As inverse of Differentiation MCQ Quiz - Objective Question with Answer for As inverse of Differentiation - Download Free PDF

Last updated on May 20, 2025

Latest As inverse of Differentiation MCQ Objective Questions

As inverse of Differentiation Question 1:

Find Y(x) if dy/dx = x3log(4x).

  1. log(4x)x3/3 + x4/4
  2. log(4x)x4/4 + x4/16
  3. log(4x)x3/3 - x4/4
  4. More than one of the above
  5. None of the above

Answer (Detailed Solution Below)

Option 5 : None of the above

As inverse of Differentiation Question 1 Detailed Solution

dy/dx = x3log(4x)

Y(x) = dydx

Y(x) = ∫x3log(4x)dx

Y(x)=log(4x)x4414x.4.x44dx=log(4x)x4414x3dx=log(4x)x44x416

As inverse of Differentiation Question 2:

If f'(x) = 3x2 - 7 such that f(0) = 2 then f(x) = ?

  1. x3 - 7x + 2
  2. x3 + 7x + 2
  3. x3 - 2x + 7
  4. None of these

Answer (Detailed Solution Below)

Option 1 : x3 - 7x + 2

As inverse of Differentiation Question 2 Detailed Solution

CONCEPT:

  • Let f(x) and g(x) be a function such that f'(x) = g(x) then g(x)dx=f(x)+C
  • [f(x)±g(x)]dx=f(x)dx±g(x)dx
  • xndx=xn+1n+1+C

CALCULATION:

Given: f'(x) = 3x2 - 7 and f(0) = 2.

As we know that, [f(x)±g(x)]dx=f(x)dx±g(x)dx

⇒ (3x27)dx=3x2dx7dx

As we know that, xndx=xn+1n+1+C

⇒ (3x27)dx=x37x+C

⇒ f(x) = x3 - 7x + C

∵ f(0) = 2

⇒ f(0) = C = 2

⇒ f(x) = x3 - 7x + 2

Hence, correct option is 1.

As inverse of Differentiation Question 3:

If f(x)=4x33x4 such that f(2) = 0 then f(x) = ?

  1. x4+1x31298
  2. x3+1x4+1298
  3. x3+1x41298
  4. x4+1x3+1298

Answer (Detailed Solution Below)

Option 1 : x4+1x31298

As inverse of Differentiation Question 3 Detailed Solution

CONCEPT:

  • Let f(x) and g(x) be a function such that f'(x) = g(x) then g(x)dx=f(x)+C
  • [f(x)±g(x)]dx=f(x)dx±g(x)dx

CALCULATION:

Given: f(x)=4x33x4 and f(2) = 0

As we know that, [f(x)±g(x)]dx=f(x)dx±g(x)dx

⇒ (4x33x4)dx=4x3dx3dxx4

⇒ 4x3dx3dxx4=x4+1x3+C

⇒ (4x33x4)dx=x4+1x3+C

⇒ f(x)=x4+1x3+C

∵ f(2) = 0

⇒ C = - 129/8

⇒ f(x)=x4+1x31298

Hence, correct option is 1.

As inverse of Differentiation Question 4:

Determine f(x) for f'(x) = 4x321x+e4x, and f(0) = 34

  1. x4 - 4xe4x4 + 1
  2. x4 - 4x+e4x4 + 1
  3. x4 - xe4x4 + 1
  4. x4 - 4x+e4x + 1

Answer (Detailed Solution Below)

Option 1 : x4 - 4xe4x4 + 1

As inverse of Differentiation Question 4 Detailed Solution

Concept:

Integral property:

 
  • ∫ xn dx = xn+1n+1+ C ; n ≠ -1
  • 1xdx=lnx + C
  • ∫ eax dx = eaxa+ C
  • ∫ adx = (ax/ln a) + C ; a > 0,  a ≠ 1
  • ∫ sin x dx = - cos x + C
  • ∫ cos x dx = sin x + C
 

Calculation:

Given f'(x) = 4x321x+e4x

f(x) = ∫ f'(x) dx

⇒ f(x) = 4x321x+e4x dx

⇒ f(x) = 4[x44]2[x1212]+[e4x4]

⇒ f(x) = x4 - 4xe4x4 + C

Now f(0) = 34

⇒  04 - 40e4(0)4 + C = 34

⇒ C - 14 = 34

⇒ C = 1

⇒ f(x) = x4 - 4\boldsymbolxe4x4 + 1

As inverse of Differentiation Question 5:

Let f(x) and g(x) be twice differentiable functions on [0, 2] satisfying f’’(x) = g”(x), f’(1) = 4, g’(1) = 6, f(2) = 3 and g(2) = 9. Then what is f(x) – g(x) at x = 4 equal to?

