Q&A - Ask Doubts and Get Answers

A massless string connects two pulley of masses ' $2 \mathrm{~kg}$ ' and ' $1 \mathrm{~kg}$ ' respectively as shown in the figure.

The heavier pulley is fixed and free to rotate about its central axis while the other is free to rotate as well as translate. Find the acceleration of the lower pulley if the system was released from the rest. [Given, $g=10 \mathrm{~m} / \mathrm{s}^2$ ]
Option: 1
$\frac{4}{3} \mathrm{~gm} / \mathrm{s}^2$

Option: 2
$\frac{3}{2} \mathrm{~gm} / \mathrm{s}^2$

Option: 3
$\frac{3}{4} \mathrm{~gm} / \mathrm{s}^2$

Option: 4
$\frac{2}{3} \mathrm{~gm} / \mathrm{s}^2$

Not understanding sir

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Posted by

Raju vittal nandi

Calculate the acceleration of block $m_1$ of the following diagram. Assume all surfaces are frictionless . Here m₁ = 100kg and m₂ = 50kg

Option: 1
0.33m/s²

Option: 2
0.66m/s²

Option: 3
1m/s²

Option: 4
1.32m/s²

1.32

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Posted by

balda gayathri

No. of transition state in given figure

Option: 1
1

Option: 2
2

Option: 3
3

Option: 4
4

2

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Posted by

Mura

When cell has stalled DNA replication fork, which checkpoint should be predominantly activated?

Option: 1
G₁/S

Option: 2
G₂/M

Option: 3
M

Option: 4
Both G₂M and M

G₂/M should be activated as the cell has stalled DNA replication fork.

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Posted by

Ajit Kumar Dubey

Given : $f(x)=\left\{\begin{matrix} x, &0\leq x< \frac{1}{2} \\ \frac{1}{2}, &x=\frac{1}{2} \\ 1-x, &\frac{1}{2} < x\leq 1 \end{matrix}\right.$ and $g(x)=\left ( x-\frac{1}{2} \right )^{2},\: x\epsilon \textbf{R}.$ Then the area (in sq. units) of the region bounded b the curves, $y=f(x)$ and $y=g(x)$ between the lines, $2x=1\: \: \: and\: \: \: 2x=\sqrt{3},$ is :
Option: 1 $\frac{\sqrt{3}}{4}-\frac{1}{3}$

Option: 2 $\frac{1}{3}+\frac{\sqrt{3}}{4}$

Option: 3 $\frac{1}{2}+\frac{\sqrt{3}}{4}$

Option: 4 $\frac{1}{2}-\frac{\sqrt{3}}{4}$

Area Bounded by Curves When Intersects at More Than One Point -

Area bounded by the curves y = f(x), y = g(x) and intersect each other in the interval [a, b]

First find the point of intersection of these curves y = f(x) and y = g(x) , let the point of intersection be x = c

Area of the shaded region

$=\int_{a}^{c}\{f(x)-g(x)\} d x+\int_{c}^{b}\{g(x)-f(x)\} d x$

When two curves intersects more than one point

rea bounded by the curves y=f(x), y=g(x) and intersect each other at three points at x = a, x = b amd x = c.

To find the point of intersection, solve f(x) = g(x).

For x ∈ (a, c), f(x) > g(x) and for x ∈ (c, b), g(x) > f(x).

Area bounded by curves,

$\\\mathrm{A=} \int_{a}^{b}\left |f(x)-g(x) \right |dx\\\\\mathrm{\;\;\;\;=} \int_{a}^{c}\left ( f(x)-g(x) \right )dx+\int_{c}^{b}\left ( g(x)-f(x) \right )dx$

-

Required area = Area of trapezium ABCD - $\int_{1/2}^{\sqrt3/2}\left ( x-\frac{1}{2} \right )^2dx$

