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Class 12th Maths Integrals

Kindly Consider the following

127. $\frac{c o s x}{(1 - s i n x) (2 - s i n x)}$

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Vishal Baghel

Contributor-Level 10

Putting sin x = t cos xdx = dt

$\begin{array}{l} \int \frac{d t}{(1 - t) (2 - t)} = \int \frac{(2 - t) - (1 - t)}{(1 - t) (2 - t)} d t . \\ = \int \frac{d t}{1 - t} - \int \frac{d t}{2 - t} \\ = \frac{l o g | 1 - t |}{(- 1)} - \frac{l o g | 2 - t |}{(- 1)} + c \\ = l o g | \frac{2 - t}{1 - t} | + c \\ = l o g | \frac{2 - s i n x}{1 - s i n x} | + c \end{array}$

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New answer posted

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Class 12th Maths Integrals

Kindly Consider the following

126. $\frac{1}{x (x^{n} + 1)}$

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Vishal Baghel

Contributor-Level 10

$\frac{1}{x (x^{n} + 1)} = \frac{x^{n - 1}}{x^{n - 1} \cdot x (x^{n} + 1)} = \frac{x^{n - 1}}{x^{n} (x^{n} + 1)}$

Putting xⁿ = t Þ n xⁿ^-¹dx = dt

$\begin{array}{l} ∴ \int \frac{d x}{x (x^{n} + 1)} = \int \frac{x^{n - 1} d x}{x^{n} (x^{n} + 1)} = \frac{1}{n} \int \frac{d t}{t (t + 1)} \\ = \frac{1}{n} \int \frac{(t + 1 - t)}{t (t + 1)} d t \\ = \frac{1}{n} \int {\frac{t + 1}{t (t + 1)} - \frac{t}{t (t + 1)}} d t . \\ = \frac{1}{n} {\int \frac{d t}{t} - \int \frac{d t}{t + 1}} \\ = \frac{1}{n} {l o g t - l o g (t + 1)} +C . \\ = \frac{1}{n} l o g ? \frac{t}{t + 1} + C \\ = \frac{1}{n} l o g | \frac{x^{n}}{x^{n} + 1} | +C . \end{array}$

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

A resistance R is to be measured using a meter bridge, a student chooses the standard resistance S to be 100 Ω He finds the null point at l₁= 2.9 cm. He is told to attempt to improve the accuracy. Which of the following is a useful way?
(a) He should measure l₁, more accurately
(b) He should change S to 1000 ? and repeat the experiment
(c) He should change S to 3 ? and repeat the experiment
(d) He should have given up hope of a more accurate measurement with a meter bridge.

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alok kumar singh

Contributor-Level 10

This is a Multiple Choice Questions as classified in NCERT Exemplar

Answer – (c)

Explanation – R/S= (l₁/100-l₁)= 100 (2.9/100-2.9)= 100/97.1=2.98ohm

So he should change S to almost 3 ohm and repeat the experiment.

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New answer posted

a year ago

Class 12th Maths Integrals

Kindly Consider the following

125. $\frac{1}{x^{4} - 1}$

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Vishal Baghel

Contributor-Level 10

Kindly go through the solution

$\begin{array}{l} \RightarrowA = \frac{1}{2} \\ A n d f r o m (1) w e g e t, - \frac{- 1}{2} \\ ∴ \int \frac{d x}{y^{2} - 1} = \frac{1}{2} \int \frac{d x}{y - 1} - \frac{1}{2} \int \frac{d x}{y + 1} \\ = \frac{1}{2} \int \frac{d x}{x^{2} - 1} - \frac{1}{2} \int \frac{d x}{x^{2} - 1} {y = x^{2}} \\ = \frac{1}{2} * \frac{1}{2 \cdot 1} l o g | \frac{x - 1}{x + 1} | - \frac{1}{2} t a n^{- 1} x + C \\ = \frac{1}{4} l o g | \frac{x - 1}{x + 1} | - \frac{1}{2} t a n^{- 1} x + C \end{array}$

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New answer posted

a year ago

Class 12th Maths Integrals

Kindly Consider the following

124. $\frac{3 x - 1}{{(x + 2)}^{2}}$

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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 Solutions Class 12th Chapter Three

Two batteries of emf e1 and e2 (e2>e1) and internal resistance r1 and r2 respectively are connected in parallel as shown in fig

(a) Two equivalent emf e_eq of the cells is between e1 and e2 i.e. e1eq

(b) The equivalent emf e_eq is samller than e₁

(c) The e_eq is given by e_eq=e₁+e₂ always

(d) E_eq is independent of internal resistance r₁ and r₂

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alok kumar singh

Contributor-Level 10

This is a Multiple Choice Questions as classified in NCERT Exemplar

Answer- (a)

Explanation- eeq= $\frac{e_{2} r_{1 +} e_{2} r_{1}}{r_{1} + r_{2}}$

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New answer posted

a year ago

Class 12th Maths Integrals

Kindly Consider the following

123. $\frac{2}{(1 - x) (1 + x^{2})}$

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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 Solutions Class 12th Chapter Three

Consider a current carrying wire (current 7) in the shape of a circle. Note that as the current progresses along the wire, the direction of j (current density) changes in an exact manner, while the current I remain unaffected. The agent that is essentially responsible for is
(a) Source of emf
(b) Electric field produced by charges accumulated on the surface of wire
(c) The charges just behind a given segment of wire which push them just the right way by repulsion
(d) The charges ahead.

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alok kumar singh

Contributor-Level 10

This is a Multiple Choice Questions as classified in NCERT Exemplar

Answer- (b)

Explanation- as we know that J=E, and current density is directly proportional to electric field, so electric field produced by charges accumulated on the surface of wire.

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New answer posted

a year ago

Class 12th Maths Integrals

Kindly Consider the following

122. $\frac{x^{3} + x + 1}{x^{2} - 1}$