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Class 12th Maths Integrals

Kindly Consider the following

74. $\frac{c o s 2 x - c o s 2 \propto}{c o s x - c o s \propto}$

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Vishal Baghel

Contributor-Level 10

$\begin{array}{l} = 4 c o s (\frac{x + \propto}{2}) c o s (\frac{x - \propto}{2}) \\ = 2 [c o s (\frac{x + \propto}{2} + \frac{x - \propto}{2}) + c o s (\frac{x - \propto}{2} - \frac{x - \propto}{2}) \\ \begin{array}{l} = 2 [c o s (x) + c o s \propto] & \begin{array}{l} = 2 c o s x + 2 c o s \propto \\ = \int \frac{c o s 2 x - c o s 2 \propto}{c o s x - c o s \propto} d x \\ \begin{array}{l} = \int (2 c o s x + 2 c o s \propto) d x \end{array} \end{array} \end{array} \end{array}$

$\begin{array}{l} \begin{array}{l} \begin{array}{l} \begin{array}{l} = 2 [s i n x + x c o s \propto] + C \end{array} \end{array} \end{array} \end{array}$

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Class 12th physics ncert exemplar solution class 12th chapter one

In figure two positive charges q₂ and q₃ fixed along the y-axis, exert a net electric force in the +x-direction on a charge q₁, fixed along the x-axis. If a positive charge Q is added at (x, 0), the force on q₁

(a) Shall increase along the positive x-axis
(b) Shall decrease along the positive x-axis
(c) Shall point along the negative x-axis
(d) Shall increase but the direction changes because of the intersection of Q with q₂ and q₃

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alok kumar singh

Contributor-Level 10

This is a Multiple Choice Questions as classified in NCERT Exemplar

Answer. (a)

Explanation-The force between (q1, q2) and (q1, q3) is must be attractive to net forces act in positive direction of x . so q1 must be negative. But if we place a positive charge Q in positive direction of x then it must be attractive towards along x-axis. So it will be attractive towards positive x axis.

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New answer posted

a year ago

Class 12th Maths Integrals

Kindly Consider the following

73. $\frac{s i n^{2} x}{1 + c o s x}$

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Vishal Baghel

Contributor-Level 10

$\frac{2 x}{1 + c o s x} = \frac{{(2 s i n \frac{x}{2} \cdot c o s \frac{x}{2})}^{2}}{2 c o s^{2} \frac{x}{2}}$

$\begin{array}{l} = \frac{4 s i n^{2} \frac{x}{2} c o s^{2} \frac{x}{2}}{2 c o s^{2} \frac{x}{2}} = \frac{2 s i n^{2} \frac{x}{2}}{2} \\ = 1 - c o s x \\ = \int \frac{s i n^{2} x}{1 + c o s x} d x \\ = \int (1 - c o s x) d x \\ = x - s i n x + C \end{array}$

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Class 12th Maths Integrals

Kindly Consider the following

72. $c o s^{4} 2 x$

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Vishal Baghel

Contributor-Level 10

Kindly go through the solution

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New answer posted

a year ago

Class 12th physics ncert exemplar solution class 12th chapter one

Figure shows the electric field lines around three point charges A, B and C.

(i) Which charges are positive?
(ii) Which charge has the largest magnitude? Why?
(iii) In which region or regions of the picture could the electric field be zero? Justify your answer.
(a) Near A (b) Near B
(c) Near C (d) Nowhere

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alok kumar singh

Contributor-Level 10

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

Explanation-The electric lines of forces always starts from a positive charge and ends at a negative charge. In case of a single isolated charge, electric lines of force start from positive charge ends at infinity.

(i) Here, in the figure, the electric lines of force starts from A and C. Therefore, charges A and C must be positive.

(ii) The number of electric lines of forces starting from charge C are maximum, so C must have the largest magnitude.

(iii) From the figure we see that a neutral point exists between charges A and

Here, more number of elect

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New answer posted

a year ago

Class 12th Maths Integrals

Kindly Consider the following

71. $s i n^{4} x$

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Vishal Baghel

Contributor-Level 10

$\begin{array}{l} s i n^{4} x = s i n^{2} x . s i n^{2} x \\ = (\frac{1 - c o s 2 x}{2}) (\frac{1 - c o s 2 x}{2}) \\ = \frac{1}{4} {(1 - c o s 2 x)}^{2} = \frac{1}{4} [1 + c o s^{2} 2 x - 2 c o s 2 x] \\ = \frac{1}{4} [1 + (\frac{1 + c o s 4 x}{2}) - 2 c o s 2 x] \\ = \frac{1}{4} [1 + \frac{1}{2} + \frac{1}{2} c o s 4 x - 2 c o s 2 x] \\ = \frac{1}{4} [\frac{3}{2} + \frac{1}{2} c o s 4 x - 2 c o s 2 x] \end{array}$

