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To Prove: 

cos ^-1 (1 – 2x ²) = 2 sin ^-1 x 

Formula Used: cos 2A = 1 – 2 sin ² A 

Proof: 

LHS = cos ^-1 (1 – 2x ²) … (1) 

Let x = sin A … (2) 

Substituting (2) in (1), 

LHS = cos ^-1 (1 – 2 sin ² A) 

= cos ^-1 (cos 2A) 

= 2A 

From (2), A = sin -1 x, 

2A = 2 sin ^-1 x 

= RHS 

Therefore, LHS = RHS 

Hence proved.

cos ^-1 (4x³ – 3x) = 3 cos ^-1 x, 1/2 ≤ x ≤ 1

Take x = cos θ

Where θ = cos ^-1 x

Here LHS = cos ^-1 (4x³ – 3x)

By substituting the value of x

= cos ^-1 (4 cos³θ – 3 cos θ)

So we get

= cos ^-1 (cos3θ)

= 3θ

By substituting the value of θ

= 3 cos ^-1 x

= RHS

Hence, it is proved.

To Prove: 

sin ^-1 (3x – 4x ³) = 3 sin ^-1 x 

Formula Used: sin 3A = 3 sin A – 4 sin ³ A 

Proof: 

LHS = sin ^-1 (3x – 4x ³) … (1) 

Let x = sin A … (2) 

Substituting (2) in (1), 

LHS = sin ^-1 (3 sin A – 4 sin ³ A) 

= sin ^-1 (sin 3A) 

= 3A 

From (2), A = sin ^-1 x, 

3A = 3 sin ^-1 x 

= RHS 

Therefore, LHS = RHS 

Hence proved.

To Prove: 

cos ^-1 (4x ³ – 3x) = 3 cos ^-1 x 

Formula Used: cos 3A = 4 cos ³ A – 3 cos A 

Proof: 

LHS = cos ^-1 (4x ³ – 3x) … (1) 

Let x = cos A … (2) 

Substituting (2) in (1), 

LHS = cos ^-1 (4 cos ³ A – 3 cos A) 

= cos ^-1 (cos 3A) 

= 3A 

From (2), A = cos ^-1 x, 

3A = 3 cos ^-1 x 

= RHS 

Therefore, LHS = RHS 

Hence proved.

To Prove: \(\tan^{-1}\left(\frac{3\mathrm x-\mathrm x^3}{1-3\mathrm x^2}\right)=3\tan^{-1}\mathrm x\)

Formula Used: \(\tan3A=\frac{3\tan A-tan^2A}{1-3\tan^2A}\)

Proof: 

LHS \(=\tan^{-1}\left(\frac{3\mathrm x-\mathrm x^2}{1-3\mathrm x^2}\right)\)… (1) 

Let x = tan A … (2) 

Substituting (2) in (1), 

LHS \(=\tan^{-1}\left(\frac{3\tan A-tan^2a}{1-3\tan^2A}\right)\)

= tan ^-1 (tan 3A) 

= 3A 

From (2), A = tan ^-1 x, 

3A = 3 tan ^-1 x 

= RHS 

Therefore, LHS = RHS 

Hence proved.

RHS = sin²A + sin²(A -B) – 2 sinA cosB sin(A -B) 

= sin²A + sin(A -B) [sin(A –B) – 2 sinA cosB] 

We know that sin(A –B) = sinA cosB – cosA sinB 

= sin²A + sin(A -B) [sinA cosB – cosA sinB – 2 sinA cosB] 

= sin²A + sin(A -B) [-sinA cosB – cosA sinB] 

= sin²A - sin(A -B) [sinA cosB + cosA sinB] 

We know that sin(A +B) = sinA cosB + cosA sinB 

= sin²A – sin(A –B) sin(A +B) 

= sin²A – sin²A + sin²B 

= sin²B = LHS 

Hence proved

Correct Answer is \(\frac{4}{5}\)

Let, x = \(cos^{-1}\frac{3}{5}\) 

⇒ cos x = \(\frac{3}{5}\)

Now , sin(\(cos^{-1}\frac{3}{5}\)) becomes sin (x)

Since we know that sin x =\(\sqrt{1-cos^2x}\)

\(=\sqrt{1-(\frac{3}{5})^2}\)

\(sin\left(cos^{-1}\frac{3}{5}\right)\)= Sin x= \(\frac{4}{5}\)

