Q: The diagram shows a uniform bar resting on two supports, P and Q.− − The weight of the bar is 4.0 N.What is the force exerted on the bar by support P?

Balancing the forces,TP+TQ=W…(1)Balancing the torque,(TP×0)+(TQ×50)=W×40(0)+(TQ×50)=4×40TQ=3.2 NSubstituting the value of TQ in eqn. (1)TP+3.2=4TP=0.80 N

Q: f(x)=7−2x, g(x)=x10,x=0, h(x)=27xFind hg(30).

hg(x)=h(g(x))=h(x10)=27x10hg(30)=273010=2731=3

Q: f(x)=7−2x, g(x)=x10,x=0, h(x)=27xSolve.g(2x+1)=4

g(2x+1)=4(2x+1)10=410=4(2x+1)10=8x+48x=10−48x=6x=86x=0.75

Q: f(x)=7−2x, g(x)=x10,x=0, h(x)=27xSimplify, giving your answer as a single fraction.f(x)1+g(x)

====f(x)1+g(x)(7−2x)1+x10x(7−2x)x+10(7−2x)7x−2x2x+70−20x7x−2x270−19x

Question 1

Which process releases heat to maintain a constant body temperature?

Accepted Answer

Thermoregulation is a mechanism by which mammals maintain body temperature with tightly controlled self-regulation independent of external temperatures. Temperature regulation is a type of homeostasis and a means of preserving a stable internal temperature in order to survive. In mammals, one way the body dissipates excess heat to maintain homeostasis is through exhalation. Air that enters the lungs is warmed by body heat and then exhaled. So, the correct answer is respiration.

Question 2

The diagram shows a uniform bar resting on two supports, P and Q.− − The weight of the bar is 4.0 N.What is the force exerted on the bar by support P?

Accepted Answer

Balancing the forces,TP​+TQ​=W…(1)Balancing the torque,(TP​×0)+(TQ​×50)=W×40(0)+(TQ​×50)=4×40TQ​=3.2 NSubstituting the value of TQ​ in eqn. (1)TP​+3.2=4TP​=0.80 N

Question 3

An aluminium block has a mass of 200 g.The specific heat capacity of aluminium is 900 J/(kg∘C).How much energy is needed to increase the temperature of the block from 20∘C to 110∘C ?

Accepted Answer

mass, m=200 g=0.2 kgspecific heat, c=900 J/(kg∘C)change in temperature, ΔT=110∘−20∘=90∘Heat energy, Q=mcΔT=0.2×900×90=16200 J

Question 4

f(x)=3x−5;g(x)=2x(a) Find fg(3).(b) Find f−1(x).

Accepted Answer

(a)  we have,f(x)=3x−5;g(x)=2xg(3)=23=8fg(3)=f(g(3))=3×8−5=19(b) let y=3x−5Interchange x and  y, x=3y−53y=x+5y=3x+5​f−1(x)=3x+5​

Question 5

f(x)=7−2x, g(x)=x10​,x=0, h(x)=27xFind f(−3)

Accepted Answer

f(x)=7−2xf(−3)=7−2×(−3)=7+6=13

Question 6

f(x)=7−2x, g(x)=x10​,x=0, h(x)=27xFind hg(30).

Accepted Answer

hg(x)=h(g(x))=h(x10​)=27x10​hg(30)=273010​=2731​=3

Question 7

f(x)=7−2x, g(x)=x10​,x=0, h(x)=27xFind f−1(x).

Accepted Answer

f(x)=7−2xlet f(x)=yy=7−2xInterchanging x and y ,x=7−2y2y=7−xy=27−x​f−1(x)=27−x​

Question 8

f(x)=7−2x, g(x)=x10​,x=0, h(x)=27xSolve.g(2x+1)=4

Accepted Answer

g(2x+1)=4(2x+1)10​=410=4(2x+1)10=8x+48x=10−48x=6x=86​x=0.75

Question 9

f(x)=7−2x, g(x)=x10​,x=0, h(x)=27xSimplify, giving your answer as a single fraction.f(x)1​+g(x)

Accepted Answer

====​f(x)1​+g(x)(7−2x)1​+x10​x(7−2x)x+10(7−2x)​7x−2x2x+70−20x​7x−2x270−19x​​

Question 10

f(x)=7−2x, g(x)=x10​,x=0, h(x)=27xFind h−1(19683).

Accepted Answer

h(x)=27xlet y= h(x)y=27xInterchanging x and yx=27yConverting into log form,log27​x=yor, y=log27​xThen,  h−1(x)=log27​x h−1(19683)=log27​19683=3

Question 11

f(x)=2x−3 g(x)=9−x2 h(x)=3xFind(i) f(4),(ii) hg(3),iii) g(2x) in its simplest form,(iv) fg(x) in its simplest form.

Accepted Answer

(i) f(4)=8−3=5(ii) hg(3)=3g(3)=39−9=1(iii) g(2x)=9−4x2(iv) fg(x)=2(9−x2)−3=15−2x2

Question 12

f(x)=2x−3 g(x)=9−x2 h(x)=3xFind f−1(x)

Accepted Answer

y=2x−3x=2y+3​

Question 13

f(x)=2x−3 g(x)=9−x2 h(x)=3xFind x when 5f(x)=3.

Accepted Answer

5f(x)=35(2x−3)=310x−15=3x=59​

Question 14

f(x)=2x−3 g(x)=9−x2 h(x)=3xSolve the equation gf(x)=−16

Accepted Answer

gf(x)=−169−(2x−3)2=−169−4x2−9+12x=−164x2+12x+16=0x=−1,4

Question 15

f(x)=2x−3 g(x)=9−x2 h(x)=3xFind x when h−1(x)=−2.

Accepted Answer

x=log3​h(x)log3​h(x)=−2h(x)=91​

Question 16

f(x)=x+23​,x=−2, g(x)=8x−5, and  h(x)=x2+6Work out g(41​).

Accepted Answer

g(x)=8x−5 g(41​)=8(41​)−5=2−5=−3

Question 17

f(x)=x+23​,x=−2 g(x)=8x−5 h(x)=x2+6Work out ff(2).

Accepted Answer

f(x)=x+23​,then,  ff(2)=f[f(2)]ff(2)=f[2+23​]=f[43​]ff(2)=f(0.75)=0.75+23​=2.753​=1112​ORf(x)=x+23​,then,  ff(x)=f[f(x)]=f(x+23​)=x+23​+23​then,  ff(2)=2+23​+23​=43​+23​=0.75+23​=1112​

Question 18

f(x)=x+23​,x=−2 g(x)=8x−5 h(x)=x2+6Find gg(x), giving your answer in its simplest form.

Accepted Answer

g(x)=8x−5gg(x)=g(8x−5)=8(8x−5)−5gg(x)=64x−40−5=64x−45gg(x)=64x−45

Question 19

f(x)=x+23​,x=−2 g(x)=8x−5 h(x)=x2+6Find g−1(x)

Accepted Answer

g(x)=8x−5let g(x)=yy=8x−5Interchanging x and y ,x=8y−58y=x+5y=8x+5​f−1(x)=8x+5​

Question 20

f(x)=x+23​,x=−2 g(x)=8x−5 h(x)=x2+6Write g(x)−f(x) as a single fraction in its simplest form.

Accepted Answer

g(x)−f(x)=8x−5−x+23​=x+2(8x−5)(x+2)−3​=x+2x(8x−5)+2(8x−5)−3​=x+28x2−5x+16x−10−3​=x+28x2+11x−13​

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