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New answer posted

7 months ago

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

Find which of the functions in Exercises 2 to 10 is continuous or discontinuous at the indicated points:

$f (x) = {\begin{matrix} 3 x + 5, & if x \geq 2 \\ x^{2}, & if x < 2 \end{matrix}$

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alok kumar singh

Contributor-Level 10

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

Sol:

$\begin{array}{l} \underset{x \to 2^{+}}{l i m} f (x) = 3 x + 5 \\ = \underset{h \to 0}{l i m} 3 (2 + h) + 5 = 11 \\ \underset{x \to 2}{l i m} f (x) = 3 x + 5 = 3 (2) + 5 = 11 \\ \underset{x \to 2^{-}}{l i m} f (x) = x^{2} = \underset{h \to 0}{l i m} {(2 - h)}^{2} \\ = \underset{h \to 0}{l i m} {(2)}^{2} + h^{2} - 4 h = {(2)}^{2} = 4 \\ Since \underset{x \to 2^{+}}{l i m} f (x) = \underset{x \to 2}{l i m} f (x) \neq \underset{x \to 2^{-}}{l i m} f (x) \\ H e n c e, f (x) i s d i s c o n t i n u o u s a t x = 2 . \end{array}$

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New answer posted

7 months ago

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

Let f(t|) $[\begin{matrix} c o s t & t & 1 \\ 2 s i n t & t & 2 t \\ s i n t & t & t \end{matrix}],$ then $\underset{t \to 0}{l i m} \frac{f (t)}{t^{2}}$ is equal to

(A) 0

(B) –1

(C) 2

(D) 3

0 Follower 7 Views

Vishal Baghel

Contributor-Level 10

This is a Objective Type Questions as classified in NCERT Exemplar

Sol:

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

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New answer posted

7 months ago

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

The direction of ray of light incident on a concave mirror is shown by PQ while directions in which the ray would travel after reflection is shown by four rays marked 1, 2, 3 and 4 (figure). Which of the four rays correctly shows the direction of reflected ray?

(a) 1

(b) 2

(c) 3

(d) 4

0 Follower 22 Views

Payal Gupta

Contributor-Level 10

This is a multiple choice answer as classified in NCERT Exemplar

(b) The PQ ray of light passes through focus F and incident on the concave mirror, after reflection, should become parallel to the principal axis and shown by ray-2 in the figure.

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New answer posted

7 months ago

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 4 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

7 months 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 $x = 1$

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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

7 months 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

7 months 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

7 months 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 4 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

7 months 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

7 months 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 10 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

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