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Given as,

f (x) = ax + b, a < 0

Suppose x₁, x₂ ∈ R and x₁ > x₂

⇒ ax₁ < ax₂ for some a > 0

⇒ ax₁ + b < ax₂ + b for some b

⇒ f (x₁) < f(x₂)

Thus, x₁ > x₂⇒ f(x₁) < f(x₂)

Therefore, f(x) is decreasing function of R

finding the slope of the tangent by differentiating the curve

\(2y\frac{dy}{dx}\)\(=\frac{(4-x)3x^2+x^4}{(4-x)^2}\)

\(\frac{dy}{dx}\)\(=\frac{(4-x)3x^2+x^4}{2y(4-x)^2}\)

m(tangent) at (2, – 2) = – 2

m(normal) at (2, -2) = \(\frac{1}{2}\)

equation of tangent is given by y – y₁ = m(tangent)(x – x₁)

y + 2 = – 2(x – 2)

y + 2x = 2

equation of normal is given by y – y₁ = m(normal)(x – x₁)

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

2y + 4 = x – 2

2y – x + 6 = 0

Given as,

f (x) = ax + b, a > 0

Suppose x₁, x₂ ∈ R and x₁ > x₂

⇒ ax₁ > ax₂ for some a > 0

⇒ ax₁ + b> ax₂ + b for some b

⇒ f (x₁) > f(x₂)

Thus, x₁ > x₂ ⇒ f(x₁) > f(x₂)

Therefore, f(x) is increasing function of R

Suppose x₁, x₂ ∈ (0, ∞)

We have, x₁ < x₂

⇒ log_e x₁ < log_e x₂

⇒ f (x₁) < f (x₂)

Hence, f(x) is increasing in (0, ∞)

Kabir is a great poet-cum-saint of Indian society. He has had appeal among Hindus and Muslims alike as it is believed that he was bom as Hindu but was brought up by a muslim couple. He wrote poems that exhorted both communities to take to social reforms. 

The major teachings of Kabir were as follows: 

1.Kabir described God as nirankar (having no shape). He used the terms drawn from Islamic tradition like Allah, Khuda, Hajrat and Peer but also used words of Vedic traditions like Alakh ( (the unseen) and nirakar ( the formless). Thus, he freely took to both traditions viz. Islamic and Vedantic. 

2.He repudiated idol worship and polytheism. 

3.He emphasised on the oneness of God though there can be many names of His. 

4.He criticised religious rituals of hindus and muslims alike. 

5.He also preached against caste discrimination. 

6.He combined the Sufi traditions of love of God with the Hindi tradition of remembrance of God.

 7.He also emphasised the dignity of labour. 

Thus, the essence of the teachings of Kabir was simple living based on love and respect all. He wrote in simple language to be understood by common man of the country. Guru Nanak and his teachings Guru Nanak was born in a Hindu family in 1469 at Nankana Saheb on the bank of the river Ravi. His birth place is now in Pakistan. He learnt Persian, Arabic , Hindi and Mathematics. He spent time in the company of Sufi saints and Bhaktas of various socio-religious movements. 

The major teachings of Guru Nanak are as follows: 

1. He rejected the religious texts of both Hindus and Muslims. 

2. He preached God is Nirakar viz. without any shape. 

3. He criticised the religious practices like ceremonial bath, sacrifices , idol worship and emphasised simplicity. 

4. He called upon his followers to connect to divine by remembering and repeating the divine name. Guru Nanak expressed himself in Punjabi, the language of the local people in a lyrical form called Shabad. Shabad can be recited in various ragas.

Given as f(x) = x⁹ + 4x⁷ + 11

Differentiate the above equation with respect to x, we get

⇒ f'(x) = (d/dx)(x⁹ + 4x⁷ + 11)

⇒ f’(x) = 9x⁸ + 28x⁶

⇒ f’(x) = x⁶(9x² + 28)

Also, as given x ϵ R

⇒ x⁶ > 0 and 9x² + 28 > 0

⇒ x⁶ (9x² + 28) > 0

⇒ f’(x) > 0

Thus, condition for f(x) to be increasing

Hence, f(x) is increasing on interval x ∈ R

Given as f(x) = x² – x + 1

Differentiate the given equation with respect to x, we get

⇒ f'(x) = (d/dx)(x² - x + 1)

