Differential Calculus MCQ -

Differential Calculus MCQ

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_{1. If y = 5x}⁵_{then find value of} $\displaystyle \frac{{dy}}{{dx}}$

(a) _25x⁵

(b) _625x

(d) ₅ $\displaystyle {{x}^{5}}$

_{y = 5}. So, $\displaystyle \frac{{dy}}{{dx}}=\frac{d}{{dx}}$ _(5x⁵) = 5 X (5 X x ^{5 – 1}) = 25x⁴

So, option (c) is correct

_{2. If y = 200 then find value of} $\displaystyle \frac{{dy}}{{dx}}$

(a) 200x

(b) 200

(d) 1

(200) = 0. So, option (c) is correct

_{3. y = + x + 1 .}Evaluate $\displaystyle \frac{{dy}}{{dx}}$

(a) 2x + 2

(b) 2x + 1

(d) 2x² + 2x + 1

y = x² + x + 1, So, $\displaystyle \frac{d}{{dx}}$ (x² + x + 1)

= $\displaystyle \frac{d}{{dx}}$ (x²) + $\displaystyle \frac{d}{{dx}}$ (x) + $\displaystyle \frac{d}{{dx}}$ (1)

= {2. x ^{2 – 1}}+ (1.x. ^1–1} + 0

= 2x+1+0 = 2x+1

So, option (b) is correct

_{4. y = (x + 1)}²_.Evaluate $\displaystyle \frac{{dy}}{{dx}}$

(a) 2(x + 1)

(b) 2x + 1

(d) x² + 1

(x + 1)² = x² + 2x + 1.

So, $\displaystyle \frac{{dy}}{{dx}}=\frac{d}{{dx}}$ (x² + 2x + 1)

= $\displaystyle \frac{d}{{dx}}$ (x² ) + $\displaystyle \frac{d}{{dx}}$ (2x) + $\displaystyle \frac{d}{{dx}}$ ( 1)

= 2x+2+0= 2(x+1)

So, option (a) is correct

5. $\displaystyle y=\sqrt{x}.\text{ Evaluate }\frac{{dy}}{{dx}}$

(a) 1 / (2 $\displaystyle \sqrt{x}$ )

(b) 1 / ( $\displaystyle \sqrt{x}$ )

(d) $\displaystyle \sqrt{x}$

y = $\displaystyle \sqrt{x}$ . So, $\displaystyle \frac{d}{{dx}}$ ( $\displaystyle \sqrt{x}$ ) = $\displaystyle \frac{d}{{dx}}$ (x^1/2) = $\displaystyle \frac{1}{2}$ {x^(1/2)-1}

= $\displaystyle \frac{1}{2}$ {x^(-1/2)} = $\displaystyle \frac{1}{2}$ $\displaystyle \left( {\frac{1}{{\sqrt{x}}}} \right)=\frac{1}{{\left( {\sqrt{x}} \right)}}$

So, option (a) is correct

6. If f(x) = x² + 1 then Evaluate f’ (x) at x = 2

(a) 8

(b) 2

(d) 4

f (x ) = x² + 1. So, f'(x) = $\displaystyle \frac{d}{{dx}}$ (x² + 1) = $\displaystyle \frac{d}{{dx}}$ (x²) + $\displaystyle \frac{d}{{dx}}$ (1)

= 2x + 0=2x. When x=2, f'(x) = 2x = 2 X 2 = 4

So, option (d) is correct

7. If f(x) = 4x⁴ + 4x³ + 2x² + 1 Evaluate f’ (1)

(a) 29

(b) 24

(d) 35

f (x) = 4x⁴ + 3x³ + 2x² + 1.