Let the age of father = x years

And the age of his son = y years

According to the question,

x = 3 + 3y ...(i)

Three year here,

Father’s age = (x + 3) years

Son’s age = (y + 3) years

According to the question,

(x + 3) = 10 + 2(y + 3)

⇒ x + 3 = 10 + 2y + 6

⇒ x = 2y + 13 …(ii)

From Eq. (i) and (ii), we get

3 + 3y = 13 + 2y

⇒ 3y – 2y = 13 – 3

⇒ y = 10

On putting the value of y = 7 in Eq. (i), we get

x = 3 + 3(10)

⇒ x = 3 + 30

⇒ x = 33

Hence, the age of father is 33 years and the age of his son is 10 years.

Let ‘x’ year be the present age of father and ‘y’ year be the present age of son.

Four years hence, it has relation by given condition,

(x+4) = 4(y+4)

x+4 = 4y+16

x- 4y-12 = 0 …(i)

and initially, x = 6y …(ii)

On putting the value of from Eq. (ii) in Eq. (i), we get

6y - 4y - 12 = 0

2y = 12

Hence, y = 6

Putting y = 6, we get x = 36.

Hence, present age of father is 36 years and age of son is 6 years.

For commutativity, condition that should be fulfilled is a * b = b * a

Consider a*b = 3ab/5 = 3ba/5 = b*a

∴ a*b = b*a

Hence, * is commutative. 

For associativity, condition is (a * b) * c = a * (b * c)

Consider (a*b)*c = (3ab/5)*c = 9ab/25

and a*(b*c) = a*(3bc/5) = 9ab/25

Hence, (a * b) * c = a * (b * c) 

∴ * is associative. 

Let e ∈ Q be the identity element, 

Then a * e = e * a = a

⇒ 3ae/5 = 3ea/5 = a ⇒ e = 5/3

Correct answer is (A) 50°

Correct answer is (D) (22.5)°

Correct answer is (B) π/6

Correct option is (A) 128/3 cm³

Volume of pyramid = 1/3 x Base area x Height

= 1/3 x 16 x 8

= 128/3 cm³

Correct option is  A) 128/3 cm³ 

Correct option is (A) 1/3 x Area of base x height

Volume of a pyramid \(=\frac{1}{3}\times\text{Area of base}\times height\)

A) 1/3 × Area of base × height

Correct option is (D) 60 cm³

Base of square pyramid is a square of edge 6 cm.

\(\therefore\) Base area \(=6^2=36\,cm^2\)

Height of pyramid = 5 cm

\(\therefore\) Volume of square pyramid \(=\frac13\times\) Base area \(\times\) Height of pyramid

\(=\frac13\times36\times5=12\times5\)

\(=60\,cm^3\)

Correct option is D) 60 cm³

Answer: (a) = 144\(​​\sqrt{2}\)  cu.cm.  

Volume of a regular tetrahedron = \(\frac{\sqrt{2}}{12} (edge)^3\)  

= \(\frac{\sqrt{2}}{12} (12)^3\) 

= 144 \(​​\sqrt{2}\)  cu.cm.

A regular tetrahedron is a solid figure whose all edges are equal and, all four faces including the base are congruent equilateral triangles.

♦ Height =\(\sqrt{\frac{2}{3}}\)edge

♦ Slant height = \(\sqrt{\frac{3}{2}}\)edge

♦ Lateral surface area =\(\frac{3\sqrt{3}}{4}(edge)^2\)

♦ Total surface area = \(\sqrt{3}\) (edge)² 

♦ Volume = \(\frac{\sqrt{2}}{12}\) (edge)³ .

Volume of a regular tetrahedron = \(\frac{\sqrt{2}}{12} (edge)^2\) = \(\frac{\sqrt{2}}{12}\times 16^2\) 

= \(\frac{\sqrt{2}\times 256}{12} \) cm³  = \(\frac{64\sqrt{2}}{3} \) cm³

Lateral surface area = \(\frac{3\sqrt{3}}{4} (edge)^2 =\frac{3\sqrt{3}}{4} 16^2\) 

= 192\(\sqrt{3}\) cm²

Total surface area =\({\sqrt{3}} (edge)^2 = {\sqrt{3}} \times 16^2\) cm² = 256\(\sqrt{3}\) cm²

Answer: (a) = 16:3

Slant surface or Lateral surface area of a pyramid

=\(\frac{1}{2}\) × Perimeter of base × Slant height

Given, 12 =  \(\frac{1}{2}\) × 4 × a × 4

where, each side of the square = a metres

⇒ \(a = \frac{24}{16} =\frac{3}{2}\) m.