  1. -10
  2. -6
  3. -4
  4. 2

Answer (Detailed Solution Below)

Option 1 : -10

As inverse of Differentiation Question 5 Detailed Solution

Concept:

  • f(x)dx=f(x)+C1
  • f(x)dx=f(x)+C2


Calculation:

Given f’’(x) = g’’(x)

Integrating both sides, we get

f(x)dx=g(x)dx

⇒ f’(x) = g’(x) + C

Using f’(1) = 4 , g’(1) = 6

f'(1) = g'(1) + C

⇒ 4 = 6 + C

⇒ C = -2

⇒ f’(x) = g’(x) - 2

Integrating both sides, we get

f(x)dx=(g(x)2)dx

⇒ f(x) = g(x) - 2x + A

Using f(2) = 3 and g(2) = 9

⇒ f(2) = g(2) - (2 × 2) + A

⇒ 3 = 9 - (2 × 2) + A

⇒ A = 3 – 9 + 4 = -2

⇒ f(x) = g(x) - 2x - 2

⇒ f(x) - g(x) = -2x - 2

So, f(x) – g(x) at x = 4:

f(4) - g(4) = -(2 × 4) – 2 = -10

Top As inverse of Differentiation MCQ Objective Questions

If f(x)=x22kx+1, such that f(0) = 0 and f(3) = 15. Find the value of k.

  1. 5 / 3
  2. 3 / 5
  3. – 5 / 3
  4. – 3 / 5

Answer (Detailed Solution Below)

Option 3 : – 5 / 3

As inverse of Differentiation Question 6 Detailed Solution

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Concept:

Integration is the inverse process of differentiation and therefore it is called anti-differentiation.

i.e If g (x) = f’(x) then g(x)dx=f(x)dx=f(x)+C

(f(x)+g(x))dx=f(x)dx+g(x)dx

axndx=a×xn+1n+1+C

Calculation:

Given: f(x)=x22kx+1, such that f(0) = 0 and f(3) = 15.

Now, by integrating f’(x), we get

f(x)=f(x)dx=(x22kx+1)dx

f(x)=x22dxkxdx+dx

f(x)=x36k×x22+x+C

As it is given that, f(0) = 0 and f(3) = 15.

f(0)=C=0

f(x)=x36k×x22+x

f(3)=9292k+3=15

⇒ k = - 5 / 3

Let f(x) and g(x) be twice differentiable functions on [0, 2] satisfying f’’(x) = g”(x), f’(1) = 4, g’(1) = 6, f(2) = 3 and g(2) = 9. Then what is f(x) – g(x) at x = 4 equal to?

  1. -10
  2. -6
  3. -4
  4. 2

Answer (Detailed Solution Below)

Option 1 : -10

As inverse of Differentiation Question 7 Detailed Solution

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Concept:

  • f(x)dx=f(x)+C1
  • f(x)dx=f(x)+C2


Calculation:

Given f’’(x) = g’’(x)

Integrating both sides, we get

f(x)dx=g(x)dx

⇒ f’(x) = g’(x) + C

Using f’(1) = 4 , g’(1) = 6

f'(1) = g'(1) + C

⇒ 4 = 6 + C

⇒ C = -2

⇒ f’(x) = g’(x) - 2

Integrating both sides, we get

f(x)dx=(g(x)2)dx

⇒ f(x) = g(x) - 2x + A

Using f(2) = 3 and g(2) = 9

⇒ f(2) = g(2) - (2 × 2) + A

⇒ 3 = 9 - (2 × 2) + A

⇒ A = 3 – 9 + 4 = -2

⇒ f(x) = g(x) - 2x - 2

⇒ f(x) - g(x) = -2x - 2

So, f(x) – g(x) at x = 4:

f(4) - g(4) = -(2 × 4) – 2 = -10

Find Y(x) if dy/dx = x3log(4x).

  1. log(4x)x3/3 + x4/4
  2. log(4x)x4/4 + x4/16
  3. log(4x)x3/3 - x4/4
  4. log(4x)x4/4 - x4/16

Answer (Detailed Solution Below)

Option 4 : log(4x)x4/4 - x4/16

As inverse of Differentiation Question 8 Detailed Solution

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dy/dx = x3log(4x)

Y(x) = dydx

Y(x) = ∫x3log(4x)dx

Y(x)=log(4x)x4414x.4.x44dx=log(4x)x4414x3dx=log(4x)x44x416

If sinx is related to cosx, then 1|x|x21 is related to:

  1. sec -1 x
  2. cosec -1 x
  3. ax loga e
  4. ax loge a

Answer (Detailed Solution Below)

Option 1 : sec -1 x

As inverse of Differentiation Question 9 Detailed Solution

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These are the basic formulae of integration and differentiation.

sinxdx=cosxanddx|x|x21=sec1x

Find the antiderivative of f(x) = 12x3 – 15x2 + 64

  1. 12x35x2+64+C
  2. 36x230x+C
  3. 12x415x3+64x+C
  4. 3x45x3+64x+C

Answer (Detailed Solution Below)

Option 4 : 3x45x3+64x+C

As inverse of Differentiation Question 10 Detailed Solution

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Given, f(x) = 12x 3 – 15x2 + 64

Antiderivative of f(x) is F(x) + C, where C is a constant.

