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[NOTE: Any functions is invertible if and only if it is bijective functions (i.e. one-one and onto)]

one-one function: A function f : A → B is said to be a one-one function or injective mapping if different elements of A have different images in B. Thus for x₁, x₂ ∈ A & f(x₁), f(x₂) ∈ B, f(x₁) = f(x₂) ↔ x₁= x₂ or x₁ ≠ x₂ ↔ f(x₁) ≠ f(x₂)

onto function: If range = co-domain then f(x) is onto functions.

So, We need to prove that the given function is one-one and onto.

Let x1, x2 ∈ R and f(x) = (9x² + 6x – 5).So f(x₁) = f(x₂) (9x₁² + 6x₁ – 5) = (9x₂² + 6x₂ – 5) on solving we get x₁=x₂

So f(x₁) = f(x₂) x₁= x₂, f(x) is one-one

Given co-domain of f(x) is [-5, ∞]

Let y = f(x) = (9x² + 6x – 5), So x = \(\frac{-1\pm\sqrt{y+6}}{3}\)[Range of f(x) = Domain of y]

So Domain of y = Range of f(x) = [-5, ∞]

Hence, Range of f(x) = co-domain of f(x) =[-5, ∞]

So, f(x) is onto function

As it is bijective function. So it is invertible

Invers of f(x) is f^-1(y) = \(\frac{-1\pm\sqrt{y+6}}{3}\)

We know that if A is a set with m elements i.e., n(A) = m, 

then, n[P(A)] = 2^m.

If A = ϕ, then n(A) = 0

∴n[P(A)] = 2^o = 1

Hence, P(A) has one element.

(i) {2, 4, ,6, 8} = {2n ; 1 \(\leq\) n \(\leq\) 5}

(ii) {2, 3, 5, 7, 11} = {P:P is a prime number less than 12}

(iii) {January , July, June} = {x: x is a a month of calendar starting with letter J}

(iv) {a, e, i, o, u} = {x : x is a vowel of english alphabet}

(v) {Tuesday, Thursday} = {x : x is a day in month starting with letter T}

(vi) {1, 4, 9, 16, 25} = {n² : 1 \(\leq\) n \(\leq\) 5}

(vii) {5, 10, 15, 20, 25, 30} = {5n : 1 \(\leq\) n \(\leq\) 6}

(p→q) ∧ ~ q → ~ q

≡ ~((p→q)∧ ~q) v ~q ( ∵ p → q ≡ ~p v q)

≡ ~ ((~p v q) ∧~q )

≡ ~((~p∧~q) v (q ∧ ~q)) v ~q

≡ ~ ((~p ∧ ~p ) v F) v ~q

≡ ~ (~p∧~p) v ~q

≡ (~(~p) v ~(~q)) v ~q

≡ (p v q) v ~q

≡ p v (q v ~q)

≡ p v T

≡ T Hence proved

i-0.6 and R is 12

In series arrangement you will get highest current output power because the current is not divided in series arrangement

Correct answer is:- D) Both B and C

Explanation:- 

Hydra has a tubular, radially symmetric body up to 10 mm (0.39 in) long when extended, secured by a simple adhesive foot known as the basal disc and also Starfish is radially symmetrical, more specifically the symmetry is pentaradial. 

Answer: B) Hydra and C) Starfish

Option (B) and (D) are correct.

Equivalent Resistance = \(1+(\frac{1}{6}+\frac{1}{3})^{-1}\)

                                         = 1 + 2 = 3 ohms

Equivalent current = V/R = 18/3 = 6 A

Power in 1 ohm resistor = V²/R = 6² =36 W

Power in 3 ohms resistor = 12²/3 = 48 W

Power in 6 ohms resistor = 12²/6 = 24 W

Therefore, power in the 6 ohms resistor is minimum and power in the 3 ohms resistor is maximum.

In the figure,

Equivalent resistance in the entire circuit = 100 ohms + 200 ohms = 300 ohms

Current flowing through the circuit = I = V/R = 12/300 = 0.04 A

Potential difference across resistance R₁ = I×R = 0.04 × 100 = 4 V

Therefore, the reading of the voltmeter will be 4 V.

\(\frac{(1+x)^{1/2}-2x(1+x){-1/2}}{(1+x)^{1/2}}\)

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

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

Given,

side of square =14cm

Area of larger Square = side²

=14²

= 196 cm²

There are four semi circle that means we have two complete circles of diameter of 2xcm then we can observe that

3+x+2x+x+3 = 14

i.e., x = 2

So radius of circle is 2 cm

Area of non shaded region = area of two circle + area of square of side 4cm

= 2.πr² + side²

= 25.12 + 16

= 41.12 cm²

Hence,

Required area of shaded region = 196 − 41.12 = 154.88 cm².

Correct option:

(B) tanx – x + c 

Correct option (A) 40 + x

Explanation:

Expanded form of two digit number = 10y + x .....(i) 

where, digit on unit place = x 

digit on ten’s place = y = 4

Putting y = 4 in equation (i) 

Number = 10 × 4 + x = 40 + x

Let radius of circular field be rm.

∴ Area of circular field is π r²m².

