MCQs' Basic Applications of Differential and Integral Calculus in Business and Economics

Q1: If f(x) = (x – 1)(x)(x + 1), then dy/dx =?
(a) 3x² – 1 
(b) 3x² + 1 
(c) x² – 3 
(d) x² + 3
Ans: (a)

Sol: 

 f(x) = (x – 1) (x) (x – 1) = x(x² – 3x + 2) = x³ – 3x² + 2x  

Differentiate:  f '(x) dy/dx = 3x²- 1

Q2: If y = x(x – 1) (x – 2), then dy/dx is:
(a) 3x² – 6x + 2
(b) –6x² + 2
(c) 3x² + 2
(d) 3x³ + 5
Ans: (a)

Sol:
 f(x) = (x – 1)(x – 2) = x(x² – 3x + 2) = x³ – 3x² + 2x  

Differentiate: dy/dx = 3x²+ 2

Q3: If a function is given by f(x) = e^3x, what is the derivative of the function?
(a) 3e^3x
(b) e^3x
(c) 3xe^3x
(d) 3e^3x+ 3
Ans: (c)

Sol:
f(x) = e^3x
f'(x) = d/dx (e^3x) = 3e^{3^3x

Q4: f(x) = x(x² – 2), then find dy/dx
(a) 3x² – 2
(b) 3x² + 2
(c) x² – 3
(d) x²
Ans: (a)

Sol:
y = f(x) = x²(x – 2) = x³ – 2x  
dy/dx = 3x² – 2

Q5: What is the differential function of
(a)
(b)
(c)
(d)
Ans: (b)

Sol: 
Let's differentiate;
f(x) = (x² + 2)^1/2
Using the chain rule:

So, the differential is;

Q6: The slope of the tangent to the curve y = x² – x at the point where the line y = 2 cuts the curve in the first quadrant is: 
(a) 2
(b) 3
(c) –3 
Ans: (b)

Sol:
Given, y = x² – x

At y = 2

2 = x² – x 
⇒ x² – x – 2 = 0

(x – 2)(x + 1) = 0 ⇒ x = 2 or x = –1

Since first quadrant ⇒ x = 2

Derivative: dy/dx = 2x – 1

At x = 2

dy/dx = 2(2) – 1 = 3

Q7: The cost function for the production of x units of a commodity is given by C(x) = 2x³ – 15x² + 36x + 15. The cost will be minimum when 'x' equal to:
(a) 3 
(b) 2 
(c) 1 
(d) 4
Ans: (a)

Sol:
The cost function is:

C(x) = 2x³ – 15x² + 36x + 15

C'(x) = 6x² – 30x + 36

Put C'(x) = 0, we get

6x² – 30x + 36 = 0

⇒ x² – 5x + 6 = 0

⇒ (x – 2)(x – 3) = 0

So, critical points are x = 2 and x = 3

Now, C''(x) = 12x – 30

At x = 2:

C''(2) = 12(2) – 30 = –6 (Negative ⇒ Maximum)

At x = 3:

C''(3) = 12(3) – 30 = 6 (Positive ⇒ Minimum)

Therefore, the cost will be minimum when x = 3

Q8: The maxima and minima of the function y = 2x³– 15x² + 36x + 10 occurs respectively at
(a) x = 2 and x = 3 
(b) x = 1 and x = 3
(c) x = 3 and x = 2 
(d) x = 3 and x = 1
Ans: (a)

Sol:
Given: y = 2x³ – 15x² + 36x + 10

On differentiating wrt x, we get

dy/dx = 6x²  – 30x + 36  ...(i)

and d² y/dx² = 12x – 30 ...(ii)

Now, substitute dy/dx = 0

⇒ 6x²  – 30x + 36 = 0

⇒ 6(x² – 5x + 6) = 0

⇒ x²  – 5x + 6 = 0

⇒ x²  – 3x – 2x + 6 = 0

⇒ x(x – 3) – 2(x – 3) = 0

⇒ (x – 3)(x – 2) = 0

⇒ x = 3 and x = 2

Now, put x = 3 in (ii), we get

d² y/dx²  = 36 – 30 = 6 > 0, attaining minima

Now, put x = 2 in (ii), we get

d² y/dx²  = 24 – 30 = –6 < 0, attaining maxima

Therefore, the given function attains maxima at x = 2 and minima at x = 3.