⇒ f’(x) = 2x – 1

On taking different region from (0, 1)

Suppose x ∈ (0, 1/2)

⇒ 2x – 1 < 0

⇒ f’(x) < 0

Hence f(x) is decreasing in (0, 1/2)

Suppose x ∈ (1/2, 1)

⇒ 2x – 1 > 0

⇒ f’(x) > 0

Hence f(x) is increasing in (1/2, 1)

So, from above condition we find that

⇒ f (x) is decreasing in (0, 1/2)  and increasing in (1/2, 1)

Thus, condition for f(x) neither increasing nor decreasing in (0, 1)

Given as f(x) = tan x

f'(x) = (d/dx)(tan x)

⇒ f’(x) = sec²x

As given

x ∈ (–π/2, π/2).

That is 4^th quadrant, where

⇒ sec²x > 0

⇒ f’(x) > 0

Thus, condition for f(x) to be increasing

Hence, f(x) is increasing on interval (–π/2, π/2).

Given as f(x) = (x – 1)e^x + 1

Differentiate the given equation with respect to x, we get

⇒ f'(x) = (d/dx)((x - 1)e^x + 1)

⇒ f’(x) = e^x + (x – 1) e^x

⇒ f’(x) = e^x(1+ x – 1)

⇒ f’(x) = x e^x

As given x > 0

⇒ e^x > 0

⇒ x e^x > 0

⇒ f’(x) > 0

Thus, condition for f(x) to be increasing

Hence f(x) is increasing on interval x > 0

Given:

The curve is x² + y² – 2x – 3 = 0

Differentiating the above w.r.t x, we get The Slope of tangent,

⇒ 2x^{2 – 1} + 2y^{2 – 1} \(\frac{dy}{dx}\)– 2 – 0 = 0

⇒ 2x + 2y \(\frac{dy}{dx}\) – 2 = 0

⇒ 2y \(\frac{dy}{dx}\) = 2 – 2x

⇒ \(\frac{dy}{dx}\) = \(\frac{2-2x}{2y}\)

⇒\(\frac{dy}{dx}\) = \(\frac{1-x}{y}\)...(1)

(i) Since, the tangent is parallel to x – axis

⇒ \(\frac{dy}{dx}\) = tan(0) = 0 ...(2)

\(\therefore\) tan(0) = 0

\(\therefore\frac{dy}{dx}\) = The Slope of the tangent = tanθ

From (1) & (2),we get,

⇒ \(\frac{1-x}{y}\) =  0

⇒ 1 – x = 0

⇒ x = 1

Substituting x = 1 in x² + y² – 2x – 3 = 0,

⇒ 1² + y² – 2×1 – 3 = 0

⇒ y² – 4 = 0

⇒ y² = 4

⇒ y = ± 2

Thus, the required point is (1,2) & (1, – 2)

Given:

The curve is \(\frac{x^2}{4}+\frac{y^2}{25}=1\)

Differentiating the above w.r.t x, we get the The Slope of a tangent,

\(\Rightarrow\)\(\frac{2x^{2-1}}{4}+\frac{2y^{2-1}\times\frac{dy}{dx}}{25}=0\)

Cross multiplying we get,

\(\Rightarrow\)\(\frac{25\times2x+4\times2y\times\frac{dy}{dx}}{100}=0\)

⇒ 50x + 8y\(\frac{dy}{dx}\) = 0

⇒ 8y \(\frac{dy}{dx}\)= – 50x

⇒ \(\frac{dy}{dx}\) = \(\frac{-50x}{8y}\)

 ⇒ \(\frac{dy}{dx}\) = \(\frac{-25x}{4y}\)......(1)

Since, the tangent is parallel to x – axis

⇒ \(\frac{dy}{dx}\) = tan(0) = 0 ...(2)

\(\therefore\) tan(0) = 0

\(\therefore\) \(\frac{dy}{dx}\) = The Slope of the tangent = tanθ

From (1) & (2),we get,

⇒ \(\frac{-25x}{4y}\) = 0

⇒ – 25x = 0

⇒ x = 0

Substituting x = 0 in \(\frac{x^2}{4}+\frac{y^2}{25}=1\) ,

\(\Rightarrow\)\(\frac{0^2}{4}+\frac{y^2}{25}=1\)

