Ncert Solutions Maths class 12th

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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

$f (x) = {\begin{matrix} \frac{x^{2}}{2}, i f & 0 \leq x \leq 1 \\ 2 x^{2} - 3 x + \frac{3}{2}, i f & 1 < x \leq 2 \end{matrix}$ at $x = 0$ , $x = 1$

7 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:

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

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New answer posted•a year ago

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

If $x = - 9$ is a root of $[\begin{matrix} x & 3 & 7 \\ 2 & x & 2 \\ 7 & 6 & x \end{matrix}] = 0,$ then the other two roots are ____.

3 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:

$\begin{array}{l} W e h a v e, | \begin{matrix} x & 3 & 7 \\ 2 & x & 2 \\ 7 & 6 & x \end{matrix} | = 0 \\ E x p a n d i n g a l o n g R_{1} \\ \Rightarrow x | \begin{matrix} x & 2 \\ 6 & x \end{matrix} | - 3 | \begin{matrix} 2 & 2 \\ 7 & x \end{matrix} | + 7 | \begin{matrix} 2 & x \\ 7 & 6 \end{matrix} | = 0 \\ \Rightarrow x (x^{2} - 12) - 3 (2 x - 14) + 7 (12 - 7 x) = 0 \\ \Rightarrow x^{3} - 12 x - 6 x + 42 + 84 - 49 x = 0 \\ \Rightarrow x^{3} - 67 x + 126 = 0 \dots (1) \\ T h e r o o t s o f t h e e q u a t i o n m a y b e t h e f a c t o r s o f 126 i . e ., 2 \times 7 \times 9 \\ 9 i s t h e g i v e n r o o t o f t h e determinant p u t x = 2 i n e q . (1) \\ {(2)}^{3} - 67 \times 2 + 126 \Rightarrow 8 - 134 + 126 = 0 \\ H e n c e, x = 2 i s t h e o t h e r r o o t . \\ N o w, p u t x = 7 i n e q . (1) \\ {(7)}^{3} - 67 \times 7 + 126 \Rightarrow 343 - 469 + 126 = 0 \\ H e n c e, x = 7 i s a l s o t h e o t h e r r o o t o f t h e determinant . \end{array}$

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New answer posted•a year ago

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

The sum of the products of elements of any row with the cofactors of corresponding elements is equal to ___.

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:

$\begin{array}{l} T h e s u m o f t h e p r o d u c t s o f e l e m e n t s o f a n y r o w w i t h t h e c o - f a c t o r s o f c o r r e s p o n d i n g e l e m e n t s \\ i s e q u a l t o t h e v a l u e o f t h e determinant o f t h e g i v e n m a t r i x . \\ L e t Δ = | \begin{matrix} a_{11} & a_{12} & a_{13} \\ a_{21} & a_{22} & a_{23} \\ a_{31} & a_{32} & a_{33} \end{matrix} | \\ E x p a n d i n g a l o n g R_{1} \\ a_{11} | \begin{matrix} a_{22} & a_{23} \\ a_{32} & a_{33} \end{matrix} | - a_{12} | \begin{matrix} a_{21} & a_{23} \\ a_{31} & a_{33} \end{matrix} | + a_{13} | \begin{matrix} a_{21} & a_{22} \\ a_{31} & a_{32} \end{matrix} | \\ \Rightarrow a_{11} M_{11} + a_{12} M_{12} + a_{13} M_{13} (w h e r e M_{11}, M_{12} a n d M_{13} a r e t h e minors o f t h e c o r r e s p o n d i n g e l e m e n t s) \end{array}$

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New answer posted•a year ago

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

If $A$ is a matrix of order $3 \times 3$ , then the number of minors in the determinant of $A$ is ____.