∴ Area of base = \(\big(\frac{3}{2}\big)^2\, m^2 = \frac{9}{4}\, m^2\)  

∴  Reqd. ratio = 12 : \(\frac{9}{4}\)  

= 16:3.

Volume of a pyramid = \(\frac{1}{3}\) × (Area of the base) × Height

= \(\frac{1}{3} \times \frac{3\sqrt{3}}{2} \times (5)^2 \times 30 \)  = 649.52 m³ ≈ 650m³  

(Area of a regular hexagon of side 'a' = \(\frac{3\sqrt{3}}{2} a^2\) )

Answer: (b)  0.577 m² 

Volume of the pyramid =\(\frac{1}{3}\) × Area of base × Height

=  \(\frac{1}{3}\times \frac{\sqrt{3}}{4} \times (1)^2 \times 4 \,m^2 = \frac{1.732}{3}\) 

= 0.577 m² (approx)

Let ‘h’ be the height of the pyramid and ‘a’ the length of each side of the base of an equilateral triangle.

Then, slant edge = \(\sqrt{h^2+\frac{a^2}{3}}\) 

⇒ 5 = \(\sqrt{h^2+\frac{16}{3}}\) ⇒ 25 - \(\frac{16}{3}\) =h² ⇒ h² = \(\frac{59}{3}\) ⇒ h = \(\sqrt{\frac{59}{3}}\) m

Slant height = \(\sqrt{h^2+\frac{a^2}{12}}\)  = \(\sqrt{\frac{59}{3}+\frac{16}{12}}\)  =  \(\sqrt{21}\) m

∴ Lateral surface area = \(\frac{1}{2}\)(Perimeter of base × Slant height) = \(\frac{1}{2}\)(4+ 4+ 4)× \( \sqrt{21}\) m²  

= \(6 \sqrt{21} \) m²  

Volume of the pyramid =   \(\frac{1}{3}\)(Area of base × height)  = \(\frac{1}{3} \times \frac{\sqrt{3}}{4} \times 4^2 \times \sqrt{\frac{59}{3}}\) m³

= \(\frac{4\sqrt{59}}{3}\) m³

Answer: (c) = 8 units  

Let a be the length of each side of the base, h be the height and l be the slant height of the pyramid. 

Here, a = 4 cm.

∴  Slant height (l) = \(\sqrt{h^2 +\frac{a^2}{12}}\)  = \(\sqrt{h^2 +\frac{16}{12}}\)  = \(\sqrt{h^2 +\frac{4}{3}}\)   

According to the given question, 

Lateral surface area + Area of the base = 3 (Volume)

⇒ \(\frac{1}{2}\times 12 \times \sqrt{h^2+\frac{4}{3}} +\frac{\sqrt{3}}{4} \times (4)^2\)   

= \(3 \times \frac{1}{3}\times \big(\frac{\sqrt{3}}{4} \times 4^2 \times h\big)\)  

⇒ \(6\sqrt{h^2+\frac{4}{3}} +4\sqrt{3}\) = \(4\sqrt{3}\) h

⇒ \(6\sqrt{h^2+\frac{4}{3}} \) = \(4\sqrt{3}(h-1)\) 

⇒ \(36\big(h^2+\frac{4}{3}\big) =48(h-1)^2\)  

⇒ \(3\big(h^2+\frac{4}{3}\big) = 4(h-1)^2\)  

⇒ 3h² + 4 = 4h² – 8h + 4 

⇒ h² – 8h = 0 ⇒ h(h – 8) = 0 

⇒ h = 8 as h ≠ 0.

Two example of: -

i. Solid fuel – Wood, Coal

ii. Liquid fuel – Kerosene, Petrol

Air dissolved in water starts escaping in the form of tiny bubbles due to heat from the sun.

Incomplete combustion of fuels produces a very poisonous gas called carbon monoxide.

(a) Mask
(b) To save himself from dirt/polluted air.
(c) Air quality of the place is not good. It is due to the smoke and gases emitted by the automobiles along with dust particles present in the air.

