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

If $[\begin{matrix} 2 x & 5 \\ 8 & x \end{matrix}] = [\begin{matrix} 6 & - 2 \\ 7 & 3 \end{matrix}]$ then the value of $x$ is

(A) 3

(B) ±3

(C) ±6

(D) 6

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This is a Objective Type Questions as classified in NCERT Exemplar

Sol:

$\begin{array}{l} G i v e n t h a t \\ \Rightarrow | \begin{matrix} 2 x & 5 \\ 8 & x \end{matrix} | = | \begin{matrix} 6 & - 2 \\ 7 & 3 \end{matrix} | \\ \Rightarrow 2 x^{2} - 40 = 18 + 14 \Rightarrow 2 x^{2} = 32 + 40 \\ \Rightarrow 2 x^{2} = 72 \Rightarrow x^{2} = 36 \\ ∴ x \pm 6 \\ H e n c e, t h e c o r r e c t x^{2} o p t i o n i s (c) . \end{array}$

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

If $x = s i n t$ and $y = s i n p t$ , prove that $(1 - x^{2}) \frac{d^{2} y}{d x^{2}} - x \frac{d y}{d x} + p^{2} y = 0$

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Asked by Alok Kumar Singh

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Alok Kumar Singh | Contributor-Level 10

a year ago

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

Sol:

$\begin{array}{l} G i v e n t h a t x = s i n t a n d y = s i n p t \\ D i f f e r e n t i a t i n g b o t h s i d e s w . r . t . t \\ \Rightarrow \frac{d x}{d t} = c o s t a n d \frac{d y}{d t} = c o s p t . p = p . c o s p t \\ \Rightarrow \frac{d y}{d x} = \frac{\frac{d y}{d t}}{\frac{d x}{d t}} = \frac{p . c o s p t}{c o s t} \\ \Rightarrow \frac{d y}{d x} = \frac{p . c o s p t}{c o s t} \\ A g a i n d i f f e r e n t i a t i n g b o t h s i d e s w . r . t . x \\ \Rightarrow \frac{d}{d x} (\frac{d y}{d x}) = p . \frac{d}{d x} (\frac{c o s p t}{c o s t}) \\ \Rightarrow \frac{d^{2} y}{d x^{2}} = p . [\frac{c o s t . \frac{d}{d x} (c o s p t) - c o s p t . \frac{d}{d x} (c o s t)}{c o s^{2} t}] \\ \Rightarrow \frac{d^{2} y}{d x^{2}} = p . [\frac{c o s t . (- s i n p t) . p \frac{d t}{d x} - c o s p t . (- s i n t) . \frac{d t}{d x}}{c o s^{2} t}] \\ \Rightarrow \frac{d^{2} y}{d x^{2}} = p . [\frac{- p c o s t . s i n p t + c o s p t . s i n t}{c o s^{2} t}] \frac{d t}{d x} \\ \Rightarrow \frac{d^{2} y}{d x^{2}} = p . [\frac{- p c o s t . s i n p t + c o s p t . s i n t}{c o s^{2} t}] \frac{1}{c o s t} \\ \Rightarrow \frac{d^{2} y}{d x^{2}} = p . (\frac{- p c o s t . s i n p t + c o s p t . s i n t}{c o s^{3} t}) \\ N o w w e h a v e t o p r o v e t h a t \\ (1 - x^{2}) \frac{d^{2} y}{d x^{2}} - x \frac{d y}{d x} + p^{2} y = 0 \\ L . H . S . = (1 - x^{2}) [p . (\frac{- p c o s t . s i n p t + c o s p t . s i n t}{c o s^{3} t})] - x (\frac{p . c o s p t}{c o s t}) + p^{2} y \\ \Rightarrow = (1 - s i n^{2} t) [p . (\frac{- p c o s t . s i n p t + c o s p t . s i n t}{c o s^{3} t})] - \frac{p . s i n t c o s p t}{c o s t} + p^{2} . s i n p t \\ \Rightarrow = c o s^{2} t [\frac{- p^{2} c o s t . s i n p t + p c o s p t . s i n t}{c o s^{3} t}] - \frac{p . s i n t c o s p t}{c o s t} + p^{2} . s i n p t \\ \Rightarrow = \frac{- p^{2} c o s t . s i n p t + p c o s p t . s i n t}{c o s t} - \frac{p . s i n t c o s p t}{c o s t} + p^{2} . s i n p t \\ \Rightarrow = \frac{- p^{2} c o s t . s i n p t + p c o s p t . s i n t - p . s i n t c o s p t + p^{2} . s i n p t c o s t}{c o s t} \\ \Rightarrow = \frac{0}{c o s t} = 0 = R . H . S . \\ H e n c e, p r o v e d . \end{array}$