$\\=\frac{1}{2}\left(\frac{\sqrt{3}-1}{2}\right)\left(\frac{1}{2}+1-\frac{\sqrt{3}}{2}\right)-\frac{1}{3}\left(\left(x-\frac{1}{2}\right)^{3}\right)_{\frac{1}{2}}^{\frac{\sqrt{3}}{2}}\\=\frac{\sqrt3}{4}-\frac{1}{3}$

Correct Option (1)

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Posted by

avinash.dongre

Let a function $f:\left [ 0,5 \right ]\rightarrow \textbf{R}$ be continuous, $f\left ( 1 \right )=3$ and F be defined as : $F(x)=\int_{1}^{x}t^{2}g\left ( t \right )dt$ , where $g(t)=\int_{1}^{t}f\left ( u \right )du.$ Then for the function F, the point $x=1$ is :
Option: 1 a point of infection.
Option: 2 a point of local maxima.
Option: 3 a point of local minima.
Option: 4 not a critical point.

Integration as Reverse Process of Differentiation -

Integration is the reverse process of differentiation. In integration, we find the function whose differential coefficient is given.

For example,

$\\\mathrm{\frac{d}{dx}(\sin x)=\cos x}\\\\\mathrm{\frac{d}{dx}\left ( x^2 \right )=2x}\\\\\mathrm{\frac{d}{dx}\left ( e^x \right )=e^x}$

In the above example, the function cos x is the derived function of sin x. We say that sin x is an anti derivative (or an integral) of cos x. Similarly, x² and e^x are the anti derivatives (or integrals) of 2x and e^x respectively.

Also note that the derivative of a constant (C) is zero. So we can write the above examples as:

$\\\mathrm{\frac{d}{dx}(\sin x+c)=\cos x}\\\\\mathrm{\frac{d}{dx}\left ( x^2+c \right )=2x}\\\\\mathrm{\frac{d}{dx}\left ( e^x +c\right )=e^x}$

Thus, anti derivatives (or integrals) of the above functions are not unique. Actually, there exist infinitely many anti derivatives of each of these functions which can be obtained by selecting C arbitrarily from the set of real numbers.

For this reason C is referred to as arbitrary constant. In fact, C is the parameter by varying which one gets different anti derivatives (or integrals) of the given function.

If the function F(x) is an antiderivative of f(x), then the expression F(x) + C is the indefinite integral of the function f(x) and is denoted by the symbol ∫ f(x) dx.

By definition,

$\\\mathrm{\int f(x)dx=F(x)+c,\;\;\;where\;\;F'(x)=f(x)\;\;and\;\;'c' \;is\;constant.}$

-

Maxima and Minima of a Function -

Maxima and Minima of a Function

Let y = f(x) be a real function defined at x = a. Then the function f(x) is said to have a maximum value at x = a, if f(x) ≤ f(a) ∀ a ≥∈ R.

And also the function f(x) is said to have a minimum value at x = a, if f(x) ≥ f(a) ∀ a ∈ R

Concept of Local Maxima and Local Minima

The function f(x) is said to have a maximum (or we say that f(x) attains a maximum) at a point ‘a’ if the value of f(x) at ‘a’ is greater than its values for all x in a small neighborhood of ‘a’ .

In other words, f(x) has a maximum at x = ‘a’, if f(a + h) ≤ f(a) and f(a - h) ≤ f(a), where h ≥ 0 (very small quantity).

The function f(x) is said to have a minimum (or we say that f(x) attains a minimum) at a point ‘b’ if the value of f(x) at ‘b’ is less than its values for all x in a small neighborhood of ‘b’ .

In other words, f(x) has a maximum at x = ‘b’, if f(a + h) ≥ f(a) and f(a - h) ≥ f(a), where h ≥ 0 (very small quantity).