Correct option is B) Toxic substances from paints

The eathe,s temperature is increasing due to global warming which is due to Increased use of fossil fuel.

Hint: It is related to global warming due to formation of carbon dioxide and some other gases.

(a) Oxygen and nitrogen together make up 99 per cent of the air.
(b) Aquatic animals use dissolved air for respiration.
(c) Air in motion is called wind.

The component of air used by green plants to make their food, is carbon dioxide.

Earthquake can be tackled by the following: 

1. Government should distribute network of instrument related to earthquake measures. 

2. Alert should be declared through TV and radio to make people aware of earthquake. 

3. At individual level people should leave the place and go in open area. 

4. Switch off the switch of electricity and regulator of cylinder. 

5. Irrespective of castae, creed, religion, etc. people should help each other on the ground or humanity during such disaster.

When production of crops and fodder fail due to less rainfall, it is called double famine.

Akal, Dvikal and Trikal are the three forms of famine.

The problems faced by the flood prone regions are: 

1. In India more than 2000 deaths occur due to flood every year. 

2. 80 lakh hectares of land is flood-prone. 

3. 35 lakh hectare of land crop destroys every year. 

4. Life in 3 crore acre land becomes unbalanced. 

5. Loss of one thousand crore rupees every year. 

6. The greatest effect of flood occurs on animal wealth. 

7. More than 12 lakh animals and houses destroy every year. 

8. 60% loss caused by flood in India held in Bihar, Uttar Pradesh, West Bengal, Assam and Odisha are main. 

9. Floods disturb communication, transport, sources of water, and destroy crops and cause epidemics.

Flood Management 

1. In 1954, National Flood Control Plan was launched to check the great loss done by flood. Under this plan bank of rivers, water flowing drains were built. 

2. Under multi-purpose projects scheme dams were built on Mahanadi, Damodar, Satluj, Vyas, Chambal, Narmada rivers. 

3. To control flood regions of river origin should be covered with forests it would check soil erosion and result in check on loss due to floods. 

4. Exploitation of forests should be ban. 

5. Water holding capacity of river should be increased by removing debris and spreading it on banks. 

6. In 1954, floods forecast/prior information organization was established. At district headquarters flood control cells were established. 

7. In 2007, the NDMA (National Disaster management Authority) of the Government of India pointed about 13 areas of flood management. These include zoning, forecasting, risk maps, structural measures, maintenance of river banks, river basin management, construction of flood shelters, etc.

8. Meterological department and irrigation department keeps a regular observation on amount of rainfall and its flow during rainy season. People are informed regularly through Television. 

9. At individual level, people should listen radio and watch television regularly in rainy season. If they are living in flood prone area and should follow the instructions and orders of government strictly. Electric appliances should switch off. Precious things food etc. should be shifted at secure places. Vehicles, animals, etc. should be shifted at secure places.

Sea storms of Arabian Sea during April to June are parallel to west to West Bengal’s cyclones. They affect Andhra Pradesh, Telangana, Odisha and West Bengal during October to December.

Correct option is B) CO₂

Correct answer is (d) Oxygen

(a) Along with government and local municipality corporations, it is also the duty of every citizen to help in garbage disposal. A clean environment is necessary to keep us healthy and also to avoid spread of diseases. We should throw garbage at proper places, such as in dustbins so that Safai Karamcharis can collect the garbage easily.

(b) It is possible to reduce the problems relating to disposal of garbage, if we adopt the following means:

• The garbage should be thrown at proper places. It should not be thrown on streets, roads, parks, etc.
• The part of the garbage that can be reused should be separated from the one that cannot be used. The
• non-useful components should be disposed off at landfill areas.

The layer of air around the earth is known as atmosphere.

The causes of flood are: 

1. Deforestation and pastures destruction 

2. Destruction of traditional water resources 

3. Building of constructions ignoring the natural drainage.

The layer of air around the earth is known as Atmosphere.

The main causes of landslides are: 

1. Formation of rocks 

2. Slope of earth

3. Vibrations caused by earthquake 

4. Prolonged rainfall and seepage 

5. Deforestation 

6. Uncontrolled human development such as road, bridges, etc. 

7. Undercutting of cliffs and banks by waves or river erosion. 

8. Volcanic eruptions

Carbon dioxide.