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

If $x^{m} y^{n} = {(x + y)}^{m + n}$ , prove that

(i) $\frac{d y}{d x} = \frac{y}{x}$

(ii) $\frac{d^{2} y}{d x^{2}} = 0$ .

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

Sol:

$\begin{array}{l} (i) G i v e n t h a t x^{m} . y^{n} = {(x + y)}^{m + n} \\ T a k i n g l o g o n b o t h s i d e s, w e g e t, \\ \Rightarrow l o g x^{m} . y^{n} = l o g {(x + y)}^{m + n} [? l o g x y = l o g x + l o g y] \\ \Rightarrow l o g x^{m} + l o g y^{n} = (m + n) l o g (x + y) \\ \Rightarrow m l o g x + n l o g y = (m + n) l o g (x + y) \\ D i f f e r e n t i a t i n g b o t h s i d e s w . r . t . x \\ \Rightarrow m . \frac{d}{d x} l o g x + n . \frac{d}{d x} l o g y = (m + n) \frac{d}{d x} l o g (x + y) \\ \Rightarrow m . \frac{1}{x} + n . \frac{1}{y} . \frac{d y}{d x} = (m + n) \frac{1}{x + y} (1 + \frac{d y}{d x}) \\ \Rightarrow \frac{m}{x} + \frac{n}{y} . \frac{d y}{d x} = \frac{m + n}{x + y} . (1 + \frac{d y}{d x}) \\ \Rightarrow \frac{m}{x} + \frac{n}{y} . \frac{d y}{d x} = \frac{m + n}{x + y} + \frac{m + n}{x + y} . \frac{d y}{d x} \\ \Rightarrow \frac{n}{y} . \frac{d y}{d x} - \frac{m + n}{x + y} . \frac{d y}{d x} = \frac{m + n}{x + y} - \frac{m}{x} \\ \Rightarrow (\frac{n}{y} - \frac{m + n}{x + y}) . \frac{d y}{d x} = \frac{m + n}{x + y} - \frac{m}{x} \\ \Rightarrow (\frac{n x + n y - m y - n y}{y (x + y)}) . \frac{d y}{d x} = (\frac{m x + n x - m x - m y}{x (x + y)}) \\ \Rightarrow (\frac{n x - m y}{y (x + y)}) . \frac{d y}{d x} = (\frac{n x - m y}{x (x + y)}) \\ \Rightarrow \frac{d y}{d x} = \frac{n x - m y}{x (x + y)} \times \frac{y (x + y)}{n x - m y} = \frac{y}{x} \\ \Rightarrow \frac{d y}{d x} = \frac{y}{x} H e n c e, p r o v e d . \end{array}$

$\begin{array}{l} (i i) G i v e n t h a t \frac{d y}{d x} = \frac{y}{x} \\ D i f f e r e n t i a t i n g b o t h s i d e s w . r . t . x \\ \Rightarrow \frac{d}{d x} (\frac{d y}{d x}) = \frac{d}{d x} (\frac{y}{x}) \\ \Rightarrow \frac{d^{2} y}{d x^{2}} = \frac{x . \frac{d y}{d x} - y . 1}{x^{2}} \\ \Rightarrow \frac{d^{2} y}{d x^{2}} = \frac{x . \frac{y}{x} - y}{x^{2}} [? \frac{d y}{d x} = \frac{y}{x}] \\ \Rightarrow \frac{d^{2} y}{d x^{2}} = \frac{y - y}{x^{2}} = \frac{0}{x^{2}} = 0 \\ H e n c e, \frac{d^{2} y}{d x^{2}} = 0 H e n c e, p r o v e d . \end{array}$

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

Find the values of $p$ and $q$ so that $f (x) = {\begin{matrix} x^{2} + 3 x + p, & x \leq 1 \\ q x + 2, & x > 1 \end{matrix}$ is differentiable at $x = 1$ .