-

$\\ \begin{aligned} &\mathrm{F}^{\prime}(\mathrm{x})=\mathrm{x}^{2} \mathrm{g}(\mathrm{x})\\ &\Rightarrow \mathrm{F}^{\prime}(1)=1.\;\;g(1)=0\;\;\;\;\;\ldots(1)\;\;\;\;\;\;\;\;(\because\;g(1)=0)\\ &\text { Now }(\mathrm{x})=2 \mathrm{xg}(\mathrm{x})+\mathrm{x}^{2} \mathrm{g}^{\prime}(\mathrm{x}) \end{aligned}$

$\\ \begin{aligned} &\Rightarrow F^{\prime \prime}(x)=2 x g(x)+x^{2} f(x)\;\;\;\;(\because\;g'(x)=f(x))\\ &\Rightarrow F^{\prime \prime}(1)=0+1 \times 3\\ &\Rightarrow F^{\prime \prime}(1)=3\;\;\;\;\;\;\;\;\;\;\ldots (2) \end{aligned}\\\text{From (1) and (2) F(x) has local minimum at x = 1 }$

Correct Option (1)

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Posted by

avinash.dongre

In the expansion of $\left ( \frac{x}{cos\theta }+\frac{1}{x\sin \theta } \right )^{16}$ , if $L_{1}$ is the least value of the term independent of $x$ when $\frac{\pi }{8}\leq \theta \leq \frac{\pi }{4}$ and $L_{2}$ is the least value of the term independent of $x$ when $\frac{\pi }{16}\leq \theta \leq \frac{\pi }{8}$ , then the ratio $L_{2}:L_{1}$ is equal to :
Option: 1 $16:1$
Option: 2 $8:1$
Option: 3 $1:8$
Option: 4 $1:16$

General Term of Binomial Expansion $\left(T_{r+1}\right)^{\mathrm{th}} \text { term is called as general term in }(x+y)^{n}\;\text{and general term is given by}$

$\mathrm{T}_{\mathrm{r}+1}=^{\mathrm{n}} \mathrm{C}_{\mathrm{r}} \mathrm{\;x}^{\mathrm{n}-\mathrm{r}} \cdot \mathrm{y}^{\mathrm{r}}$

Term independent of x: It means term containing x⁰,

Now,

$\\\mathrm{T}_{\mathrm{r}+1}=^{16} \mathrm{C}_{\mathrm{r}}\left(\frac{\mathrm{x}}{\cos \theta}\right)^{16-\mathrm{r}}\left(\frac{1}{\mathrm{x} \sin \theta}\right)^{\mathrm{r}}\\\text{for r = 8 term is free from 'x' }\\\begin{aligned} &\mathrm{T}_{9}=^{16} \mathrm{C}_{8} \frac{1}{\sin ^{8} \theta \cos ^{8} \theta}\\ &\mathrm{T}_{9}=^{16} \mathrm{C}_{8} \frac{2^{8}}{(\sin 2 \theta)^{8}}\\ &\text { in } \theta \in\left[\frac{\pi}{8}, \frac{\pi}{4}\right], L_{1}=^{16} \mathrm{C}_{8} 2^{8} \end{aligned}$

$\\\because \text{Min value of L}_1\;\text{at }\theta=\pi/4\\\text { in } \theta \in\left[\frac{\pi}{16}, \frac{\pi}{8}\right], L_{2}=16 \mathrm{C}_{8} \frac{2^{8}}{\left(\frac{1}{\sqrt{2}}\right)^{8}}=^{16} \mathrm{C}_{8} \cdot 2^{8} \cdot 2^{4}\\\\\because \text{Min value of L}_2\;\text{at }\theta=\pi/8\\\frac{L_{2}}{L_{1}}=\frac{16 \mathrm{C}_{8} \cdot 2^{8} 2^{4}}{^{16} \mathrm{C}_{8} \cdot 2^{8}}=16$

Correct option 1

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Posted by

avinash.dongre

The value of $\int_{0}^{2x}\frac{x\sin ^{8}x}{\sin ^{8}x+\cos ^{8}x}dx$ is equal to :
Option: 1 $2\pi$

Option: 2 $4\pi$

Option: 3 $2\pi ^{2}$

Option: 4 $\pi ^{2}$

Properties of the Definite Integral (Part 2) - King's Property -

Property 4 (King's Property)

This is one of the most important properties of definite integration.

