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7 months ago

Class 12th Maths Integrals

Kindly Consider the following

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

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Vishal Baghel

Contributor-Level 10

Putting x² = t such that 2xdx = dt

$\begin{array}{l} \int \frac{2 x d x}{(x^{2} + 1) (x^{2} + 3)} = \int \frac{d t}{(t + 1) (t + 3)} = \frac{1}{2} \int \frac{2 d t}{(t + 1) (t + 3)} \\ = \frac{1}{2} \int \frac{(t + 3) - (t + 1) d t}{(t + 1) (t + 3)} \\ = \frac{1}{2} {\int \frac{d t}{t + 1} - \frac{d t}{t + 3}} \\ = \frac{1}{2} {l o g | t + 1 | - l o g | t + 3 |} + c \\ = \frac{1}{2} l o g | \frac{t + 1}{t + 3} | + c \end{array}$

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New answer posted

7 months ago

Class 12th Maths Integrals

Kindly Consider the following

128. $\frac{(x^{2} + 1) (x^{2} + 2)}{(x^{2} + 3) (x^{2} + 4)}$

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Vishal Baghel

Contributor-Level 10

Kindly go through the solution

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7 months ago

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

Two cells of emfs approximately 5 V and 10 V are to be accurately compared using a potentiometer of length 400 cm.4.
(a) The battery that runs the potentiometer should have voltage of 8 V.
(b) The battery of potentiometer can have a voltage of 15 V and R adjusted so that the potential drop across the wire slightly exceeds 10 V.

(c) The first portion of 50 cm of wire itself should have a potential drop of 10 V.
(d) Potentiometer is usually used for comparing resistances and not voltages.

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alok kumar singh

Contributor-Level 10

This is a Multiple Choice Questions as classified in NCERT Exemplar

Answer- (b)

Explanation-The potential drop along the wires of potentiometer should be greater than emfs of cells.
In a potentiometer experiment, the emf of a cell can be measured if the potential drop along the potentiometer wire is more than the emf of the cell to be determined. Here, values of emfs of two cells are given as 5 V and 10 V, therefore, the potential drop along the potentiometer wire must be more than 10 V.

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New answer posted

7 months ago

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

7 months ago

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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New answer posted

7 months ago

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

7 months 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

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