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Alok Kumar Singh | Contributor-Level 10

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This is a Long Answer Type Questions as classified in NCERT Exemplar $\begin{array}{l} G i v e n t h a t f (x) = {\begin{matrix} x^{2} + 3 x + p, & x \leq 1 \\ q x + 2, & x > 1 \end{matrix} a t x = 1 . \\ L . H . L . f' (c) = \underset{x \to 1^{-}}{l i m} \frac{f (x) - f (c)}{x - c} \\ f' (1) = \underset{x \to 1^{-}}{l i m} \frac{f (x) - f (1)}{x - 1} \\ = \underset{x \to 1^{-}}{l i m} \frac{(x^{2} + 3 x + p) - (1 + 3 + p)}{x - 1} \\ = \underset{h \to 0}{l i m} \frac{[{(1 - h)}^{2} + 3 (1 - h) + p] - (1 + 3 + p)}{1 - h - 1} \\ = \underset{h \to 0}{l i m} \frac{[1 + h^{2} - 2 h + 3 - 3 h + p] - (4 + p)}{- h} \\ = \underset{h \to 0}{l i m} \frac{[h^{2} - 5 h + 4 + p] - [4 + p]}{- h} \\ = \underset{h \to 0}{l i m} \frac{h^{2} - 5 h + 4 + p - 4 - p}{- h} \\ = \underset{h \to 0}{l i m} \frac{h^{2} - 5 h}{- h} = \underset{h \to 0}{l i m} \frac{h [h - 5]}{- h} = 5 \\ R . H . L . f' (1) = \underset{x \to 1^{+}}{l i m} \frac{f (x) - f (1)}{x - 1} \\ = \underset{x \to 1^{+}}{l i m} \frac{(q x + 2) - (1 + 3 + p)}{x - 1} \\ = \underset{h \to 0}{l i m} \frac{[q (1 + h) + 2] - [4 + p]}{1 + h - 1} \\ = \underset{h \to 0}{l i m} \frac{q + q h + 2 - 4 - p}{h} = \underset{h \to 0}{l i m} \frac{q + q h - 2 - p}{h} \\ F o r e x i s t i n g t h e limit \\ q - 2 - p = 0 \Rightarrow q - p = 0 \dots (1) \\ \underset{h \to 0}{\Rightarrow l i m} \frac{q h - 0}{h} = q \\ I f L . H . L . f' (1) = R . H . L . f' (1) t h e n q = 5 . \\ N o w p u t t \end{array}$

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

Kindly Consider the following

Find $A^{- 1}$ if $A = [\begin{matrix} 0 & 1 & 1 \\ 1 & 0 & 1 \\ 1 & 1 & 0 \end{matrix}]$ and show that $A^{- 1} = \frac{3 I - A^{2}}{2}$

