rightarrow k = \frac{7}{2} \\ A s \underset{x \to 1^{-}}{l i m} f (x) = \underset{x \to 1^{+}}{l i m} f (x) = \underset{x \to 1}{l i m} f (x) \\ H e n c e, t h e v a l u e o f k i s \frac{7}{2} . \end{array}$

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New answer posted•a year ago

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

${(A^{3})}^{- 1} = {(A^{- 1})}^{3}$ , where $A$ is a square matrix and $? A ? \neq 0$ .

5 Views|Posted a year ago

Asked by Vishal Baghel

Answered by

Vishal Baghel | Contributor-Level 10

a year ago

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

Sol:

$\begin{array}{l} Since {(A^{K})}^{- 1} = {(A^{- 1})}^{K} w h e r e K \in N \\ S o, {(A^{3})}^{- 1} = {(A^{- 1})}^{3} i s t r u e \end{array}$

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New answer posted•a year ago

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

If $f (x) = [\begin{matrix} {(1 + x)}^{17} & {(1 + x)}^{19} & {(1 + x)}^{23} \\ {(1 + x)}^{23} & {(1 + x)}^{29} & {(1 + x)}^{34} \\ {(1 + x)}^{41} & {(1 + x)}^{43} & {(1 + x)}^{47} \end{matrix}] = A + B x + C x^{2} + \dots$ then $A =$ ____.

6 Views|Posted a year ago

Asked by Vishal Baghel

Answered by

Vishal Baghel | Contributor-Level 10

a year ago

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

Sol:

$\begin{array}{l} G i v e n t h a t [\begin{matrix} {(1 + x)}^{17} & {(1 + x)}^{19} & {(1 + x)}^{23} \\ {(1 + x)}^{23} & {(1 + x)}^{29} & {(1 + x)}^{34} \\ {(1 + x)}^{41} & {(1 + x)}^{43} & {(1 + x)}^{47} \end{matrix}] = A + B x + C x^{2} + \dots \\ T a k i n g {(1 + x)}^{17}, {(1 + x)}^{23} a n d {(1 + x)}^{41} c o m m o n f r o m R_{1}, R_{2} a n d R_{3} r e s p e c t i v e l y \\ {(1 + x)}^{17} . {(1 + x)}^{23} . {(1 + x)}^{41} | \begin{matrix} 1 & {(1 + x)}^{2} & {(1 + x)}^{6} \\ 1 & {(1 + x)}^{6} & {(1 + x)}^{11} \\ 1 & {(1 + x)}^{2} & {(1 + x)}^{6} \end{matrix} | \\ \Rightarrow {(1 + x)}^{17} . {(1 + x)}^{23} . {(1 + x)}^{41} . 0 (R_{1} a n d R_{3} a r e i d e n t i c a l) \\ ∴ 0 = A + B x + C x^{2} + \dots \\ B y c o m p a r i n g t h e l i k e t e r m s, w e g e t A = 0 . \end{array}$

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New answer posted•a year ago

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

Kindly consider the following

$f (x) = | x | + | x - 1 |$ at $x = 1$

4 Views|Posted a year ago

Asked by Alok Kumar Singh

Answered by

Alok Kumar Singh | Contributor-Level 10

a year ago

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

Sol:

$\begin{array}{l} \underset{x \to 1^{-}}{l i m} f (x) = | x | + | x - 1 | = \underset{h \to 0}{l i m} | 1 - h | + | 1 - h - 1 | \\ = | 1 - 0 | + | 1 - 0 - 1 | = 1 + 0 = 1 \\ \underset{x \to 1}{l i m} f (x) = | x | + | x - 1 | = | 1 | + | 1 - 1 | = 1 + 0 = 1 \\ \underset{x \to 1^{+}}{l i m} f (x) = | x | + | x - 1 | = \underset{h \to 0}{l i m} | 1 + h | + | 1 + h - 1 | \\ = | 1 + 0 | + | 1 + 0 - 1 | = 1 + 0 = 1 \\ A s \underset{x \to 1^{-}}{l i m} f (x) = \underset{x \to 1^{+}}{l i m} f (x) = \underset{x \to 1}{l i m} f (x) \\ H e n c e, f (x) i s c o n t i n u o u s a t x = 1 . \end{array}$

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New answer posted•a year ago

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

Kindly Consider the following

$[\begin{matrix} 0 & - x y z & x - z \\ - y & x & - y - z \\ z & x - z & y \end{matrix}]$ =____.

7 Views|Posted a year ago

Asked by Vishal Baghel

Answered by

Vishal Baghel | Contributor-Level 10

a year ago

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

Sol:

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

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Ncert Solutions Maths class 12th

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