∴ Cost of ploughing circular field at the rate of Rs 0.75 per 

m² = πr²x Rs 0.75 = Rs (3/4 πr²)m²

∴ 3/4 πr² = 4158 (∵ Given cost = Rs 4158)

= r² = 4158 x 4/3 x 7/22 ( ∵ π = 22/7)

= r² = 63 x 4 x 7

= 9 x 7 x 4 x7

= 3² x 2² x 7²

= (3x2x7)²

= 42²

= r = 42 m

∴ Perimeter of circular field is P = 2 πr

= 2 x 22/7 x 42 = 264 m

∴ Cost of fencing the field = Rs 30 per meter

∴ Cost of fencing the circular field 

= Rs (264 x 30)

= Rs 7920

Correct option (A) (3, 1)

Explanation:

Solution of 2x – y = 5 ........(i) 

putting points (3, 1) in equation (1) R.H.S 

2(3) – (1) ⇒ 6 – 1 = 5

∴ L.H.S. = R.H.S. 

So solution is (3, 1)

Correct option:

(B) cotθ

Correct option:

(C) π/3

If every edge of the prism is 1 meter it means height and base are equal to 1 meter, i.e. a=h=1 m 

Formula of volume of prism 

V = 3√3/2 × a² × h 

V = 3√3/2 × 1 × 1

V = 3√3/2 m³

One atmospheric pressure is equal to 10 meters of water column or 760 mm of mercury. 

Manometer, Diaphragm gauges, Bellows, Strain gauges etc. 

Bubbler method is used to measure the level.

Orifice, Venturi tubes, flowrator (rotameter) etc. 

Mechanical (float) type and ultrasonic type methods are used to measure the flow. 

Barometer is used to measure atmospheric pressure. 

It is used for aluminium conductors while crimping to a lug or ferrule. It prevents corrosion of aluminium conductor due to oxidation and due to saline atmosphere. 

Pounds / inch² and kg / cm² .

(b) 10 MHz For beyond-the-horizon communication, it is necessary for the signal waves to travel a large distance. 10 KHz signals cannot be radiated efficiently because of the antenna size. The high energy signal waves (1GHz − 1000 GHz) penetrate the ionosphere. 10 MHz frequencies get reflected easily from the ionosphere. Hence, signal waves of such frequencies are suitable for beyond-the-horizon communication.

a. Information gathering instrumentation. 

b. Regulating instrumentation. 

c. Protective instrumentation. 

a. Less holdup. 

b. No support needed. 

c. Any actuator can be used. 

d. Quick opening and closing. 

e. Less space required. 

f. Used for low-pressure low temperature and large pipelines. 

Cavitation: Bubbles form in the liquids whenever there is pressure reduction inside the pump. These bubbles collapse when they approach high-pressure areas damaging pump internals. This is called the cavitation.

NPSH: Net Positive Suction Heat. It is the head available at the eye of the impeller corrected to vapour pressure. 

A. Centrifugal pumps – 

a. Single volute 

b. Double volute 

c. Diffuser type 

d. Mixed flow 

e. Axial flow 

f. Turbine or regenerative type 

B. Rotating pumps 

a. Screw type 

b. Gear type 

c. Vane type 

C. Reciprocating pumps 

a. Piston 

b. Plunger 

c. Bucket 

a. ON and OFF service. 

b. Throttling or regulating liquid flow. 

c. Avoid back flow. 

d. Regulating pressure. 

e. Relieving pressure. 

Baffle plates are used in heat exchanger so that the maximum heat can be transferred and to avoid tube sagging. 

Cable trays are grounded because to avoid any shocks to personnel incase of leakage. A grounding wire runs at the side of tray through a parallel groove clamp through out the length of the cable tray. If trays are one above another we can loop up the grounding wire to the tray below. This saves extra ground wire. 

Copper is less resistive than aluminium. 

Power cable: It is used for supplying current to load. It is of larger current carrying capacity available in single core, 2 cores, 3 cores, 3 cores, and 4 cores. Single core is available upto 1000 mm² . Usually power cables are of aluminium. These cables are graded for higher voltages and possess more cross section area. 

Control cable: Control cables are used for control purposes for logics, indication or annunciation etc. These are of lower current carrying capacity and voltage grading is also less. These are of less cross sectional area are available in pairs of 2,5,10,25,50 etc

Aluminium armouring for single core 1000 mm² is used so that heating will not take place due to the flux around the conductor, as the aluminium is a non-magnetic material. 

Armouring is required to protect the cores from mechanical damage. 

Feed valves are used to prevent back flow. 

Plant Emergency: Excessive release of radioactive ma : terial or high radiation fields in a section of the plant .

Site Emergency: Uncontrolled release of radioactive material or high radiation fields with in the site boundary 

Off-Site Emergency: High release of radioactive material from the plant resulting in significantly increased radiation fields and/or contamination levels extending to areas outside the site 

Emergency Planning Zones (EPZ): Emergency planning zone, defined around the plant up to 16 km, provides a basic geographic frame work for decision making on implementing measures as part of a graded response in the event of an emergency. The area around the Kaiga generating station is divided into the following Zones up to 16 km radius. 

Exclusion Zone: The exclusion Zone extends up to a distance of 1.6 km around the central plant zone of 0.7 km where no public habitation is permitted. This zone is physically isolated from out side areas by plant fencing and is under the control of Kaiga Generating Station. 

Sterilised Zone: Sterilised zone is an area where no new growth of population is permitted. Natural growth is however allowed in this Zone. This are extends up to a radius of 5 km from the central plant Zone. This Zone is defined to restrict the population to an easily transportable number in case of an emergency. 

Primary Zone: The primary Zone extends up to 8 km from central part Zone where protective measures like evacuation and sheltering are required against possible plume exposures during an Emergency. 

Secondary Zone: The secondary Zone extends up to 16 km from central plant Zone protective measures like sheltering control on food stuff are required against possible exposure from ingestion of radioactivity.