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

$\begin{array}{l} H e r e, A = [\begin{matrix} 0 & 1 & 1 \\ 1 & 0 & 1 \\ 1 & 1 & 0 \end{matrix}] \\ | A | = 0 | \begin{matrix} 0 & 1 \\ 1 & 0 \end{matrix} | - 1 | \begin{matrix} 1 & 1 \\ 1 & 0 \end{matrix} | + 1 | \begin{matrix} 1 & 0 \\ 1 & 1 \end{matrix} | \\ = 0 - 1 (0 - 1) + 1 (1 - 0) \\ = 1 + 1 = 2 \neq 0 (n o n - singular m a t r i x) \\ N o w, c o - f a c t o r s, \\ a_{11} = + | \begin{matrix} 0 & 1 \\ 1 & 0 \end{matrix} | = - 1, a_{12} = - | \begin{matrix} 1 & 1 \\ 1 & 0 \end{matrix} | = 1, a_{13} = + | \begin{matrix} 1 & 0 \\ 1 & 1 \end{matrix} | = 1 \\ a_{21} = - | \begin{matrix} 1 & 1 \\ 1 & 0 \end{matrix} | = 1, a_{22} = + | \begin{matrix} 0 & 1 \\ 1 & 0 \end{matrix} | = - 1, a_{23} = - | \begin{matrix} 0 & 1 \\ 1 & 1 \end{matrix} | = 1 \\ a_{31} = + | \begin{matrix} 1 & 1 \\ 0 & 1 \end{matrix} | = 1, a_{32} = - | \begin{matrix} 0 & 1 \\ 1 & 1 \end{matrix} | = 1, a_{33} = + | \begin{matrix} 0 & 1 \\ 1 & 0 \end{matrix} | = - 1 \\ A d j (A) = {[\begin{matrix} - 1 & 1 & 1 \\ 1 & - 1 & 1 \\ 1 & 1 & - 1 \end{matrix}]}^{'} = [\begin{matrix} - 1 & 1 & 1 \\ 1 & - 1 & 1 \\ 1 & 1 & - 1 \end{matrix}] \\ ∴ A^{- 1} = \frac{1}{| A |} A d j (A) = \frac{1}{2} [\begin{matrix} - 1 & 1 & 1 \\ 1 & - 1 & 1 \\ 1 & 1 & - 1 \end{matrix}] \\ N o w, A^{2} = A . A = [\begin{matrix} 0 & 1 & 1 \\ 1 & 0 & 1 \\ 1 & 1 & 0 \end{matrix}] [\begin{matrix} 0 & 1 & 1 \\ 1 & 0 & 1 \\ 1 & 1 & 0 \end{matrix}] \\ = [\begin{matrix} 0 + 1 + 1 & 0 + 0 + 1 & 0 + 1 + 0 \\ 0 + 0 + 1 & 1 + 0 + 1 & 1 + 0 + 0 \\ 0 + 1 + 0 & 1 + 0 + 0 & 1 + 1 + 0 \end{matrix}] = [\begin{matrix} 2 & 1 & 1 \\ 1 & 2 & 1 \\ 1 & 1 & 2 \end{matrix}] \\ H e n c e, A^{2} = [\begin{matrix} 2 & 1 & 1 \\ 1 & 2 & 1 \\ 1 & 1 & 2 \end{matrix}] \\ N o w, w e h a v e t o p r o v e t h a t A^{- 1} = \frac{A^{2} - 3 I}{2} \\ R . H . S . = \frac{[\begin{matrix} 2 & 1 & 1 \\ 1 & 2 & 1 \\ 1 & 1 & 2 \end{matrix}] - 3 [\begin{matrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{matrix}]}{2} \\ = \frac{[\begin{matrix} 2 & 1 & 1 \\ 1 & 2 & 1 \\ 1 & 1 & 2 \end{matrix}] - [\begin{matrix} 3 & 0 & 0 \\ 0 & 3 & 0 \\ 0 & 0 & 3 \end{matrix}]}{2} = \frac{1}{2} [\begin{matrix} - 1 & 1 & 1 \\ 1 & - 1 & 1 \\ 1 & 1 & - 1 \end{matrix}] \\ = A^{- 1} = L . H . S . \\ H e n c e, p r o v e d . \end{array}$

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

Kindly Consider the following

Show that the $? A B C$ is an isosceles triangle if the determinant

$[\begin{matrix} 1 & 1 & 1 \\ 1 + c o s A & 1 + c o s B & 1 + c o s C \\ c o s^{2} A + c o s A & c o s^{2} B + c o s B & c o s^{2} C + c o s C \end{matrix}] = 0$

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Vishal Baghel | Contributor-Level 10

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

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

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

Kindly Consider the following

Show that the points $(a + 5, a - 4)$ , $(a - 2, a + 3)$ , and $(a, a)$ do not lie on a straight line for any value of $a$ .