$\\\mathbf{\int_{a}^{b}f(x)\;dx=\int_{a}^{b}f(a+b-x)\;dx}$

-

Application of Periodic Properties in Definite Integration -

Property 9

If f(x) is a periodic function with period T, then the area under f(x) for n periods would be n times the area under f(x) for one period, i.e.

$\mathbf{\int_{0}^{n T} f(x) d x=n \int_{0}^{T} f(x) d x}$

-

$\\I=\int _0^{2\pi }\frac{x\sin ^8x\:}{\sin ^8x+\cos ^8x\:\:}dx\\I=\int _0^{2\pi }\frac{(2\pi-x)\sin ^8(2\pi-x)\:}{\sin ^8(2\pi-x)+\cos ^8(2\pi-x)\:\:}dx\\2I=\int _0^{2\pi }\frac{2\pi \sin ^8x\:}{\sin ^8x+\cos ^8x\:\:}dx\\I=\int _0^{2\pi }\frac{\pi \sin ^8x\:}{\sin ^8x+\cos ^8x\:\:}dx$

$\\I=\int _0^{2\pi }\frac{\pi \sin ^8x\:}{\sin ^8x+\cos ^8x\:\:}dx\\I=4\int _0^{\pi/2 }\frac{\pi \sin ^8x\:}{\sin ^8x+\cos ^8x\:\:}dx\\I=4\int _0^{\pi/2 }\frac{\pi \sin ^8(\pi/2-x)\:}{\sin ^8(\pi/2-x)+\cos ^8(\pi/2-x)\:\:}dx\\I=2\int_{0}^{\pi/2}\pi dx\\I=\pi^2$

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Posted by

avinash.dongre

At 300 K and 1 atm, 15 mL of a gaseous hydrocarbon requires 375 mL air containing 20% O₂ by volume for complete combustion. After combustion, the gases occupy 330 mL. Assuming that the water formed is in liquid form and the volumes were measured at the same temperature and pressure, the formula of the hydrocarbon is :
Option: 1 C₄H₈
Option: 2 C₄H₁₀
Option: 3 C₃H₆
Option: 4 C₃H₈

Volume of N₂ in air = 375 × 0.8 = 300 ml

Volume of O₂ in air = 375 × 0.2 = 75 ml

$C_{x}H_{y} +\left ( x +\frac{y}{4} \right )O_{2} \; \rightarrow \; xCO_{2}(g) + \frac{y}{2} H_{2}O(l)$

15ml $15\left ( x +\frac{y}{4} \right )$

0 0 15x -

After combustion total volume

$330 =V_{N_{2}} + V_{CO_{2}}$

330 = 300 + 15x

x = 2

Volume of O₂used

$15\left ( x +\frac{y}{4} \right ) = 75$

$\left ( x +\frac{y}{4} \right ) = 5$

y = 12

So hydrocarbon is = C₂H₁₂

None of the options matches it therefore it is a BONUS.

----------------------------------------------------------------------

Alternatively Solution

$C_{x}H_{y} +\left ( x +\frac{y}{4} \right )O_{2} \; \rightarrow \; xCO_{2}(g) + \frac{y}{2} H_{2}O(l)$

15ml $15\left ( x +\frac{y}{4} \right )$

0 0 15x -

Volume of O₂ used

$15\left ( x +\frac{y}{4} \right ) = 75$

$\left ( x +\frac{y}{4} \right ) = 5$

If further information (i.e., 330 ml) is neglected, option (C₃H₈ ) only satisfy the above equation.

