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• Bananas. Bananas may be one of the best foods for energy.
• Fatty fish. Fatty fish like salmon and tuna are good sources of protein, fatty acids, and B vitamins, making them great foods to include in your diet.
• Brown rice. Brown rice is a very nutritious food.
• Eggs

Eggs are not only a tremendously satisfying food but also full of energy that can help fuel your day.

They’re packed with protein, which can give you a steady and sustained source of energy.

Additionally, leucine is the most abundant amino acid in eggs, and it’s known to stimulate energy production in several ways.

Leucine can help cells take in more blood sugar, stimulate the production of energy in the cells, and increase the breakdown of fat to produce energy.

Moreover, eggs are rich in B vitamins. These vitamins help enzymes perform their roles in the process of breaking down food for energy

Semen is made up of sperm cells, as well as a number of bodily secretions. These secretions include: prostatic fluid, which neutralizes the acidity of the vagina. seminal fluid, which contains proteins, fatty acids, and fructose to nourish the sperm.

Conventional methods :
The common methods of conventional propagation are cutting, grafting and layering.
(a) Cutting :
(i) It is the method of producing a newplant by cutting the plant parts such as root, stem and leaf from the parent plant.
(ii)The cut part is placed in a suitable medium for growth.
(iii) It produces root and grows into a new plant. Depending upon the part used it is called as root cutting (Malus), stem cutting (Hibicus, Bougainvillea and Moringa) and leaf cutting (Begonia, Bryophullum).
Stem cutting is widely used for propagation.
(b) Grafting :
(i)In this, parts of two different plants are joined so that they continue to grow as one plant.
(ii)Of the two plants, the plant which is in contact with the soil is called stock and plant used for grafting is called scion. examples are Citrus, Mango and Apple. There are different types of grafting based on the method of uniting the scion and stock.
(iii)They are bud grafting, approach grafting, togue grafting, crown grafting and wedge grafting.
(c ) Layering :
(i) In this method, the stem of a parent plant is allowed to develop roots while still intact.
(ii) When the root develops, the rooted part is cut and planted to grow as a new plant. Examples : Ixora and Jasminum.
(iii) Mound layering and Air layering are few types of layering.

Moon is a non-luminous body because it shines by reflecting the sunlight falling on it. Moon does not have its own light.

There are three types of syngamy : isogamy, heterogamy, and oogamy.

• Use latex condoms every time you have sex.
• Avoid sharing towels or underclothing.
• Wash before and after intercourse.
• Get a vaccination for hepatitis B.
• Get tested for HIV.

Syngamy is the fusion of two cells, resulting in a cell that has twice as many chromosomes. The two cells which are fused together are called gametes, and the resulting cell is a zygote. The goal of syngamy is the renovation of genetic material.

Yes, human beings are also a threat to animals. Due to continuous poaching, many animals have either become extinct or have come to the stage of extinction. Elephants are killed for their tusks; rhinoceros for their hones, tigers; crocodiles, and snakes for their skins and so on. Musk deer are killed to prepare scent from their musk. Further, growing human interference and destruction of forests have only aggravated dangers to these animals. 

• Fatty fish, like tuna, mackerel, and salmon.
• Foods fortified with vitamin D, like some dairy products, orange juice, soy milk, and cereals.
• Beef liver.
• Cheese.
• Egg yolks.

Cells are similar to factories with different labourers and departments that work towards a common objective. Various types of cells perform different functions. Based on cellular structure, there are two types of cells:

• Prokaryotes
• Eukaryotes

The young 's Modulus for a perfect rigid body as the perfectly rigid body is one for which strain is zero whatever may be the stress. 

Young's modulus is defined as = stress / strain

Since strain is zero, Young's modulus for a perfectly rigid body is infinity and young modulus for a perfect infinity thereof.

ratio of stress to starin is called youngs modulus.

hence Youngs Modulus for a perfectly rigid body is INFINITE

The atoms of X lose electrons whereas the atoms of Y gain electrons. Thus, there is transfer of electrons from atoms of X to atoms of Y. The bond formed by the transfer of electrons is called ionic bond. Therefore, the nature of bond in the compound XY is ionic.

