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New answer posted

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

If $A, B a n d C$ are angles of a triangle, then the determinant $[\begin{matrix} - 1 & c o s C & c o s B \\ c o s C & - 1 & c o s A \\ c o s B & c o s A & - 1 \end{matrix}]$ is equal to

(A) 0

(B) –1

(C) 1

(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:

$\begin{array}{l} L e t Δ = | \begin{matrix} - 1 & c o s C & c o s B \\ c o s C & - 1 & c o s A \\ c o s B & c o s A & - 1 \end{matrix} | \\ C_{1} \to a C_{1} + b C_{2} + c C_{3} \\ \Rightarrow | \begin{matrix} - a + b c o s C + c c o s B & c o s C & c o s B \\ a c o s C - b + c c o s A & - 1 & c o s A \\ a c o s B + b c o s A - C & c o s A & - 1 \end{matrix} | \\ \Rightarrow | \begin{matrix} - a + a & c o s C & c o s B \\ - b + b & - 1 & c o s A \\ - c + c & c o s A & - 1 \end{matrix} | [\begin{array}{l} ? F r o m p r o t e c t i o n f o r m u l a \\ a = b c o s C + c c o s B \\ b = a c o s C + c c o s A \\ c = b c o s A + a c o s B \end{array}] \\ \Rightarrow | \begin{matrix} 0 & c o s C & c o s B \\ 0 & - 1 & c o s A \\ 0 & c o s A & - 1 \end{matrix} | = 0 \\ H e n c e, t h e c o r r e c t o p t i o n i s (a) . \end{array}$

0 0

New answer posted

a year ago

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

Examine the continuity of the function
$f (x) = x^{3} + 2 x^{2} - 1$ at $x = 1$

0 Follower 4 Views

alok kumar singh

Contributor-Level 10

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

Sol:

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

0 0

New answer posted

a year ago

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

The number of distinct real roots of $[\begin{matrix} s i n x & c o s x & c o s x \\ c o s x & s i n x & c o s x \\ c o s x & c o s x & s i n x \end{matrix}] = 0$ in the interval $- \frac{π}{4} \leq x \leq \frac{π}{4}$ is

(A) 0

(B) 2

(C) 1

(D) 3

0 Follower 4 Views

Vishal Baghel

Contributor-Level 10

This is a Objective Type Questions as classified in NCERT Exemplar

Sol:

$\begin{array}{l} G i v e n t h a t | \begin{matrix} s i n x & c o s x & c o s x \\ c o s x & s i n x & c o s x \\ c o s x & c o s x & s i n x \end{matrix} | = 0 \\ C_{1} \to C_{1} + C_{2} + C_{3} \\ \Rightarrow | \begin{matrix} 2 c o s x + s i n x & c o s x & c o s x \\ 2 c o s x + s i n x & s i n x & c o s x \\ 2 c o s x + s i n x & c o s x & s i n x \end{matrix} | = 0 (T a k i n g 2 c o s x + s i n x c o m m o n f r o m C_{1}) \\ \Rightarrow (2 c o s x + s i n x) | \begin{matrix} 1 & c o s x & c o s x \\ 1 & s i n x & c o s x \\ 1 & c o s x & s i n x \end{matrix} | = 0 \\ R_{1} \to R_{1} - R_{2}, R_{2} \to R_{2} - R_{3} \\ \Rightarrow (2 c o s x + s i n x) | \begin{matrix} 0 & c o s x - s i n x & 0 \\ 0 & s i n x - c o s x & c o s x - s i n x \\ 1 & c o s x & s i n x \end{matrix} | = 0 \\ \Rightarrow (2 c o s x + s i n x) [1 | \begin{matrix} c o s x - s i n x & 0 \\ s i n x - c o s x & c o s x - s i n x \end{matrix} |] \\ \Rightarrow (2 c o s x + s i n x) {(c o s x - s i n x)}^{2} = 0 \\ 2 c o s x + s i n x = 0 a n d {(c o s x - s i n x)}^{2} = 0 \\ 2 + t a n x = 0 a n d (c o s x - s i n x) = 0 \\ ∴ t a n x = - 2 a n d \Rightarrow t a n x = 1 \\ B u t - \frac{π}{4} \leq x \leq \frac{π}{4} a n d \Rightarrow t a n x = t a n \frac{π}{4} \\ ∴ x = \frac{π}{4} \in [- \frac{π}{4}, \frac{π}{4}] \\ S o, x &thins \end{array}$