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Posted by

Ritika Jonwal

$\lim_{n\rightarrow \infty }\left ( \frac{(n+1)(n+2).....3n}{n^{2n}} \right )^{\frac{1}{n}}$ is equal to
Option: 1 $\frac{18}{e^{4}}$
Option: 2 $\frac{27}{e^{2}}$
Option: 3 $3\log ^{3-2}$
Option: 4 $3\log ^{3-2}$

As we learnt

Walli's Method -

Definite integral by first principle

$\int_{a}^{b}f(x)dx= \left ( b-a \right )\lim_{n \to \infty }\frac{1}{n}\left [ f(a) +f(a+h)+f(a+2h)....\right ]$

where

$h=\frac{b-a}{n}$

- wherein

$y=\lim_{n\rightarrow \infty }\left [ \frac{(n+1)(n+2)...(n+2n)}{n^{2n}} \right ]^{1/n}$

$=\lim_{n\rightarrow \infty }\left [ \left ( 1+\frac{1}{n} \right )\left ( 1+\frac{2}{n} \right ) ...\left ( 1+\frac{2n}{n} \right )\right ]^{1/n}$

$\Rightarrow \log y=\lim_{n\rightarrow \infty }\frac{1}{n}\left [ \left ( 1+\frac{1}{n} \right )\left ( 1+\frac{2}{n} \right ) ...\left ( 1+\frac{2n}{n} \right )\right ]$

$=\lim_{n\rightarrow \infty }\frac{1}{n}\sum_{r=1}^{2n}\log\left ( 1+\frac{r}{n} \right )$

$=\int_{0}^{2}\log\left ( 1+x \right )dx$

$=\left [ \log(1+x).x \right ]-\int_{0}^{2}\frac{1}{1+x}.xdx$

$= \left [ x\log(1+x)-x +\log(1+x)\right ]^{2}_{0}$

$=2\log 3-2+\log 2$

$\Rightarrow y=\frac{27}{e^{2}}$

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Get Answers to all your Questions

All Questions

Posted by

Raju vittal nandi

Calculate the acceleration of block of the following diagram. Assume all surfaces are frictionless . Here m1 = 100kg and m2 = 50kg Option: 1 0.33m/s2Option: 2 0.66m/s2Option: 3 1m/s2Option: 4 1.32m/s2

Posted by

balda gayathri

Crack CUET with india's "Best Teachers"

No. of transition state in given figure Option: 1 1Option: 2 2Option: 3 3Option: 4 4

Posted by

Mura

When cell has stalled DNA replication fork, which checkpoint should be predominantly activated? Option: 1 G1/S Option: 2 G2/M Option: 3 M Option: 4 Both G2 M and M

Posted by

Ajit Kumar Dubey

JEE Main high-scoring chapters and topics

Given : and Then the area (in sq. units) of the region bounded b the curves, and between the lines, is : Option: 1 Option: 2 Option: 3 Option: 4

Posted by

Calculate the acceleration of block $m_1$ of the following diagram. Assume all surfaces are frictionless . Here m₁ = 100kg and m₂ = 50kg

Option: 1
0.33m/s²

Option: 2
0.66m/s²

Option: 3
1m/s²

Option: 4
1.32m/s²

No. of transition state in given figure

Option: 1
1

Option: 2
2

Option: 3
3

Option: 4
4

When cell has stalled DNA replication fork, which checkpoint should be predominantly activated?

Option: 1
G₁/S

Option: 2
G₂/M

Option: 3
M

Option: 4
Both G₂M and M

The value of $\int_{0}^{2x}\frac{x\sin ^{8}x}{\sin ^{8}x+\cos ^{8}x}dx$ is equal to :
Option: 1 $2\pi$

Option: 2 $4\pi$

Option: 3 $2\pi ^{2}$

Option: 4 $\pi ^{2}$

$\lim_{n\rightarrow \infty }\left ( \frac{(n+1)(n+2).....3n}{n^{2n}} \right )^{\frac{1}{n}}$ is equal to
Option: 1 $\frac{18}{e^{4}}$
Option: 2 $\frac{27}{e^{2}}$
Option: 3 $3\log ^{3-2}$
Option: 4 $3\log ^{3-2}$