 Properties of ionic compound XY:
(i) The compound will be soluble in water.
(ii) The compound will conduct electricity when dissolved in water or inmolten state.

Gold and platinum.
(i) A thin impervious layer of aluminium oxide forms a protective layer which protects the aluminium metals underneath from further damage.
(ii) Corrosion of iron is a serious problem. Every year enormous amount of money is spent to replace damaged iron and steel structures.

Double circulation is a process during which blood passes twice through the heart during one complete cycle. This type of circulation is found in birds, and mammals as in them the heart is completely divided into four chambers – the right atrium, the right ventricle, the left atrium, and the left ventricle.

The majority of mammals (including humans) utilize a double circulatory system. This means we have two loops in our body in which blood circulates. One is oxygenated, meaning oxygen rich, and the other is deoxygenated, which means it has little to no oxygen, but a lot of carbon dioxide.

Double circulatory systems are important because they ensure that we are giving our tissues and muscles blood full of oxygen, instead of a mixture of oxygenated and deoxygenated blood. While it may take a bit more energy than a single circulatory system, this system is much more efficient!

The movement of blood in an organism is divided into two parts:
(i) Systemic circulation
(ii) Pulmonary circulation

Systemic circulation involves the movement of oxygenated blood from the left ventricle of the heart to the aorta. It is then carried by blood through a network of arteries, arterioles, and capillaries to the tissues. From the tissues, the deoxygenated blood is collected by the venules, veins, and vena cava, and is emptied into the right auricle.

Pulmonary circulation involves the movement of deoxygenated blood from the right ventricle to the pulmonary artery, which then carries blood to the lungs for oxygenation. From the lungs, the oxygenated blood is carried by the pulmonary veins into the left atrium.

Hence, in double circulation, blood has to pass alternately through the lungs and the tissues.

Significance of double circulation:

The separation of oxygenated and deoxygenated blood allows a more efficient supply of oxygen to the body cells. Blood is circulated to the body tissues through systemic circulation and to the lungs through pulmonary circulation.

Maximum circumference of the junior football be 60 cm. (Given).

∴ 2πR = 60 cm

⇒ 2R = \(\frac{60}{\pi}\) = \(\frac{60}{3.14}\) 

= 19.11 cm (approx)

Hence, 

The maximum diameter of the junior football can be 19.11 cm (approx).

The definition of a prime number is a number that is divisible by only one and itself. A prime number can't be divided by zero, because numbers divided by zero are undefined. The smallest prime number is 2, which is also the only even prime.

I = \(\int\limits_0^1x^3(log x)^3dx\)

let log x = -t ⇒ x = e^-t

1/x dx = -dt

Limits converts info from t = -log (0) = -(-\(\infty\)) = \(\infty\)

to f = -log 1 = -0 = 0

∴ I = \(\int\limits_{\infty}^0e^{-4t}(-t)^3dt\) = \(-\int\limits_0^{\infty}t^3e^{-4t}dt\) 

= \(\frac{\Gamma(4)}{4^4}\) ( ∵ \(\int\limits_0^{\infty}\)x^n-1e^-axdx = \(\frac{\Gamma(n)}{a^n}\), n = 4)

 ∴ \(\int\limits_0^1\)(x log x)³dx = 3/128

a. \(\frac{15}{30}=\frac{1}{2}\)

b. Multiple of 3 = 10

∴ The probability to be a multiple of 3 = \(\frac{10}{30}=\frac{1}{30}\)

c. The multiples of 3 and 5 are 15 and 30 only ∴ Required probability \(\frac{2}{30}=\frac{1}{15}\)

d. The probability to be a natural number = \(\frac{30}{30}=1\)

\(\int sin^{-1}2xdx\)

 = sin^-12x \(\int 1dx\) - \(\int(\frac d{dx}sin^{-1}2x\int 1dx)dx\) 

 = x sin^-12x - \(\int\frac{2x}{\sqrt{1-4x^2}}dx\)

 = x sin^-12x + \(\frac14\)\(\int\frac{-8x}{\sqrt{1-4x^2}}dx\)

 = x sin^-12x + \(\frac14\)\(\int\frac{2tdt}t\) (By taking 1 - 4x² = t² ⇒ -8x dx = 2tdt)