⇒ y² = 25

⇒ y = ± 5

Thus, the required point is (0,5) & (0, – 5)

Since, the tangent is parallel to y – axis, its slope is not defined, then the normal is parallel to x – axis whose slope is zero.

i.e, \(\cfrac{-1}{\frac{dy}{dx}}\) = 0

⇒ \(\cfrac{-1}{\frac{1-x}{y}}\) = 0

⇒ \(\cfrac{-y}{1-x}\) = 0

⇒ y = 0

Substituting y = 0 in x² + y² – 2x – 3 = 0,

⇒ x² + 0² – 2×x – 3 = 0

⇒ x² – 2x – 3 = 0

Using factorization method, we can solve above quadratic equation

⇒ x² – 3x + x – 3 = 0

⇒ x(x – 3) + 1(x – 3) = 0

⇒ (x – 3)(x + 1) = 0

⇒ x = 3 & x = – 1

Thus, the required point is (3,0) & ( – 1,0)

Given:

 The curve \(\frac{x^2}{9}+\frac{y^2}{16}=1\)

Differentiating the above w.r.t x, we get the Slope of tangent,

\(\Rightarrow\)\(\frac{2x^{2-1}}{9}+\frac{2y^{2-1}\times\frac{dy}{dx}}{16}=0\)

\(\Rightarrow\)\(\frac{2x}{9}+\frac{y\times\frac{dy}{dx}}{8}=0\)

Cross multiplying we get,

⇒ \(\frac{(8\times2x)+(9\times y)\frac{dy}{dx}}{72}=0\)

\(\Rightarrow 16x+9y\frac{dy}{dx}=0\)

⇒ \(9y\frac{dy}{dx}=-16x\)

⇒ \(\frac{dy}{dx}=\frac{-16}{9y}\)..........(1)

Since, the tangent is parallel to x – axis

⇒ \(\frac{dy}{dx}\)= tan(0) = 0 ...(2)

\(\therefore\) tan(0) = 0

\(\therefore\) \(\frac{dy}{dx}\) = The Slope of the tangent = tanθ

From (1) & (2),we get,

⇒ \(\frac{-16x}{9y}\) = 0

⇒ – 16x = 0

⇒ x = 0

Substituting x = 0 in \(\frac{x^2}{9}+\frac{y^2}{16}=1\) ,

\(\Rightarrow\)\(\frac{0^2}{9}+\frac{y2}{16}=1\)

⇒ y² = 16

⇒ y =  ± 4

Thus, the required point is (0,4) & (0, – 4)

Given:

The curve y = 7x³ + 11

Differentiating the above w.r.t x

⇒ \(\frac{dy}{dx}\) = 3 x 7x ^{3 – 1} + 0

⇒ \(\frac{dy}{dx}\) = 21x²

when x = 2

⇒ \(\cfrac{dy}{dx_{x=2}}\) = 21×(2)²

 ⇒ \(\cfrac{dy}{dx_{x=2}}\) = 21 x 4

 ⇒ \(\cfrac{dy}{dx_{x=2}}\)= 84

when x = – 2

  ⇒ \(\cfrac{dy}{dx_{x=2}}\) = 21 x (-2)²

  ⇒ \(\cfrac{dy}{dx_{x=2}}\) = 21 x 4

  ⇒ \(\cfrac{dy}{dx_{x=2}}\) = 84

Let y = f(x) be a continuous function and P(x₀,y₀) be point on the curve, then,

The Slope of the tangent at P(x,y) is f'(x) or \(\frac{dy}{dx}\)

Since, the Slope of the tangent is at x = 2 and x = – 2 are equal, the tangents at x = 2 and x = – 2 are parallel.