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

$\begin{array}{l} I f t h e g i v e n points l i e o n a s t r a i g h t l i n e, t h e n t h e a r e a o f t h e t r i a n g l e f o r m e d b y j o i n i n g t h e \\ points p a i r w i s e i s z e r o . \\ S 0, | \begin{matrix} a + 5 & a - 4 & 1 \\ a - 2 & a + 3 & 1 \\ a & a & 1 \end{matrix} | \\ R_{1} \to R_{1} - R_{2}, R_{2} \to R_{2} - R_{3} \\ \Rightarrow | \begin{matrix} 7 & - 7 & 0 \\ - 2 & 3 & 0 \\ a & a & 1 \end{matrix} | \\ E x p a n d i n g a l o n g C_{3} \\ \Rightarrow 1 . | \begin{matrix} 7 & - 7 \\ - 2 & 3 \end{matrix} | = 21 - 14 = 7 u n i t s \\ A s 7 \neq 0 . H e n c e, t h e t h r e e points d o n o t l i e o n a s t r a i g h t l i n e f o r a n y v a l u e o f a . \end{array}$

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

Kindly Consider the following

If $a_{1}, a_{2}, a_{3}, . . ., a_{r}$ are in G.P., then prove that the determinant $[\begin{matrix} a_{r + 1} & a_{r + 5} & a_{r + 9} \\ a_{r + 7} & a_{r + 11} & a_{r + 15} \\ a_{r + 11} & a_{r + 17} & a_{r + 21} \end{matrix}]$ is

independent of $r$ .

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Vishal Baghel | Contributor-Level 10

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

$\begin{array}{l} I f a_{1}, a_{2}, a_{3}, \dots a_{r} b e t h e t e r m s o f G . P ., t h e n \\ a_{n} = A R^{n - 1} (w h e r e A i s t h e f i r s t t e r m a n d R i s t h e c o m m o n r a t i o o f t h e G . P .) \\ ∴ a_{r + 1} = A R^{r + 1 - 1} = A R^{r}; a_{r + 5} = A R^{r + 5 - 1} = A R^{r + 4} \\ a_{r + 9} = A R^{r + 9 - 1} = A R^{r + 8}; a_{r + 7} = A R^{r + 7 - 1} = A R^{r + 6} \\ a_{r + 11} = A R^{r + 11 - 1} = A R^{r + 10}; a_{r + 15} = A R^{r + 15 - 1} = A R^{r + 14} \\ a_{r + 17} = A R^{r + 17 - 1} = A R^{r + 16}; a_{r + 21} = A R^{r + 21 - 1} = A R^{r + 20} \\ ∴ T h e determinant b e c o m e s \\ | \begin{matrix} A R^{r} & A R^{r + 4} & A R^{r + 8} \\ A R^{r + 6} & A R^{r + 10} & A R^{r + 14} \\ A R^{r + 10} & A R^{r + 16} & A R^{r + 20} \end{matrix} | \\ T a k i n g A R^{r}, A R^{r + 6} a n d A R^{r + 10} c o m m o n f r o m R_{1,} R_{2,} a n d R_{3} r e s p e c t i v e l y . \\ A R^{r} . A R^{r + 6} . A R^{r + 10} | \begin{matrix} 1 & R^{4} & R^{8} \\ 1 & R^{4} & R^{8} \\ 1 & R^{6} & R^{10} \end{matrix} | \\ = A R^{r} . A R^{r + 6} . A R^{r + 10} | 0 | [? R_{1} a n d R_{2} a r e i d e n t i c a l r o w s] \\ = 0 \\ H e n c e, t h e g i v e n determinant i s i n d e p e n d e n t o f r . \end{array}$

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

Kindly Consider the following

If $[\begin{matrix} 4 - x & 4 + x & 4 + x \\ 4 + x & 4 - x & 4 + x \\ 4 + x & 4 + x & 4 - x \end{matrix}] ? = 0,$ then find values of $x$ .