0 0

New answer posted

a year ago

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

The determinant $[\begin{matrix} b^{2} - a b & b - c & b c - a c \\ b - a^{2} & a - b & b^{2} - a b \\ b c - a c & c - a & a b - a^{2} \end{matrix}]$ equals

(A) $a b c (b - c) (c - a) (a - b)$

(B) $(b - c) (c - a) (a - b)$

(C) $(a + b + c) (b - c) (c - a) (a - b)$

(D) None of these

0 Follower 3 Views

Vishal Baghel

Contributor-Level 10

This is a Objective Type Questions as classified in NCERT Exemplar

Sol:

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

0 0

New answer posted

a year ago

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

The phenomena involved in the reflection of radio waves by ionosphere is similar to

(a) Reflection of light by a plane mirror

(b) Total internal reflection of light in air during a mirage

(c) Dispersion of light by water molecules during the formation of a rainbow

(d) Scattering of light by the particles of air

0 Follower 5 Views

Payal Gupta

Contributor-Level 10

This is a multiple choice answer as classified in NCERT Exemplar

(b) The phenomenon involved in the reflection of radio waves by ionosphere is similar to total internal reflection of light in air during a mirage i.e., angle of incidence is greater than critical angle.

0 0

New answer posted

a year ago

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

Kindly consider the following:

0 Follower 3 Views

alok kumar singh

Contributor-Level 10

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

Sol:

0 0

New answer posted

a year ago

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

The area of a triangle with vertices $(- 3, 0)$ , $(3, 0)$ , and $(0, k)$ is 9 sq. units. The value of $k$ will be

(A) 9

(B) 3

(C) –9

(D) 6

0 Follower 17 Views

Vishal Baghel

Contributor-Level 10

This is a Objective Type Questions as classified in NCERT Exemplar

Sol:

$\begin{array}{l} A r e a o f t r i a n g l e w i t h v e r t i c e s (x_{1}, y_{1}), (x_{2}, y_{2}) a n d (x_{3}, y_{3}) w i l l b e : \\ Δ = \frac{1}{2} | \begin{matrix} x_{1} & y_{1} & 1 \\ x_{2} & y_{2} & 1 \\ x_{3} & y_{3} & 1 \end{matrix} | \Rightarrow Δ = \frac{1}{2} | \begin{matrix} - 3 & 0 & 1 \\ 3 & 0 & 1 \\ 0 & k & 1 \end{matrix} | \\ \Rightarrow = \frac{1}{2} [- 3 | \begin{matrix} 0 & 1 \\ k & 1 \end{matrix} | - 0 | \begin{matrix} 3 & 1 \\ 0 & 1 \end{matrix} | + 1 | \begin{matrix} 3 & 0 \\ 0 & k \end{matrix} |] \\ \Rightarrow = \frac{1}{2} [- 3 (- k) - 0 + 1 (3 k)] \\ \Rightarrow = \frac{1}{2} (3 k + 3 k) \Rightarrow = \frac{1}{2} (6 k) = 3 k \\ 3 k = 9 \Rightarrow k = 3 \\ H e n c e, t h e c o r r e c t o p t i o n i s (b) . \end{array}$

0 0

New answer posted

a year ago

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

The value of determinant $[\begin{matrix} a - b & b + c & a \\ b - a & c + a & b \\ c - a & a + b & c \end{matrix}]$ is

(A) $a^{3} + b^{3} + c^{3}$

(B) $3 b c$

(C) $a^{3} + b^{3} + c^{3} - 3 a b c$

(D) None of these

0 Follower 62 Views

Vishal Baghel

Contributor-Level 10

This is a Objective Type Questions as classified in NCERT Exemplar

Sol:

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

0 0

New answer posted

a year ago

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

The radius of curvature of the curved surface of a plano-convex lens is 20 cm. If the refractive index of the material of the lens be 1.5, it will

(a) Act as a convex lens only for the objects that lie on its curved side

(b) Act as a concave lens for the objects that lie on its curved side

(c) Act as a convex lens irrespective of the side on which the object lies

(d) Act as a concave lens irrespective of side on which the object lies

0 Follower 4 Views

Payal Gupta

Contributor-Level 10

This is a multiple choice answer as classified in NCERT Exemplar

0 0

New answer posted

a year ago

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

0 Follower 6 Views

Vishal Baghel

Contributor-Level 10

This is a Objective Type Questions as classified in NCERT Exemplar

Sol:

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

0 0

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