Since the tangent is parallel to y–axis, its slope is not defined, then the normal is parallel to x–axis whose The Slope is zero.

i.e., \(\cfrac{-1}{\frac{dy}{dx}}\) = 0

⇒ \(\cfrac{-1}{\frac{-16x}{9y}}\) = 0

⇒ \(\frac{-9y}{16x}\) = 0

⇒ y = 0

Substituting y = 0 in \(\frac{x^2}{9}+\frac{y^2}{16}=1\),

\(\Rightarrow\)\(\frac{x^2}{9}+\frac{0^2}{16}=1\)

⇒ x² = 9

⇒ x = ± 3

Thus, the required point is (3,0) & ( – 3,0)

Given:

The curve is y = x³

y = x³

Differentiating the above w.r.t x

⇒ \(\frac{dy}{dx}\) = 3x^{2 – 1}

⇒ \(\frac{dy}{dx}\) = 3x² ...(1)

Also given the The Slope of the tangent is equal to the x – coordinate,

\(\frac{dy}{dx}\) = x ...(2)

From (1) & (2),we get,

i.e, 3x² = x

⇒ x(3x – 1) = 0

⇒ x = 0 or x = \(\frac{1}{3}\)

Substituting x = 0 or x = \(\frac{1}{3}\)this in y = x³,we get,

when x = 0

⇒ y = 0³

⇒ y = 0

when x = \(\frac{1}{3}\)

⇒ y = ( \(\frac{1}{3}\) )³

⇒ y = \(\frac{1}{27}\)

Thus, the required point is (0,0) & ( \(\frac{1}{3}\), \(\frac{1}{27}\))

Given as f(x) = cos x

f'(x) = (d/dx)(cos x)

⇒ f’(x) = –sin x

On taking different region from 0 to 2π

Suppose x ∈ (0, π).

⇒ Sin(x) > 0

⇒ –sin x < 0

⇒ f’(x) < 0

Hence, f(x) is decreasing in (0, π)

Suppose x ∈ (–π, o).

⇒ Sin (x) < 0

⇒ –sin x > 0

⇒ f’(x) > 0

Hence f(x) is increasing in (–π, 0).

So, from above condition we find that

⇒ f (x) is decreasing in (0, π) and increasing in (–π, 0).

Thus, condition for f(x) neither increasing nor decreasing in (–π, π)

Given:

The curve is y = x²

Differentiating the above w.r.t x

⇒ y = x²

⇒ \(\frac{dy}{dx}\) = 2x^{2 – 1}

⇒ \(\frac{dy}{dx}\) = 2x ...(1)

\(\therefore\frac{dy}{dx}\)= The Slope of the tangent = tanθ 

Since, the tangent make an angle of 45^o with x – axis

i.e,

⇒ \(\frac{dy}{dx}\) = tan(45°) = 1 ...(2)

\(\therefore\) tan(45°) = 1

From (1) & (2), we get

⇒ 2x = 1

⇒ x = \(\frac{1}{2}\)

Substituting x = \(\frac{1}{2}\) in y = x², we get,

⇒ y = ( \(\frac{1}{2}\) )²

⇒ y = \(\frac{1}{4}\)

Thus, the required point is ( \(\frac{1}{2}\), \(\frac{1}{4}\))

Given:

The curve is x² + y² – 2x – 4y + 1 = 0

Differentiating the above w.r.t x

⇒ x² + y² – 2x – 4y + 1 = 0

⇒ 2x^{2 – 1} + 2y^{2 – 1} x \(\frac{dy}{dx}\)– 2 – 4 x \(\frac{dy}{dx}\)+ 0 = 0

⇒ 2x + 2y \(\frac{dy}{dx}\)– 2 – 4\(\frac{dy}{dx}\)= 0

⇒ \(\frac{dy}{dx}\) (2y – 4) = – 2x + 2

⇒ \(\frac{dy}{dx}\) \(= \frac{-2(x-1)}{2(y-2)}\)

⇒ \(\frac{dy}{dx}\) \(= \frac{-(x-1)}{(y-2)}\)......(1)

\(\therefore\) \(\frac{dy}{dx}\) = The Slope of the tangent = tanθ 

Since, the tangent is parallel to x – axis

i.e,

⇒ \(\frac{dy}{dx}\) = tan(0) = 0 ...(2)

\(\therefore\) tan(0) = 0

From (1) & (2),we get,

⇒\(=\frac{-(x-1)}{(y-2)}=0\)