 = x sin^-12x + \(\frac12\sqrt{1-4x^2}+c\) (\(\because t=\sqrt{1-4x^2}\))

\(\therefore\) \(\int sin^{-1}2xdx\) = xsin^-12x + \(\frac{\sqrt{1-4x^2}}2+c\)

∫\(\sqrt{1+cos\,2x}\) dx

= ∫\(\sqrt{1+2\,cos^2x-1}\) dx (∵ cos 2x = 2 cos²x - 1)

= ∫√2 cos x dx

= √2 sin x+c (∵ ∫cos x dx = sin x)

∫sin²x cos⁴x dx

= 1/4 ∫(2 sin x cos x)² cos²x dx

= 1/8 ∫sin²2x x 2cos²x dx (∵ 2 sinθ cosθ = sin2θ)

= 1/8 ∫sin²2x(1 + cos2x) dx (∵ cos2θ = 2cos²θ - 1 ⇒ 2 cos²θ = cos2θ + 1)

= 1/8 ∫sin²2x cos2x dx + 1/8 ∫sin²2x dx

= 1/8 ∫t² dt/2 + 1/8 ∫\(\frac{1-cos4x}{2}dx\)

(By taking sin 2x = t ⇒ cos2x dx = dt/2 And sin²θ \(=\frac{1-cos2\theta}{2})\)

= 1/16 x t³/3 + 1/16 (x - sin4x/4) + c

= 1/48 sin³2x + x/16 - 1/64 sin4x + c (By putting t = sin 2x)

∫(sec²x - secx. tanx)dx = tanx – secx + c

∫(Cosec²x + Cosec xCotx)dx 

= cot x – Cosec x + c

\(\int\limits_0^{\pi/2}\frac{dx}{1+cos^2x}=\int\limits_{0}^{\pi/2}\frac{sec^2xdx}{sec^2x+1}\)

= \(\int\limits_0^{\pi/2}\frac{sec^2xdx}{2+tan^2x}\)

Let tan x = t ⇒ sec² x dx = dt

x = 0 then t = 0 & when x = \(\frac{\pi}2\) then t = \(\infty\)

= \(\int\limits_0^{\infty}\frac{dt}{2+t^2}\) = \(\frac1{\sqrt2}[tan^{-1}\frac{x}{\sqrt2}]_0^{\infty}\) = \(\frac1{\sqrt2}(tan^-1\infty - tan^{-1}0)\)

 = \(\frac1{\sqrt2}(\frac{\pi}2-0)=\frac{\pi}{2\pi}\)

 \(\int\frac{x^2+a^2}{x^4+a^4}dx\)

 = \(\int\frac{x^2+a^2}{(x^2+a^2)^2-4a^2x^2}dx\)

 = \(\int\frac{x^2+a^2}{(x^2+a^2-2ax)(x^2+a^2+2ax)}dx\) (\(\because\) a² - b² = (a + b) (a - b))

 = \(\int\frac{x^2+a^2}{(x-a)^2(x+a)^2}dx\) 

 = \(\frac12\int(\frac1{(x-a)^2}\frac1{(x+a)^2})dx\) 

 = \(\frac12(-\frac1{x-a}-\frac1{x+a})+c\) 

 = \(-\frac12(\frac1{x+a}+\frac1{x-a})+c\) 

 = \(-\frac12(\frac{2x}{x^2-a^2})+c\)

 = \(\frac{x}{a^2-x^2}+c\)

\(\int\limits^1_0\int\limits^x_0(x^2 + y^2 )\,dx\,dy\)

\(= \int\limits^1_0\left[\int\limits^x_0x^2+ y^2 \,dy\right]dx\)

\(= \int\limits^1_0\left[x^2y+\frac{y^3}{3}\right]^x_0\, dx\)

\(= \int\limits^1_0\left(x^3 + \frac{x^3}{3}\right)dx\)

\(= \frac43\int\limits^1_0x^3\,dx\)

\(= \frac43 \left[\frac{x^4}{4}\right]^1_0\)

\(= \frac43\times\frac14\)

\(= \frac13\)

\(\because\) Revolution is about y - axis (or x = 0 line)