$\begin{array}{l} = \int s i n^{4} x d x = \frac{1}{4} \int [\frac{3}{2} + \frac{1}{2} c o s 4 x - 2 c o s 2 x] d x \\ = \frac{1}{4} [\frac{3}{2} + \frac{1}{2} (\frac{s i n 4 x}{4}) - \frac{2 s i n 2 x}{2}] + C \\ = \frac{1}{8} [3 x + \frac{s i n^{4} x}{4} - 2 s i n 2 x] + C \\ = \frac{3 x}{8} - \frac{1}{4} s i n 2 x + \frac{1}{32} s i n 4 x + C \end{array}$

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New answer posted

a year ago

Class 12th Maths Integrals

Kindly Consider the following

70. $\frac{c o s x}{1 + c o s x}$

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Vishal Baghel

Contributor-Level 10

$\begin{array}{l} \frac{c o s x}{1 + c o s x} = \frac{c o s^{2} \frac{x}{2} - s i n^{2} \frac{x}{2}}{2 c o s^{2} \frac{x}{2}} \\ = \frac{1}{2} [1 - t a n^{2} \frac{x}{2}] \\ = \int \frac{c o s x}{1 + c o s x} d x = \frac{1}{2} \int (1 - t a n^{2} \frac{x}{2}) d x \\ = \frac{1}{2} (1 - s e c^{2} \frac{x}{2} + 1) d x \\ = \frac{1}{2} (2 - s e c^{2} \frac{x}{2}) d x = \frac{1}{2} [2 x - \frac{t a n \frac{x}{2}}{\frac{1}{2}}] + C \\ = x - t a n \frac{x}{2} + C \end{array}$

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New answer posted

a year ago

Class 12th Maths Integrals

Kindly Consider the following

69. $\frac{1 - c o s x}{1 + c o s x}$

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Vishal Baghel

Contributor-Level 10

$\begin{array}{l} \frac{1 - c o s x}{1 + c o s x} = \frac{2 s i n^{2} \frac{x}{2}}{2 c o s^{2} \frac{x}{2}} \\ = t a n^{2} \frac{x}{2} = s e c^{2} \frac{x}{2} - 1 \\ = \int \frac{1 - c o s x}{1 + c o s x} d x = \int (s e c^{2} \frac{x}{2} - 1) d x \\ = [\frac{t a n \frac{x}{2}}{\frac{1}{2}}] + C \\ = 2 t a n \frac{x}{2} + C \end{array}$

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New answer posted

a year ago

Class 12th Maths Integrals

Kindly Consider the following

68. $s i n 4 x s i n 8 x$

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Vishal Baghel

Contributor-Level 10

$\begin{array}{l} s i n A s i n B = \frac{- 1}{2} {c o s (A+B) - c o s (A-B)} \\ = \int s i n 4 x s i n 8 x d x = \int \frac{1}{2} {c o s (4 x - 8 x) - c o s (4 x + 8 x)} d x \\ = \frac{1}{2} \int c o s (- 4 x) - c o s 12 x d x \\ = \frac{1}{2} \int (c o s 4 x - c o s 12 x) d x \\ = \frac{1}{2} [\frac{s i n 4 x}{4} - \frac{s i n 12 x}{12}] + C \end{array}$

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New answer posted

a year ago

Class 12th physics ncert exemplar solution class 12th chapter one

Two charges q and -3q are placed fixed on x-axis separated by distance d. where should a third charge 2q be placed such that it will not experience any force?

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alok kumar singh

Contributor-Level 10

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

Answer = $\frac{d}{2}$ (1+ $\sqrt[]{3}$ ) towards left.

Explanation -lets consider 2q charge is placed in between q and -3q, here it will definitely experience some force q charge repel 2q charge and -3q charge will attract 2q charge. So 2q charge will move towards -3q charge.

Now lets consider it to the left of q at some distance x, here the force experience by q is repulsive and force experience by -3q is attractive . so they cancel out each other and no net force is experience by 2q.

Thus, force of attraction by -3q = force of repulsion by q

$?$ k 2q q/x² = k 2q 3q/ (x+d)²

$?$ (x+d)² = 3x

...more

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

Answer = $\frac{d}{2}$ (1+ $\sqrt[]{3}$ ) towards left.

Thus, force of attraction by -3q = force of repulsion by q

$?$ k 2q q/x² = k 2q 3q/ (x+d)²

$?$ (x+d)² = 3x²

$?$ x² + d² + 2xd = 3x²

$?$ 2x² - 2dx – d² = 0

$?$ x = $\frac{d}{2}$ + $\frac{\sqrt[]{3} d}{2}$

$?$ x= $\frac{d}{2}$ (1+ $\sqrt[]{3}$ ) towards left.

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