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Vishal Baghel | Contributor-Level 10

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

$\begin{array}{l} L e t A = [\begin{matrix} 4 - x & 4 + x & 4 + x \\ 4 + x & 4 - x & 4 + x \\ 4 + x & 4 + x & 4 - x \end{matrix}] = 0 \\ | A | = | \begin{matrix} 4 - x & 4 + x & 4 + x \\ 4 + x & 4 - x & 4 + x \\ 4 + x & 4 + x & 4 - x \end{matrix} | = 0 \\ R_{1} \to R_{1} + R_{2} + R_{3} \\ \Rightarrow | \begin{matrix} 12 + x & 12 + x & 12 + x \\ 4 + x & 4 - x & 4 + x \\ 4 + x & 4 + x & 4 - x \end{matrix} | = 0 \\ T a k i n g (12 + x) c o m m o n f r o m R_{1} \\ \Rightarrow (12 + x) | \begin{matrix} 1 & 1 & 1 \\ 4 + x & 4 - x & 4 + x \\ 4 + x & 4 + x & 4 - x \end{matrix} | = 0 \\ C_{1} \to C_{1} - C_{2}, C_{2} \to C_{2} - C_{3} \\ \Rightarrow (12 + x) | \begin{matrix} 0 & 0 & 1 \\ 2 x & - 2 x & 4 + x \\ 0 & 2 x & 4 - x \end{matrix} | = 0 \\ E x p a n d i n g a l o n g R_{1} \\ \Rightarrow (12 + x) [1 . | \begin{matrix} 2 x & - 2 x \\ 0 & 2 x \end{matrix} |] = 0 \\ \Rightarrow (12 + x) (4 x^{2} - 0) = 0 \Rightarrow 12 + x = 0 o r 4 x^{2} = 0 \\ \Rightarrow x = - 12 o r x = 0 \end{array}$

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

Kindly Consider the following

Find the value of $θ$ satisfying

$[\begin{matrix} 1 & 1 & s i n 3 θ \\ - 4 & 3 & c o s 2 θ \\ 7 & - 7 & - 2 \end{matrix}] = 0$

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Vishal Baghel | Contributor-Level 10

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

$\begin{array}{l} L e t A = [\begin{matrix} 1 & 1 & s i n 3 θ \\ - 4 & 3 & c o s 2 θ \\ 7 & - 7 & - 2 \end{matrix}] = 0 \\ | A | = | \begin{matrix} 1 & 1 & s i n 3 θ \\ - 4 & 3 & c o s 2 θ \\ 7 & - 7 & - 2 \end{matrix} | = 0 \\ C_{1} \to C_{1} - C_{2} \\ \Rightarrow = | \begin{matrix} 0 & 1 & s i n 3 θ \\ - 7 & 3 & c o s 2 θ \\ 14 & - 7 & - 2 \end{matrix} | = 0 \\ T a k i n g 7 c o m m o n f r o m C_{1} \\ \Rightarrow = 7 | \begin{matrix} 0 & 1 & s i n 3 θ \\ - 1 & 3 & c o s 2 θ \\ 2 & - 7 & - 2 \end{matrix} | = 0 \\ \Rightarrow = | \begin{matrix} 0 & 1 & s i n 3 θ \\ - 1 & 3 & c o s 2 θ \\ 2 & - 7 & - 2 \end{matrix} | = 0 \\ E x p a n d i n g a l o n g C_{1} \\ \Rightarrow 1 | \begin{matrix} 1 & s i n 3 θ \\ - 7 & - 2 \end{matrix} | + 2 | \begin{matrix} 1 & s i n 3 θ \\ 3 & c o s 2 θ \end{matrix} | = 0 \\ \Rightarrow - 2 + 7 s i n 3 θ + 2 (c o s 2 θ - 3 s i n 3 θ) = 0 \\ \Rightarrow - 2 + 7 s i n 3 θ + 2 c o s 2 θ - 6 s i n 3 θ = 0 \\ \Rightarrow - 2 + 2 c o s 2 θ + s i n 3 θ = 0 \\ \Rightarrow - 2 + 2 (1 - 2 s i n^{2} θ) + 3 s i n θ - 4 s i n^{3} θ = 0 \\ \Rightarrow - 2 + 2 - 4 s i n^{2} θ + 3 s i n θ - 4 s i n^{3} θ = 0 \\ \Rightarrow - 4 s i n^{3} θ - 4 s i n^{2} θ + 3 s i n θ = 0 \\ \Rightarrow - s i n θ (4 s i n^{2} θ + 4 s i n θ - 3) = 0 \\ - s i n θ = 0 o r 4 s i n^{2} θ + 4 s i n θ - 3 = 0 \\ ∴ θ = n π o r 4 s i n^{2} θ + 6 s i n θ - 2 s i n θ - 3 = 0 w h e n n \in I \\ \Rightarrow 2 s i n θ (2 s i n θ + 3) - 1 (2 s i n θ + 3) = 0 \\ \Rightarrow (2 s i n θ + 3) (2 s i n θ - 1) = 0 \\ \Rightarrow &thinsp \end{array}$