Hence, the correct option is (a).

Q9: Find the gradient of the curve y = 3x² – 6x + 4 at the point (1, 2). 
(a) 1 
(b) –1
(c) 0 
(d) 2
Ans: (c)

Sol: 
dy/dx = 6x – 6
At x = 1 
dy/dx = 6(1) – 6 = 0

Q10: The gradient of the curve y = 2x³ – 5x² – 3x at x = 0 is 
(a) 3 
(b) –3 
(c) 1/3 
(d) None of these
Ans: (b)

Sol: 
Given: y =2x³ – 5x² – 3x 

Therefore, the gradient of the curve is given by:

dy/dx = 6x² – 10x – 3

At x = 0,

y'(0) = –3

Q11: Determine f(x), given that f'(x) = 12x² – 4x and f(-3) = 17
(a) f(x) = 4x³– 2x² + 143
(b) f(x) = 6x² – x⁴ + 137
(c) f(x) = 3x⁴ – x³ – 137
(d) f(x) = 4x³ – 2x² – 143
Ans: (a)

Sol:
f(x) = ∫ f'(x) dx

= ∫ (12x² – 4x) dx

f(x) = 4x³ – 2x² + C

f(–3) = 4(–3)³ – 2(–3)² + C

17 = –108 – 18 + C

C = 143

⇒ f(x) = 4x³ – 2x² + 143

Q12:
(a) log |x| + c 
(b) log |log x| + c 
(c) (log x)² + c 
(d) None of these
Ans: (b)

Sol:
= log |log x| + c

Q13:  The equation of the curve in the form y = f(x) if the curve passes through the point (1, 0) and f'(x) = 2x – 1, then f(x) is:   
(a) y = x² – x 
(b) x = y² – y 
(c) y = x²
(d) None of these
Ans: (a)

Sol: 
f(x) = ∫(2x−1)dx = x² − x + k
Using point (1, 0): 
0 = 1² − 1 + k 
⇒ k = 0 
So, y = x² − x

Q14: The equation of the curve which passes through the point (1, 2) and has the slope 3x – 4 and the point of (x, y) is
(a) 2y = 3x³ – 8x + 9 
(b) y = 6x³ – 8x + 9 
(c) y = x³ – 8x + 9 
(d) 2y = 3x³ – 8x + c
Ans: (a)

Sol:  
Given slope dy/dx = 3x − 4 
Integrate:  
Multiply by 2: 2y = 3x² − 8x + 2c 
Use point (1, 2): 4 = 3 − 8 + 2c 
⇒ 2c = 9 
So, 2y = 3x² − 8x + 9

Q15: Evaluate; ∫ 2^x x² dx
(a)
(b)
(c)
(d)  None of these
Ans: (d)

Sol: 
∫ 2^x x² dx
According to the rule, 
Here, u = x² and v = 2^{x
⇒∫ 2x x2 dx}

Q16:
(a) 104 
(b) 100
(c) 10 
(d) 52
Ans: (a)

Sol: 
Using the identity:

Therefore,

Q17:

(a) 7 
(b) –8
(c) 8 
(d) –7
Ans: (c)

Sol: 

Q18: Evaluate  and the value is
(a) 3
(b) 10 
(c) 30 
(d) None of these
Ans: (c)

Sol: 

(16 + 20) – (1 + 5) = 36 – 6 = 30

Q19: If 1 2 0 (3 2 ) 0 x x k dx ++ = ∫ , find k
 (a) 0 
(b) –1
(c) –2 
(d) 1
(d) None of these
Ans: (c)

Sol: 

1 + 1 + k = 0 
2 + k = 0 
⇒ k = –2

Q20:
(a) –1 
(b) 1 
(c) e1
(d) 1
Ans: (d)

Sol:

Apply 'LAE' rule, we get

= (1 · e¹ - 0) -
= e¹ - (e¹ - e⁰) 
= e¹ - e¹ + 1 = 1