So, f'(x) = $\displaystyle \frac{d}{{dx}}$ (4x⁴ + 3x³ + 2x² + 1)

= $\displaystyle \frac{d}{{dx}}$ (4x⁴) + $\displaystyle \frac{d}{{dx}}$ (3x³)+ $\displaystyle \frac{d}{{dx}}$ (2x²)+ $\displaystyle \frac{d}{{dx}}$ (1)

=4.(4).(x³ ) + 3.(3).(x²)+ 2.( 2).(x⁰)+ 0

=16x³ + 9x² + 4x. So, f'(1)

= 16.(1)³ + 9.(1)³ + 4.(1) = 16+9+4=29

So, option (a) is correct

8. f(x) = 2x³ + 3x² – 12x, for what values of x, f'(x) = 0.

(a) 4,1

(b) – 1, – 2

(d) 6, 3

f (x) = 2x³ + 3x² – 12x. f'(x) = $\displaystyle \frac{d}{{dx}}$ (2x³ + 3x² – 12x)

= $\displaystyle \frac{d}{{dx}}$ (2x³ ) + $\displaystyle \frac{d}{{dx}}$ (3x² ) – $\displaystyle \frac{d}{{dx}}$ (12x).

=2.(3). (x² )+ 3. (2). x – 12.(1) = 6x² + 6x – 12.

Now 6x² + 6x – 12 =0. Or, 6 (x² + x – 2) = 0,

Or, x² + x – 2 = 0, Or, x² + 2x – x – 2 = 0

Or, x(x + 2) – 1 (x + 2) = 0, Or, (x + 2) (x – 1) = 0,

Or, x = 1, Or, x = – 2

So, option (c) is correct

9. y = (x + 1) (2×3 – 21). Evaluate $\displaystyle \frac{{dy}}{{dx}}$

(a) _8x³_–_6x²_{+ 21}

(b) _8x³_{+ 6x}²_–₂₁

(d) _6x²_–_4x

Let (x + 1) =u and (2x³ – 21)=v. So, y=uv.

We know $\displaystyle \frac{d}{{dx}}$ (uv)=v $\displaystyle \frac{d}{{dx}}$ (u)+u $\displaystyle \frac{d}{{dx}}$ (v)

So, $\displaystyle \frac{d}{{dx}}$ {(x + 1) (2x³ – 21)}

= {(2x³ – 21) X $\displaystyle \frac{d}{{dx}}$ ((x + 1)} + {(x + 1) X $\displaystyle \frac{d}{{dx}}$ (2x³ – 21)}

= (2x³ – 21) (1.x ^{1 – 1}+ 0) + (x + 1) (2. 3. x ^{3 – 1}– 0)

= (2x³ – 21) (x⁰ + 0) + (x + 1) (6x²)

= {(2x³ – 21) X 1} + 6x³ + 6x² = 8x³ + 6x² – 21

So, option (b) is correct

10. y = (x + 1) (x + 2). Evaluate $\displaystyle \frac{{dy}}{{dx}}$

(a) 2x + 3

(b) 2(x + 3)

(d) x + 1

Let (x + 1) =u and (x+2)=v. So, y=uv.

We know $\displaystyle \frac{d}{{dx}}$ (uv)=v $\displaystyle \frac{d}{{dx}}$ (u)+u $\displaystyle \frac{d}{{dx}}$ (v)

$\displaystyle \frac{{dy}}{{dx}}$ = $\displaystyle \frac{d}{{dx}}$ {(x+ 1) (x + 2)}

={(x+ 1) X $\displaystyle \frac{d}{{dx}}$ (x + 2)}+ {(x + 2)x $\displaystyle \frac{d}{{dx}}$ (x+1)}

= {(x + 1) (x⁰ + 0)} + {(x + 2) (x⁰ + 0)}

= (x + 2) .1 + (x + 1).1 = x + 2 + x + 1 = 2x+3

So, option (a) is correct

11. y = x .e^x. Evaluate $\displaystyle \frac{{dy}}{{dx}}$ _.