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Ncert Solutions Maths class 12th

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If $[\begin{matrix} 2 x & 5 \\ 8 & x \end{matrix}] = [\begin{matrix} 6 & - 2 \\ 7 & 3 \end{matrix}]$ then the value of $x$ is

(A) 3

(B) ±3

(C) ±6

(D) 6

If $x = s i n t$ and $y = s i n p t$ , prove that $(1 - x^{2}) \frac{d^{2} y}{d x^{2}} - x \frac{d y}{d x} + p^{2} y = 0$

If $x^{m} y^{n} = {(x + y)}^{m + n}$ , prove that

(i) $\frac{d y}{d x} = \frac{y}{x}$

(ii) $\frac{d^{2} y}{d x^{2}} = 0$ .

Find the values of $p$ and $q$ so that $f (x) = {\begin{matrix} x^{2} + 3 x + p, & x \leq 1 \\ q x + 2, & x > 1 \end{matrix}$ is differentiable at $x = 1$ .

Kindly Consider the following

Find $A^{- 1}$ if $A = [\begin{matrix} 0 & 1 & 1 \\ 1 & 0 & 1 \\ 1 & 1 & 0 \end{matrix}]$ and show that $A^{- 1} = \frac{3 I - A^{2}}{2}$

Kindly Consider the following

Show that the $? A B C$ is an isosceles triangle if the determinant

$[\begin{matrix} 1 & 1 & 1 \\ 1 + c o s A & 1 + c o s B & 1 + c o s C \\ c o s^{2} A + c o s A & c o s^{2} B + c o s B & c o s^{2} C + c o s C \end{matrix}] = 0$

Kindly Consider the following

Show that the points $(a + 5, a - 4)$ , $(a - 2, a + 3)$ , and $(a, a)$ do not lie on a straight line for any value of $a$ .

Kindly Consider the following

If $a_{1}, a_{2}, a_{3}, . . ., a_{r}$ are in G.P., then prove that the determinant $[\begin{matrix} a_{r + 1} & a_{r + 5} & a_{r + 9} \\ a_{r + 7} & a_{r + 11} & a_{r + 15} \\ a_{r + 11} & a_{r + 17} & a_{r + 21} \end{matrix}]$ is

independent of $r$ .

Kindly Consider the following

If $[\begin{matrix} 4 - x & 4 + x & 4 + x \\ 4 + x & 4 - x & 4 + x \\ 4 + x & 4 + x & 4 - x \end{matrix}] ? = 0,$ then find values of $x$ .

Kindly Consider the following

Find the value of $θ$ satisfying

$[\begin{matrix} 1 & 1 & s i n 3 θ \\ - 4 & 3 & c o s 2 θ \\ 7 & - 7 & - 2 \end{matrix}] = 0$