Volume = \(\int\limits^{y_1}_{y_2}\pi x^2\,dy\) 

Given curve is x² - 2x + y = 3

⇒ y = 3 - x² + 2x

= 4 - (x² - 2x + 1)

= 4 - (x - 1)²

At x = 0 on y-axis, 

y = 4 - 1 = 3

\(\therefore\) y₁ = 3, y₂ = 4

\(\therefore\) Generated volume = \(\int\limits^4_3\pi x^2\,dy\)

\(=\int\limits^4_3\pi (-\sqrt{4-y}+1)\,dy\)

\(=\, ^\pi \left(\frac23(4-y)^{\frac32}+y\right)^4\)

\(= \pi \left(\frac23 \times0+4-\frac23 \times 1 -3\right)\)

\(= \pi \left(1 - \frac23\right)\)

\(= \frac {\pi}3\) cubic units.

 \(\because\) Revolution is about y - axis (or x = 0 line)

Volume = \(\int\limits^{y_1}_{y_2}\pi x^2\,dy\) 

Given curve is x² - 2x + y = 3

⇒ y = 3 - x² + 2x

= 4 - (x² - 2x + 1)

= 4 - (x - 1)²

At x = 0 on y-axis, 

y = 4 - 1 = 3

\(\therefore\) y₁ = 3, y₂ = 4

\(\therefore\) Generated volume = \(\int\limits^4_3\pi x^2\,dy\)

\(=\int\limits^4_3\pi (-\sqrt{4-y}+1)\,dy\)

\(=\, ^\pi \left(\frac23(4-y)^{\frac32}+y\right)^4\)

\(= \pi \left(\frac23 \times0+4-\frac23 \times 1 -3\right)\)

\(= \pi \left(1 - \frac23\right)\)

\(= \frac {\pi}3\) cubic units.

∫\(\frac{(1+x^2)}{\sqrt x}dx=\int\frac{x^2+2x+1}{\sqrt x}dx\)

= ∫(x^3/2 + 2 x^1/2 + x^-1/2)dx

= 2/5 x^5/2 + 4/3 x^3/2 + 2x^1/2 + c (∵ ∫x^x dx \(=\frac{x^n+1}{n+1})\)

∵ \(\frac{cos\,x-cos\,2x}{1-cos\,x}=\frac{cos\,x-(2cos^2x-1)}{1-cos\,x}\) (∵ cos 2x = 2 cos²x - 1)

\(=\frac{-2cos^2x+cos\,x+1)}{1-cos\,x}\)

\(=\frac{-2cos^2x+2\,cos\,x-cos\,x+1)}{1-cos\,x}\)

\(=\frac{2\,cos\,x(1-cos\,x)+(1-cos\,x)}{1-cos\,x}\)

\(=\frac{(1-cos\,x)(2\,cos\,x+1)}{1-cos\,x}\)

= 2 cos x+1

∴ ∫\(\frac{cos\,x-cos\,2x}{1-cos\,x}dx\) = ∫(2 cos x+1)dx

= 2∫cos x dx + ∫dx

= 2 sin x+x+c

Given that u, v, w are roots of the equation (x-a)³ + (x-b)³ + (x-c)³ = 0

⇒ 3x³ - 3(a+b+c)x² + 3(a²+b²+c²)x - (a³+b³+c³) = 0

∵ u, v and w are roots of given equation.

∴ Sum of roots = u+v+w = -b/a \(=\frac{-(-3(a+b+c))}{3}\) = a+b+c

⇒ u + v + w - a - b - c = 0

⇒ f(u, v, w) = u+v+w - (a+b+c) = 0

Now, \(\frac{\partial u}{\partial a}=1,\frac{\partial u}{\partial b}=1,\frac{\partial u}{\partial c}=1,\)

\(\frac{\partial v}{\partial a}=1,\frac{\partial v}{\partial b}=1,\frac{\partial v}{\partial c}=1,\)

\(\frac{\partial w}{\partial a}=1,\frac{\partial w}{\partial b}=1,\frac{\partial w}{\partial c}=1\)