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:

$\begin{array}{l} T h e o r d e r o f a m a t r i x i s 3 \times 3 \\ ∴ T o t a l n u m b e r o f e l e m e n t s = 3 \times 3 = 9 \\ H e n c e, t h e n u m b e r o f minors i n t h e determinant i s 9 . \end{array}$

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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

$f (x) = {\begin{matrix} \frac{e^{\frac{1}{x}}}{1 + e^{\frac{1}{x}}}, & i f x \neq 0 \\ 0, & i f x = 0 \end{matrix}$

4 Views|Posted a year ago

Asked by Alok Kumar Singh

Answered by

Alok Kumar Singh | Contributor-Level 10

a year ago

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

Sol:

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

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New answer posted•a year ago

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

If $A$ is a matrix of order $3 \times 3$ , then ${(A^{2})}^{- 1} =$ ____.

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: $F o r a n y s q u a r e m a t r i x A, {(A^{2})}^{- 1} = {(A^{- 1})}^{2} .$

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New answer posted•a year ago

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

If $c o s^{2} θ = 0$ , then _. $[\begin{matrix} 0 & c o s θ & s i n θ \\ c o s θ & s i n θ & 0 \\ s i n θ & 0 & c o s θ \end{matrix}] =$ _________

1 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:

Upvote

New answer posted•a year ago

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

Kindly consider the following

$f (x) = {\begin{matrix} | x - a | s i n \frac{1}{x - a}, & x \neq a \\ 0, & x = a \end{matrix}$ at $x = 0$ , $x = a$

3 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:

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

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New answer posted•a year ago

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

If $x, y, z \in R$ , then the value of the determinant $[\begin{matrix} {(2^{x} + 2^{- x})}^{2} & {(2^{x} - 2^{- x})}^{2} & 1 \\ {(3^{x} + 3^{- x})}^{2} & {(3^{x} - 3^{- x})}^{2} & 1 \\ {(4^{x} + 4^{- x})}^{2} & {(4^{x} - 4^{- x})}^{2} & 1 \end{matrix}]$ is equal to ___.

18 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:

$\begin{array}{l} W e h a v e, | \begin{matrix} {(2^{x} + 2^{- x})}^{2} & {(2^{x} - 2^{- x})}^{2} & 1 \\ {(3^{x} + 3^{- x})}^{2} & {(3^{x} - 3^{- x})}^{2} & 1 \\ {(4^{x} + 4^{- x})}^{2} & {(4^{x} - 4^{- x})}^{2} & 1 \end{matrix} | \\ C_{1} \to C_{1} - C_{2} \\ \Rightarrow | \begin{matrix} {(2^{x} + 2^{- x})}^{2} - {(2^{x} - 2^{- x})}^{2} & {(2^{x} - 2^{- x})}^{2} & 1 \\ {(3^{x} + 3^{- x})}^{2} - {(3^{x} - 3^{- x})}^{2} & {(3^{x} - 3^{- x})}^{2} & 1 \\ {(4^{x} + 4^{- x})}^{2} - {(4^{x} - 4^{- x})}^{2} & {(4^{x} - 4^{- x})}^{2} & 1 \end{matrix} | \\ \Rightarrow | \begin{matrix} 4 . 2^{x} . 2^{- x} & {(2^{x} - 2^{- x})}^{2} & 1 \\ 4 . 3^{x} 3^{- x} & {(3^{x} - 3^{- x})}^{2} & 1 \\ 4 . 4^{x} 4^{- x} & {(4^{x} - 4^{- x})}^{2} & 1 \end{matrix} | [a p p l y i n g {(a + b)}^{2} - {(a - b)}^{2} = 4 a b] \\ \Rightarrow | \begin{matrix} 4 & {(2^{x} - 2^{- x})}^{2} & 1 \\ 4 & {(3^{x} - 3^{- x})}^{2} & 1 \\ 4 & {(4^{x} - 4^{- x})}^{2} & 1 \end{matrix} | \Rightarrow | \begin{matrix} 1 & {(2^{x} - 2^{- x})}^{2} & 1 \\ 1 & {(3^{x} - 3^{- x})}^{2} & 1 \\ 1 & {(4^{x} - 4^{- x})}^{2} & 1 \end{matrix} | (T a k i n g 4 c o m m o n f r o m C_{1}) \\ \Rightarrow 4.0 = 0 (? C_{1} a n d C_{3} a r e i d e n t i c a l c o l u m n s) \end{array}$