Here,      2f(x) = kx – 3,     13g(x) = 1x+2 and      f og-1(3) = -14 ∴ f (x) = kx – 32,     g(x) = 3x+2 For g-1(x);  Let y = g(x)   or, y = 3x+2 Interchanging .....

Here,  f(x) = 3x – 7 and g(x) = 4x-23 Let       y = f(x)  or, y = 3x – 7  Interchanging x and y, then x = 3y – 7  or, x + 7 = 3y  or, y = x+73 ∴ f-1(x) = x+73 We have given,   .....

Here,         f(x) = 3x + 4 and g(x) = 2(x + 1)                = 2x + 2. Now, f og(x) = f (g(x))              = f(2x + 2)        .....

Here,       f (x)=3x-2 and  f og(x) = 6x-2 .            So,       f (g (x))= 6x-2 or, 3g (x)-2 =6x-2          or, 3g (x) =6x   or, g(x) = .....

Here, f(x) = y = x + 3 Under the f-1,        x = y + 3 or, y = x - 3  So, f-1(x) = x - 3. Finally, f-1(1) = 1 - 3 = .....

Given function is cubic .....

If h-1(x) = {(a, x), (b, y), (c, z)},then  h(x) = {(x, a), (y, b), (z, .....

The graph of the  function is identity .....

The amplitude of the function y = cosx  is .....

The nature of the graph of constant function f(x) = c is a straight line // to x - .....

Tthe period of the function y = tan x  .....

The period of the function y = cosx  is .....

The y-intercept of function y = sinx is .....

The nature of graph of a quadratic function is .....

If f(x) = x, the value of f-1(x) is also .....

The y-intercept of function y = cosx is .....

Function which can be expressed in the form of f(x) = x is called identity .....

If f : P⟶Q and g : Q⟶R, then the composite function  from P to R is given by .....

If g(x) = {(a, 1), (b, 2), (c, 3)}, then g-1(x) = {(1, a), (2, b), (3, .....

The function from B to A is called inverse function and it is denoted by .....

Axis of symmetry of parabola y = ax2 + bx + c is x = .....

The x - coordinate of  vertex of parabola y = ax2 + bx + c is .....

The amplitude of the function y = -2 sin x   is |-2| = .....

The period of the function y = sin2x  .....

Here,  f(x) = x+12,  g(x) = x-52 and f og(x) = 6 Taking g(x) = x-52 Let y = g(x) or,  y = x-52 Interchanging the position of x and y       x = y-52 or, 2x = y – 5 ∴ y = 2x + 5  ∴ g-1(x) = .....

Here,       f(x) = 3x – 1 and      fog(x) = 6x + 5 or, f(g(x)) = 6x + 5 or, 3g(x) – 1 = 6x + 5 or, 3g(x) = 6x + 6  ∴   g(x) = 2x + 2. Now,  gof(x) = g(f(x)) = g(3x – .....

Here, f = {x, 5x – 13}, g = {x, 2x+73 } and g-1(x) = f of(x) Let y = g(x) or, y = 2x+73 Interchanging the position of x and y, then x = 2y+73 or, 3x = 2y + 7 or, 3x – 7 = 2y or, y = 3x-72 ∴ g-1(x) = 3x-72 Again,   f .....

Here,       f (x) = 2x+32 Let y = f(x) or,  y= 2x+32 Interchanging the position of x and y,       x = 2y+32 or, 2x= 2y+3 or, 2x- 3=2y or, y = 2x-32 ∴ f-1(x) =2x-32 Thus, the value of f-1(x) is .....

Here,       f(x) = 3x – 4 and      f-1g(x) = 3x+23 Let,       y = f(x)   or, y = 3x – 4  Interchanging the position of x and y then,       x = 3y .....

Here, f(x) = x/(2 - x) Taking     g(x) + f(1) = 5, or, g(x) + 1/(2-1) = 5 or, g(x) = 4. Again, taking  gof(4) = -8, or, g{4/(2-4)} = -8 or, g(-2)= -8 or, -2b - 1 = - 8 or, b = 7/2. Thus, g(x) = 7x/2 - 1 But g(x) = .....

Here,  f(x) = 2x + 3,  g(x) = x+52 ff(x) = g-1(x) So,        ff(x) = f(f(x))                = f(2x + 3)                = 2(2x + 3) + .....

Here, f(x) = 2x + 3 and g(x) = 2x + 5. Now, gof(x) = g(2x + 3)                      = 2(2x + 3) + 5                    = 4x + 11. Let y = .....

Here, f(x) = a/(3 - x) and  fof(5) = 4/5 or, f{a/(5 - 3)} = 4/5 or, f(a/2) = 4/5 or, a/(3 - a/2) = 4/5 or, 5a = 4(6 - a)/2 or, 5a + 2a = 12 or, a = 12/5. Similarly, fog(2) = 1/2 or, f{11 - 4b} = 1/2 or, (12/5)/{3 - (11 - 4b)} = .....