(a) 1

(b) x² e^x + e^x

(d) e^x + xe^x

$\displaystyle \frac{{dy}}{{dx}}$ = $\displaystyle \frac{d}{{dx}}~$ (x . e^x) = {(x . $\displaystyle \frac{d}{{dx}}~$ (e^x)} + e^x . (x $\displaystyle \frac{d}{{dx}}~$ )

= e^x. 1 + x. e^x = e^x + xe^x

So, option (d) is correct

12. y=2x. Evaluate $\displaystyle \frac{{{{d}^{2}}y}}{{d{{x}^{2}}}}$ (y)

(a) 2x

(b) $\displaystyle \sqrt{{{{x}^{3}}}}$

(d) 0

y = 2x. So, $\displaystyle \frac{{dy}}{{dx}}$ (2y)=2. $\displaystyle \frac{{dy}}{{dx}}$ (x)+x. $\displaystyle \frac{{dy}}{{dx}}$ (2)=2. 1 + x. 0 = 2.

So, 2^nd derivative $\displaystyle \frac{{{{d}^{2}}y}}{{d{{x}^{2}}}}$ (y) = $\displaystyle \frac{{dy}}{{dx}}$ (2)=0

So, option (d) is correct

13. y = x .log x_.Evaluate $\displaystyle \frac{{{{d}^{2}}y}}{{d{{x}^{2}}}}$ (y)

(a) log x

(b) 1 + log x.

(d) 3x

y = x. log x. So, $\displaystyle \frac{{dy}}{{dx}}$ (y)= $\displaystyle \frac{{dy}}{{dx}}$ (x.logx.)

={x. $\displaystyle {\frac{{dy}}{{dx}}}$ (logx)}+ {log x. $\displaystyle {\frac{{dy}}{{dx}}}$ (x)}

=x.(1/x) + lox x. (1) = 1+log x

The second order differential $\displaystyle \frac{{{{d}^{2}}y}}{{d{{x}^{2}}}}$ (y) = dy/dx (1+log x)

= dy/dx (1) + dy/dx (log x) = 0+ 1/x = 1/x

So, option (c) is correct

14. y = 2x⁴ Evaluate $\displaystyle \frac{{{{d}^{2}}y}}{{d{{x}^{2}}}}$ is

(a) 36x³

(b) 12x⁴

(d) 24x²

y = 2x ⁴. $\displaystyle {\frac{{dy}}{{dx}}}$ (y)= $\displaystyle {\frac{{dy}}{{dx}}}$ (2x ⁴) = 2. $\displaystyle {\frac{{dy}}{{dx}}}$ (x ⁴) + x ⁴. $\displaystyle {\frac{{dy}}{{dx}}}$ (2)

=2 x 4 x (x³) + x ⁴. 0 = 8x³

So, the second derivative d²y/dx² (y) = $\displaystyle {\frac{{dy}}{{dx}}}$ (8x³)

= 3 X 8 X (x^3-1) = 24x².

So, option (d) is correct

15. y = log x_.Evaluate $\displaystyle \frac{{{{d}^{2}}y}}{{d{{x}^{2}}}}$ (y)

(a) 1/ x²

(b) -4x²

(d) x

y = log x. So, $\displaystyle \frac{d}{{dx}}$ (y)= $\displaystyle \frac{{dy}}{{dx}}$ (logx)= $\displaystyle \frac{1}{x}$

So, the second derivative d²y / dx² (y) = $\displaystyle \frac{{dy}}{{dx}}$ ( $\displaystyle \frac{1}{x}$ )

= $\displaystyle \frac{{dy}}{{dx}}$ (x^-1) = -1 Xx^-1-1) = – (x ^-2) = -1 / x²

So, option (c) is correct

16. y = x (1 – x)² . Evaluate $\displaystyle \frac{{{{d}^{2}}y}}{{d{{x}^{2}}}}$

(a) 2(x – 2)

(b) 6x – 4

(d) 4x – 3

y = x (1 – x)² = x (1 – 2x + x²).

