Category Archives: IIT JEE

A horizontal force F is applied at the center of mass …

August 26, 2023 Manish Verma Leave a comment

A horizontal force F is applied at the center of mass of a cylindrical object of mass m and radius R, perpendicular to its axis as shown in the figure. The coefficient of friction between the object and the ground is μ. The center of mass of the object has an acceleration a. The acceleration due to gravity is g. Given that the object rolls without slipping, which of the following statement(s) is(are) correct?

(A) For the same F, the value of a does not depend on whether the cylinder is solid or hollow
(B) For a solid cylinder, the maximum possible value of a is 2μg
(C) The magnitude of the frictional force on the object due to the ground is always μmg
(D) For a thin-walled hollow cylinder, $𝑎 = \frac {𝐹}{2𝑚}$ Continue reading A horizontal force F is applied at the center of mass … →

IIT JEE

$f(x) = \frac{{{x^2} – 3x – 6}}{{{x^2} + 2x + 4}}$
Domain/Range/Onto/Into

August 25, 2023 Manish Verma Leave a comment

Let $f:R \to R$ be defined by $f(x) = \frac{{{x^2} – 3x – 6}}{{{x^2} + 2x + 4}}$. Then which of the following statements is(are) correct?

(A) f is onto
(B) Range of f is $\left[ { – \frac{3}{2},2} \right]$
(C) f is into
(D) Domain of f is [-4, 0] Continue reading $f(x) = \frac{{{x^2} – 3x – 6}}{{{x^2} + 2x + 4}}$

Domain/Range/Onto/Into →

IIT JEE

$I = \int\limits_{ – \pi }^\pi {\frac{{{{\sin }^2}x}}{{1 + {e^x}}}dx} = ?$

August 15, 2023 Manish Verma Leave a comment

Let $x=-t$, so $dx=-dt$

$I = \int\limits_\pi ^{ – \pi } { – \frac{{{{\sin }^2}t}}{{1 + {e^{ – t}}}}dt} = \int\limits_{ – \pi }^\pi {\frac{{{e^t}{{\sin }^2}t}}{{1 + {e^t}}}dt} = \int\limits_{ – \pi }^\pi {\frac{{{e^x}{{\sin }^2}x}}{{1 + {e^x}}}dx} $

$I + I = \int\limits_{ – \pi }^\pi {\frac{{{{\sin }^2}x}}{{1 + {e^x}}}dx} + \int\limits_{ – \pi }^\pi {\frac{{{e^x}{{\sin }^2}x}}{{1 + {e^x}}}dx} $

$ \Rightarrow 2I = \int\limits_{ – \pi }^\pi {\frac{{(1 + {e^x}){{\sin }^2}x}}{{1 + {e^x}}}dx} = \int\limits_{ – \pi }^\pi {{{\sin }^2}xdx} $

$ \Rightarrow 2I = \int\limits_0^\pi {2{{\sin }^2}xdx} = \int\limits_0^\pi {(1 – \cos 2x)dx} $

$ \Rightarrow 2I = \left. {\left( {x – \frac{{\sin 2x}}{2}} \right)} \right|_0^\pi = \pi $

$ \Rightarrow I = \frac{\pi }{2}$

IIT JEE

$I = \int\limits_0^1 {\frac{{\ln (1 + x)}}{{1 + {x^2}}}dx} = ?$

August 14, 2023 Manish Verma Leave a comment

Let, $x = \tan \theta $

$ \Rightarrow dx = {\sec ^2}\theta d\theta $

$I = \int\limits_0^{\pi /4} {\frac{{\ln (1 + \tan \theta )}}{{1 + {{\tan }^2}\theta }}{{\sec }^2}\theta d\theta } $

$ = \int\limits_0^{\pi /4} {\ln \left( {\frac{{\sin \theta + \cos \theta }}{{\cos \theta }}} \right)d\theta } $

$ = \int\limits_0^{\pi /4} {\ln (\sin \theta + \cos \theta )d\theta – \int\limits_0^{\pi /4} {\ln \cos \theta d\theta } } $

$ = \int\limits_0^{\pi /4} {\ln \left\{ {\sqrt 2 \left( {\cos \theta .\frac{1}{{\sqrt 2 }} + \sin \theta .\frac{1}{{\sqrt 2 }}} \right)} \right\}d\theta – \int\limits_0^{\pi /4} {\ln \cos \theta d\theta } } $

$ = \int\limits_0^{\pi /4} {\ln \left\{ {\sqrt 2 \left( {\cos \theta .\cos \frac{\pi }{4} + \sin \theta .\sin \frac{\pi }{4}} \right)} \right\}d\theta – \int\limits_0^{\pi /4} {\ln \cos \theta d\theta } } $

$ = \int\limits_0^{\pi /4} {\ln \left\{ {\sqrt 2 \cos \left( {\theta – \frac{\pi }{4}} \right)} \right\}d\theta – \int\limits_0^{\pi /4} {\ln \cos \theta d\theta } } $

$ = \int\limits_0^{\pi /4} {\ln \sqrt 2 d\theta + } \int\limits_0^{\pi /4} {\ln \cos \left( {\theta – \frac{\pi }{4}} \right)d\theta – \int\limits_0^{\pi /4} {\ln \cos \theta d\theta } } $

Let, $\theta – \frac{\pi }{4} = – \phi $ for the second integral.

$I = \frac{\pi }{4}\ln \sqrt 2 – \int\limits_{\pi /4}^0 {\ln \cos ( – \phi )d\phi – \int\limits_0^{\pi /4} {\ln \cos \theta d\theta } } $

$ = \frac{\pi }{4}\ln \sqrt 2 + \int\limits_0^{\pi /4} {\ln \cos \phi d\phi – \int\limits_0^{\pi /4} {\ln \cos \theta d\theta } } $

$ = \frac{\pi }{4}\ln \sqrt 2 = \frac{\pi }{8}\ln 2$

IIT JEE

A projectile is thrown from a point O ….

August 11, 2023 Manish Verma Leave a comment

A projectile is thrown from a point O on the ground at an angle 45° from the vertical and with a speed 5√2 m/s. The projectile at the highest point of its trajectory splits into two equal parts. One part falls vertically down to the ground, 0.5 s after the splitting. The other part, t seconds after the splitting, falls to the ground at a distance x meters from the point O. The acceleration due to gravity $g = 10 m/s^2$.

Q.1 The value of t is ___ .
Q.2 The value of x is ___ .

Solution

The part that falls vertically down to the ground has initial speed = 0. So, it falls freely taking 0.5 s to reach the ground. Continue reading A projectile is thrown from a point O …. →

IIT JEE

In the circuit shown, the switch S is connected to …

August 10, 2023 Manish Verma Leave a comment

In the circuit shown, the switch S is connected to position P for a long time so that the charge on the capacitor becomes $q_1$ μC. Then S is switched to position Q. After a long time, the charge on the capacitor is $q_2$ μC.