Now, \(\frac{\partial(u,v,w)}{\partial(a,b,c)}=\begin{bmatrix}​​\frac{\partial u}{\partial a}&​​\frac{\partial u}{\partial b}&​​\frac{\partial u}{\partial c}\\​​\frac{\partial v}{\partial a}&​​\frac{\partial v}{\partial b}&​​\frac{\partial v}{\partial c}\\​​\frac{\partial w}{\partial a}&​​\frac{\partial w}{\partial b}&​​\frac{\partial w}{\partial c}\end{bmatrix}\)

\(=\begin{bmatrix}1&1&1\\1&1&1\\1&1&1\end{bmatrix}=0\)

Hence, \(\frac{\partial(u,v,w)}{\partial(a,b,c)}=0\)

∫sin³x dx

= ∫\(\frac{3sin\,x-sin\,3x}{4}dx\) (∵ sin 3θ = 3 sin θ - 4 sin³θ ⇒ sin³θ  \(=\frac{3sin\,\theta-sin\,3\theta}{4}\))

= 1/4 (-3cos x + \(\frac{cos3x}3\)) + c (∵ ∫sin ax dx \(= \frac{-cos\,ax}{a})\)

= -3/4 cos x + 1/12 cos 3x + c

I = \(\int\cfrac{d\text x}{(2\text x+5)\sqrt{4\text x^2+20\text x+16}}\) 

= \(\int\cfrac{d\text x}{(2\text x+5)\sqrt{(2\text x+5)^2+9}}\) 

= \(\int\cfrac{d\text x}{(2\text x+5)\sqrt{(2\text x+5)^2+3^2}}\) 

= \(\cfrac12\times\cfrac13\) sec^-1\(\left(\cfrac{2\text x+5}3\right)+c\)

\(\left(\because\quad\int\cfrac{d\text x}{\text x\sqrt{\text x^2-a^2}}=\cfrac1asec^{-1}\cfrac{\text x}a+c\right)\)

= \(\cfrac16\) sec^-1\(\left(\cfrac{2\text x+5}3\right)+c\)

Hence, m = 6, p = 3

The correct answer is 12.

• The rivers of India can be classified into four groups viz., Himalayan rivers, Deccan rivers, Coastal rivers, and Rivers of the inland drainage basin.
  
• The Himalayan Rivers are formed by melting snow and glaciers and therefore, continuously flow throughout the year.
• During the monsoon months, Himalayas receive very heavy rainfall and rivers swell, causing frequent floods.
• The Deccan Rivers on the other hand are rain-fed and therefore fluctuate in volume. Many of these are non-perennial.
• The Coastal streams, especially on the west coast are short in length and have limited catchment areas.
• Most of them are non-perennial. The streams of the inland drainage basin of western Rajasthan are few.
• Most of them are of an ephemeral character.

The correct answer is 1565 AD.

• Battle of Talikota, confrontation in the Deccan region of southern India between the forces of the Hindu raja of Vijayanagar and the four allied Muslim sultans of Bijapur, Bidar, Ahmadnagar, and Golconda.
  • The battle was fought on January 23, 1565, at a site southeast of Bijapur, in what is now northern Karnataka state.

•  Krishna Deva Raya, king of the Vijaynagar empire at its zenith, stormed the Raichur Fort after a bloodily victorious battle against the forces of Ismail Adil Shah, ruler of the Bijapur Sultanate.
  • The Battle of Raichur (1520), though celebrated widely through the streets of Vijayanagara, signaled the slow-yet-steady downfall of the Vijayanagara Empire.
  • Ultimately, the Battle of Talikota (1565) brought about the fated decline of a kingdom over two hundred years old.
  • After this decisive battle, the Vijayanagara Empire could never reach the same heights of the past. 

Jagannathdev of Puri

Conches of the Pandavas

Correct answer is (b) 2

Correct answer is (a) conservative

(b) Charge added/Potential difference

Correct answer is (b) energy 

Let the breadth of the wall be x metres. 

Then, Height = 5x metres and Length = 40x metres. 

x * 5x * 40x = 12.8 

x³ =12.8/200 = 128/2000 = 64/1000 

So, x = (4/10) m = ((4/10)*100)cm = 40 cm

No other dramatist was as great as Shakespeare.