⇒ – (x – 1) = 0

⇒ x = 1

Substituting x = 1 in x² + y² – 2x – 4y + 1 = 0,we get,

⇒ 1² + y² – 2 x1 – 4y + 1 = 0

⇒ 1 – y² – 2 – 4y + 1 = 0

⇒ y² – 4y = 0

⇒ y(y – 4) = 0

⇒ y = 0 & y = 4

Thus, the required point is (1,0) & (1,4)

Given:

The curve is y = 3x² – 9x + 8

Differentiating the above w.r.t x

⇒ y = 3x² – 9x + 8

⇒\(\frac{dy}{dx}\) = 2 x 3x^{2 – 1} – 9 + 0

⇒ \(\frac{dy}{dx}\) = 6x – 9 ...(1)

Since, the tangent are equally inclined with axes

i.e, θ = \(\frac{\pi}{4}\) or θ \(=\frac{-\pi}{4}\)

\(\therefore\) \(\frac{dy}{dx}\) = The Slope of the tangent = tanθ

⇒ \(\frac{dy}{dx}\) = tan( \(\frac{\pi}{4}\) ) or tan( \(\frac{-\pi}{4}\) )

⇒ \(\frac{dy}{dx}\) = 1or – 1 ...(2)

\(\therefore\) tan( \(\frac{\pi}{4}\) ) = 1

From (1) & (2),we get,

⇒ 6x – 9 = 1 0r 6x – 9 = – 1

⇒ 6x = 10 0r 6x = 8

⇒ x = \(\frac{10}{6}\) or x = \(\frac{8}{6}\)

⇒ x = \(\frac{5}{3}\) or x = \(\frac{4}{3}\)

Substituting x = \(\frac{5}{3}\) or x = \(\frac{4}{3}\) in y = 3x² – 9x + 8,we get,

When x = \(\frac{5}{3}\)

⇒ y = 3( \(\frac{5}{3}\) )² – 9( \(\frac{5}{3}\) ) + 8

⇒ y = 3( \(\frac{25}{9}\) ) – ( \(\frac{45}{3}\) ) + 8

⇒ y = ( \(\frac{75}{9}\) ) – ( \(\frac{45}{3}\) ) + 8

taking LCM = 9

⇒ y = ( \(\frac{(75\times1)-(45\times3)+(8\times9)}{9}\) )

⇒ y = ( \(\frac{75-135+72}{9}\) )

⇒ y = ( \(\frac{12}{9}\) )

⇒ y = ( \(\frac{4}{3}\) )

⇒ y = 3( \(\frac{4}{3}\) )2 – 9( \(\frac{4}{3}\) ) + 8

⇒ y = 3( \(\frac{16}{9}\) ) – ( \(\frac{26}{3}\) ) + 8

⇒ y = ( \(\frac{48}{9}\) ) – ( \(\frac{36}{3}\) ) + 8

taking LCM = 9

⇒ y = ( \(\frac{(48\times1)-(36\times3)+(8\times9))}{9}\) )

⇒ y = ( \(\frac{48-108+72}{9}\) )

⇒ y = ( \(\frac{12}{9}\) )

⇒ y = ( \(\frac{4}{3}\) )

Thus, the required point is ( \(\frac{5}{3}\), \(\frac{4}{3}\)) & ( \(\frac{4}{3}\), \(\frac{4}{3}\))

The objectives of structural design may be summarised as follows:

1. Safety: The structure should be able to carry all expected loads safely, without failure, that is, without breaking or collapsing under the loads.

2. Stability: The structure should not move (by sliding or overturning) under the expected loads.

3. Serviceability: The structure should not deform or crack unreasonably under the expected loads.

4. Durability: The structure should last for a reasonable period of time.

Given : 

V₁ = Initial volume = 2800 m³, 

T₁ = Initial temperature = 99°C 

= 99 + 273.15 

= 372.15 K,

T₂ = Final temperature = 80°C 

= 80 + 273.15 K 

= 353.15 K

To find : 

V₂ = Final volume

Formula :

\(\frac{V_1}{T_1}\) = \(\frac{V_2}{T_2}\) (at constant n and P)

Calculation : 

According to Charles’ law,

\(\frac{V_1}{T_1}\) = \(\frac{V_2}{T_2}\) (at constant n and P)

∴ V₂ = \(\frac{V_1T2}{T_1}\) 

= \(\frac{2800\times 353.15}{372.15}\) 

= 2657 m³

∴ The volume of the balloon when the air cools to 80°C is 2657 m³.

