IIT JEE

Prove that,

$\int\limits_0^1 {\frac{1}{{x(1 + x)}}.\ln \left[ {\frac{1}{{1 + x(1 + x)}}} \right]dx} = $

$\frac{4}{3}\left( {1 + \frac{1}{{{2^2}}} + \frac{1}{{{3^2}}} + \frac{1}{{{4^2}}} + ……..\infty \, terms} \right)$

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The given integral =$ – \int\limits_{ – 1}^0 {\frac{{\ln (1 + x + {x^2})}}{{x(1 + x)}}dx} $

$ = \int\limits_{ – 1}^0 {\frac{{\ln (1 + x + {x^2})}}{{1 + x}}dx} – \int\limits_{ – 1}^0 {\frac{{\ln (1 + x + {x^2})}}{x}dx} $

$ = \int\limits_{ – 1}^0 {\frac{{\ln \left[ {1 + ( – 1 – x) + {{( – 1 – x)}^2}} \right]}}{{1 + (-1-x)}}dx} – \int\limits_{ – 1}^0 {\frac{{\ln (1 + x + {x^2})}}{x}dx} $

$ = \int\limits_{ – 1}^0 {\frac{{\ln (1 + x + {x^2})}}{{ – x}}dx} – \int\limits_{ – 1}^0 {\frac{{\ln (1 + x + {x^2})}}{x}dx} $

$ = – 2\int\limits_{ – 1}^0 {\frac{{\ln (1 + x + {x^2})}}{x}dx} $

$ = – 2\int\limits_{ – 1}^0 {\frac{{\ln \frac{{(1 – x)(1 + x + {x^2})}}{{(1 – x)}}}}{x}dx} $ Continue reading Prove that,

$\int\limits_0^1 {\frac{1}{{x(1 + x)}}.\ln \left[ {\frac{1}{{1 + x(1 + x)}}} \right]dx} = $

$\frac{4}{3}\left( {1 + \frac{1}{{{2^2}}} + \frac{1}{{{3^2}}} + \frac{1}{{{4^2}}} + ……..\infty \, terms} \right)$ →

Math

$f(x) + f\left( {\frac{1}{{1 – x}}} \right) = x,\forall x \in \mathbb{R}$

$f(x) = ?$

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$x \to 1 – \frac{1}{x}$ yields,

$f\left( {1 – \frac{1}{x}} \right) + f\left[ {\frac{1}{{1 – \left( {1 – \frac{1}{x}} \right)}}} \right] = 1 – \frac{1}{x}$

$ \Rightarrow f\left( {\frac{{x – 1}}{x}} \right) + f(x) = \frac{{x – 1}}{x}……..(A)$ Continue reading $f(x) + f\left( {\frac{1}{{1 – x}}} \right) = x,\forall x \in \mathbb{R}$

$f(x) = ?$ →

Math

Circular Permutations ~ 3 Boys & 3 Girls

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In how many ways can 3 boys and 3 girls be seated at a round table such that exactly 2 boys sit together?

Solution

Circular permutations without any restriction = (6 – 1)! = 120

Circular permutations when no 2 boys are together which is B G B G B G situation is a circle = 2! $\times$ 3! = 12

Circular permutations when all the 3 boys are together = (4 – 1)! $\times $ 3! = 36

Circular permutations for exactly 2 boys sitting together = 120 – (12 + 36) = 72

Physics

Conducting Bar Across $R_1$ & $R_2$

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A conducting bar PQ is free to slide on two parallel conducting rails as shown in figure. Two resistors $R_1$ and $R_2$ are connected across the ends of the rails. There is a uniform magnetic field B pointing into the page. An external agent pulls the bar to the left at a constant speed v. The correct statement about the directions of induced currents $I_1$ and $I_2$ flowing through $R_1$ and $R_2$ respectively is:

(A) $I_1$ is in anticlockwise direction and $I_2$ is in clockwise direction
(B) Both $I_1$ and $I_2$ are in anticlockwise direction
(C) Both $I_1$ and $I_2$ are in clockwise direction
(D) $I_1$ is in clockwise direction and $I_2$ is in anticlockwise direction

Solution

As PQ moves leftwards, the area of loop ABQP decreases causing reduction in flux. So, the direction of current through $R_1$ should be such that the flux can be increased. An upward current through $R_1$ helps to strengthen the magnetic field. Upward current through $R_1$ is the clockwise current through the left loop.

As PQ moves leftwards, the area of loop A’B’QP increases causing increase in flux. So, the direction of current through $R_2$ should be such that the flux can be decreased. An upward current through $R_2$ helps to weaken the magnetic field. Upward current through $R_2$ is the anticlockwise current through the right loop.

Hence, Option (D).

IIT JEE

2-D Collision

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A ball of mass 10 kg moving with a velocity $10\sqrt 3$ m/s along x-axis, hits another ball of mass 20 kg which is at rest kept at the origin. After collision, the first ball comes to rest and the second one disintegrates into two pieces. One of the pieces moves along y-axis at a speed of 10m/s. The second piece of mass m (say) moves at a speed of 20m/s at an angle $\theta$ (degree) with respect to the x–axis. Find the values of m and $\theta $.

Solution

Using conservation of linear momentum (COLM) along x direction,

$10\times 10 \sqrt 3 + 20 \times 0 = 10 \times 0 + m \times 20 \cos \theta + (20-m) \times 0$ Continue reading 2-D Collision →

IIT JEE

System of Masses $m_1 + m_2 + m_3 $

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Three masses $m_1 , m_2 , m_3 $ kept on a smooth horizontal surface under the influence of force F have got certain acceleration (refer figure). Find the force that mass $m_1$ exerts on mass $m_2$.

Solution

Let a be the acceleration.

Taking $m_1 + m_2 + m_3 $ as system,

$F = (m_1 + m_2 + m_3)a$

$\Rightarrow a = \frac {F}{m_1 + m_2 + m_3 }$

Let N be the force that $m_1 $ exerts on $m_2 $. Taking $m_2 + m_3$ as system,

$N = (m_2 + m_3 ) a$

$\Rightarrow N = F. \frac {m_2 + m_3 }{m_1 + m_2 + m_3 } $

Interestingly, as $m_1 \to 0$, $N \to F$

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