F(x)=12x315x2+64dx=12x4415x33+64x+C=3x45x3+64x+C

Find f(x) if f'(x) = x + m, f(1) = 5 and f(2) = 10.

  1. x227x2+1
  2. x22+7x2+1
  3. x227x21
  4. x22+7x21

Answer (Detailed Solution Below)

Option 2 : x22+7x2+1

As inverse of Differentiation Question 11 Detailed Solution

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Concept:

Integration is the inverse process of differentiation and therefore it is called anti-differentiation.

i.e If g (x) = f’(x) then g(x)dx=f(x)dx=f(x)+C

(f(x)+g(x))dx=f(x)dx+g(x)dx

axndx=a×xn+1n+1+C

Calculation:

Given: f'(x) = x + m, such that f(1) = 5 and f(2) = 10.

Now, by integrating f’(x), we get

f(x) = ∫f'(x)dx

f(x) = ∫(x + m)dx

f(x)=x22+mx+C

As it is given that, f(1) = 5 and f(2) = 10

f(1) = 1/2 + m + C = 5

2m + 2C = 9           ----(1)

f(2) = 2 + 2m + C = 10

2m + C = 8           ----(2)

Solving equations (1) and (2),

⇒ C = 1 and m = 7/2

So, f(x) = x22+7x2+1

If f'(x) = 3x2 - 7 such that f(0) = 2 then f(x) = ?

  1. x3 - 7x + 2
  2. x3 + 7x + 2
  3. x3 - 2x + 7
  4. None of these

Answer (Detailed Solution Below)

Option 1 : x3 - 7x + 2

As inverse of Differentiation Question 12 Detailed Solution

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CONCEPT:

  • Let f(x) and g(x) be a function such that f'(x) = g(x) then g(x)dx=f(x)+C
  • [f(x)±g(x)]dx=f(x)dx±g(x)dx
  • xndx=xn+1n+1+C

CALCULATION:

Given: f'(x) = 3x2 - 7 and f(0) = 2.

As we know that, [f(x)±g(x)]dx=f(x)dx±g(x)dx

⇒ (3x27)dx=3x2dx7dx

As we know that, xndx=xn+1n+1+C

⇒ (3x27)dx=x37x+C

⇒ f(x) = x3 - 7x + C

∵ f(0) = 2

⇒ f(0) = C = 2

⇒ f(x) = x3 - 7x + 2

Hence, correct option is 1.

If f(x) = sech x, then its antiderivative is

  1. log (cosh x) + C
  2. tan-1 (sin h x) + C
  3. 2 tanh-1 ex + C
  4. sinh x (½ (ex – e-x)) + C

Answer (Detailed Solution Below)

Option 2 : tan-1 (sin h x) + C

As inverse of Differentiation Question 13 Detailed Solution

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Given f(x) = sech x,

Antiderivative, F(x)=f(x)dx+C

sech x dx=1cosh xdx

⇒ Substitute u = sinh x

du = cosh x dx

 ducosh2 x=du1+sinh2 x

=du1+u2

⇒ substitute u = tan θ

du = sec2θ dθ

sec2θdθ1+tan2θ

=dθ

= θ + c

= tan-1(u) + c

= tan-1 (sin hx) + c

Comprehension:

For the next two (02) items that follow:

Consider f(x)=x22kx+1 such that f(0) = 0 and f(3) = 15

The value of k is

  1. 53
  2. 35
  3. 53
  4. 35

Answer (Detailed Solution Below)

Option 3 : 53

As inverse of Differentiation Question 14 Detailed Solution

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Concept:

f(x)dx=f(x)

 

Calculation:

f(x)=x22kx+1

f(x)dx=(x22kx+1)dx        (Integrate)

f(x)=x36kx22+x+c

 

Now, f (0) = 0

∴ 06k(0)2+0+c=0

c=0

f (3) = 15

336k(3)22+3=15

2769k2+3=15

9k2=9212

=152

k=159

=53

Hence, option (3) is correct

Comprehension:

For the next two (02) items that follow:

Consider f(x)=x22kx+1 such that f(0) = 0 and f(3) = 15

f(23) is equal to

  1. -1
  2. 13
  3. 12
  4. 1

Answer (Detailed Solution Below)

Option 4 : 1

As inverse of Differentiation Question 15 Detailed Solution

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Calculation:

We have, f(x)=x22kx+1 

f(x)=ddx(f(x))

=2x2k

⇒ x - k

And k = -5/3

f(23)=(23)(53)

=23+53

= 1 

Hence, option (4) is correct.

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