Here, fog(x) = 8x + 13 or, f(2x + 2) = 8x + 13 Replacing x by (x - 2)/2, we get f{2(x - 2)/2 + 2} = 8(x - 2)/2 + 13 or f(x) = 4x + 5 Again,  gof(x) = 20 or, g(4x + 5) = 20 or, 2(4x + 5) + 2 = 20 or, 8x + 12 = 20 or, 8x = 8 or, x = .....

Here, f(x) = x/(x - 2) Replacing x by 2 we get f(2) = 2/(2 - 2)       = 2/0      = ∞. Again,  gof(4) = - 8 or, g{f(4)} = - 8 or, g{4/(4 -2)} = - 8 or, g(2) = -8 or, 2b - 2 = - 8 or, 2b = - .....

Here,  gof(x) = g(3x - 2)  = 4(3x - 2) = 12x - 8 Again,  g(x) = 4x or, y = 4x Under the f-1,  x = 4y or, y = x/4 = g-1(x) Again,  f(x) = 3x - 2 or, y = 3x - 2 Under the f-1, x = 3y - 2 or, y = (x + 2)/3  = .....

Here, f(x) = 3x - 2 and  fog(x) = 6x - 2 or, 3(g(x)) - 2 = 6x - 2 or, 3g(x) = 6x or, g(x) = 2x. Now,  gof(x) = 8 or, g(3x - 2) = 8 or, 2(3x - 2) = 8 or, 3x - 2 = 4 or, 3x = 6 or, x = 2. Thus, x = .....

Here,       f-1(x) =x+32 Let y =x+32 Interchanging x and y then, x = y+32 or, 2x = y+3 or, 2x -3= y ∴    y =2x-3. Thus, f(x) = .....

Here, LHS = 2sin 75°×sin15° = 2sin(90 - 15)°×sin15° = 2cos15°×sin15°   [ Since, sin(90 - A) = cosA.] = sin(2.15o)    [ Since, sin2A = 2sinA cosA.] = .....

Here,        f (x) = 4x+3          or, y = 4x+3         Interchanging the position of x and y then,       x =4y+3 or, x-3= 4y  or, y = (x-3)/4   .....

Here, g(x) = x-23 and h (x)=3x+2. We have, gh(x) = g (h(x))= g(3x+2)                           = 3x+2-23                       .....

Here, f(x) = 3x – 7 and g(x) = 5x+23 Taking g(x) = 5x3 Let y = g(x) = 5x+23 Interchanging the position of x and y.        x = 5y+23 or, 3x = 5y + 2  or, 3x – 2 = 5y  or, y = 3x-25 ∴ g-1(x) .....

Solution: Steps:  1. Draw line BC = 6 cm. 2. From point B, mark an arc BA = 5.6 cm. 3. From point C, mark an arc CA = 6 cm. 4. Two arcs from 2. And 3. Intersect at A. 5. Join AB and AC. 6. ∆ABC is formed. 7. From point A, draw a line .....

Given, Steps i. Draw a line LM = 6.3 cm ii. From point M, make a line which makes ∡NML= 45°. iii. From point L, make a line which makes ∡NLM = 105° iv. The two lines from ii. and iii. Intersect at point N. v. From point M make a .....

Solution: Steps: i. Draw a line BC = 6.4 cm ii. From point B, mark an arc BA = 5.6 cm From point C, mark an arc CA = 6 cm. iii. Join AB and AC line. iv. From point A, draw a line parallel to side BC. v. From point B, mark an arc BD = 6.4 cm .....

Solution: 1. Draw a horizontal line. 2. Draw a line perpendicular to line 1. 3. Mark intersecting point of line 1 and line 2 as O, 4. From point O, mark an arc OS = OQ = 4 cm on horizontal line. 5. From point O, mark an arc OP = OR = 3 cm on .....

Solution: Steps: 1. Draw a line PQ= 5.9 cm 2. Draw a line making angle ∡SPQ = 75°. 3. From point P, mark an arc PS = 6.1 cm on the line 2. 4. From point S, mark an arc SR= 6.1 cm. From point Q, mark an arc QR = 5.9 cm. 5. Two arcs .....

Solution: Steps: 1. Draw a line AB = 5.5 cm 2. Draw a line making angle ∡DAB = 60°. 3. From point A, mark an arc AD = 4.3 cm on the line 2. 4. From point D, mark an arc DC = 4.7 cm. From point B, mark an arc BC = 5.7 cm. 5. Two arcs .....

Solution: Steps: i.   Draw line AB = 6 cm. ii. From point A, make a line which makes ∡BAD=60° with line AB.   iii. Mark a point D, with arc AD = 4 cm from A on the line. iv. From B, draw an arc BC = 4 cm ,From D, draw an arc CD .....

Solution: Steps: 1. Draw line AB = 4 cm. 2. From point A, mark an arc AC = 3.5 cm. 3. From point B, mark an arc BC = 3.2 cm. 4. Two arcs from 2. And 3. Intersect at C. 5. Join AB and AC. 6. ∆ABC is formed. 7. From point C, draw a line .....