The Kelvin scale is based on absolute zero which means it will never go negative unlike Celsius scale. That is why it is convenient for all scientific calculations.

Given : 

V₁ = Initial volume = 20 L, 

n₁ = Initial number of moles = 0.650 mol 

P₁ = Initial pressure = 628.3 bar 

T₁ = Initial temperature = 37°C 

= 37 + 273.15 K 

= 310.15 K 

n₂ = Final number of moles = 0.650 + 1.25 

= 1.90 mol, 

V₂ = Final volume = 12 L 

T₂ = Final temperature = 177°C 

= 177 + 273.15 K 

= 450.15 K, 

R = 0.0821 L atm K^-1 mol^-1 

To find : 

P = Final pressure 

Formula : 

PV = nRT

Calculation : 

According to ideal gas equation, 

P₂V₂ = n₂RT₂.

∴ P₂ = \(\frac{n_2RT_2}{V_2}\) 

= \(\frac{1.90\times 0.0821 \times 450.15}{12}\) 

= 5.852 atm.

∴ The final pressure of the gas is 5.852 atm.

[Note : In the above numerical, converting the pressure value to different units, we get : 5.852 atm = 4447.52 torr = 5.928 bar]

Bulliform cells are found in the upper epidermis of monocot leaves.

Maharaja Raj Singh constructed in 1662 the Rajsamand lake.

Hirakud dam is constructed on the river Mahanadi.

Shari’a is the Islamic law that is applied in a truly Islamic country. The Shari’a law owes its origin to the Holy book of Quran, Hadis (Law book of Islam) and teachings of Prophet Muhammad.

 In the medieval ages the Islamic world witnessed a big social and religious movement called Sufi movement. Sufi movement was the people-centric and not God-centric. It believed serving people was the real form of worship. Sufi movement has had many branches too. One group of Sufi preachers took very radical path. They were mystics who renounced material world took to the life of asceticism. Further they also rejected the supremacy of the Shari’a laws. Such sufis werecalled be-shari‘a.

 On the other hand , there were sufi saints who criticised the extravagant lifestyle of monarchs and Khaliphates but did not reject Shari’a laws. For them Shari’a laws were sacrosant. These Sufi saints have been called be-shari‘a.

(b) If mortality is more than natality, the density of population declines.

Correct answer is (a) 0.2

Rate of mortality of fishes per month = \(\cfrac{Number\,of\,individuals\,dead}{Total\,number\,of\,individual}\) 

=\(\cfrac{30}{150}\) 

= 0.2

These are the chemicals which help to transmitting, nerve impulses across the synapse e.g., acetylcholine.

Energy pumps are used against a concentrations gradient; in case of active transport. Active transport is carried out by membrane proteins. Pumps are proteins which use energy to carry substances across the cell membrane. The rate of transport reaches the maximum when all the protein transporters are being used or are saturated.

It performs two functions: 

(i) The stimuli are passed from and to the brain through the spinal cord. 

(ii) It is the centre of spinal reflex action.

b. The number of people present in 2015 is predicted as 17millions.

Vagus nerve is the longest cranial nerve.

A stimulus is sudden change in the environment which is strong enough to excite the nerve or muscle or organism as a whole.

i. The resources become limited at certain point of time, so no population can grow exponentially. 

ii. Every ecosystem or environment or habitat has limited resources to support a particular maximum number of individuals called its carrying capacity (K)

The nature of the nerve impulse is electrochemical.

Small birds have larger surface area relative to their volume, so they lose heat much faster, spend more energy to generate body heat.

Eornea can be transplanted as it non-vascular laver.

An aquenous fluid called ‘aqueous humour’ supplies nutrients to the cornea and thus maintain its shape.

Factors affecting diffusion are; Concentration gradient, Permeability of membrane; separating the substance, Temperature and Pressure.

When the antigens are coming in the form of living or dead microbes, the body of organism produce antibodies. This type of immunity is called active immunity.

When ready – made antibodies are directly injected to protect the body against foreign agents, it is called passive immunity.

Eg - Anti-venom.

A) Both ‘A’ and ‘R’ are correct and ‘A’ is supported by ‘R’.

Correct answer is 1:4