So, $\displaystyle \frac{{dy}}{{dx}}$ (y)= $\displaystyle \frac{{dy}}{{dx}}$ {x(1 – 2x + x²) }= $\displaystyle \frac{{dy}}{{dx}}$ { x (1 – 2x + x²) }

= (1 – 2x + x²). $\displaystyle \frac{{dy}}{{dx}}$ (x) + x $\displaystyle \frac{{dy}}{{dx}}$ (1 – 2x + x²)

= {(1 – 2x + x²) . 1 } + x {(0 – 2. 1. x⁰ + 2. x. ^{2 – 1}}

= (1 – 2x + x²) + x ( – 2 + 2. x) =1 – 2x + x² – 2x + 2x²

= 3x² – 4x + 1

So, the second derivative d²y/dx² (y) = $\displaystyle \frac{{dy}}{{dx}}$ = $\displaystyle \frac{d}{{dx}}$ (3x² – 4x + 1)

= (3. 2. x^{2 – 1} ) – ( 4. x ^{1 – 1} ) + 0

= 6x – 4. x⁰= 6x – 4. 1 = 6x – 4

So, option (b) is correct

17. If y = 5x – 9 Evaluate $\displaystyle \frac{{{{d}^{2}}y}}{{d{{x}^{2}}}}$

(a) 5

(b) 6

(d) 12

y = 5x – 9. So, $\displaystyle \frac{{dy}}{{dx}}$ (y)=So, $\displaystyle \frac{d}{{dx}}$ (5x-9)

= $\displaystyle \frac{d}{{dx}}$ (5x) – $\displaystyle \frac{d}{{dx}}$ (9) = 5. $\displaystyle \frac{d}{{dx}}$ (x) – 0 = 5 X 1 – 0 = 5 – 0=5

So, the second derivative d²y/dx² (y) = $\displaystyle \frac{{dy}}{{dx}}$ = $\displaystyle \frac{{dy}}{{dx}}$ (5) = 0

So, option (c) is correct

18. If y = 2x³– 3x² -36x + 10 then y is maximum at
(a) 85
(b) 107
(c) 95
(d) 54

y = 2x³ – 3x² – 36x + 10, So, $\displaystyle \frac{{dy}}{{dx}}$ (y)= $\displaystyle \frac{{dy}}{{dx}}$ (2x³ – 3x² – 36x + 10)

= (2 X3 X x²) – (3 X 2 X x) – (36 X 0) +0

= 6x² – 6x – 36.

For stationery values, $\displaystyle \frac{{dy}}{{dx}}$ (y)=0, So, 6x² – 6x – 36 = 0,

Or, 6x² + 12x – 18x – 36 = 0 Or,

6x (x + 2) – 18 (x + 2) = 0,

Or, (x + 2) (6x – 18) = 0, Or, (x + 2) (6x – 18) = 0,

Or, (x + 2) X 6X (x – 3) = 0, x = 3 or, x = – 2

Now d²y/dx² (y) = $\displaystyle \frac{{dy}}{{dx}}$ (6x² – 6x – 36) . = 12x -6

For x=3, d²y/dx² (y) = (12X3) – 6 = 36-6=30.

For x=-2, d²y/dx² (y) = (12X(-2)) – 6

= -24 -6 =-30 (negative).

So, value of y is maximum at x = – 2.

So, y (maximum) = 2x³ – 3x² – 36x + 10

= 2 (– 2)³ – 3. (– 2)² – 36. (– 2) + 10

= 2. (– 8) – 3. 4 + 72 + 10 = -16 -12 +72 +10 = 54

So, option (d) is correct

19. y = (2 – x)². Evaluate $\displaystyle y={{x}^{3}}$ .

(a) 2(2 – x)

(b) 2x – 4

(d) 12 – x

y=(2 – x)² = 4 – 4x + x².

So $\displaystyle \frac{{dy}}{{dx}}~$ (4 – 4x + x²) = 0 -4 +2x = 2x-4

So, option (b) is correct

20. If $\displaystyle y={{x}^{3}}$ , evaluate (1+ $\displaystyle \frac{{dy}}{{dx}}$ ) when x= 1.

(a) 12

(b) 16

(d) 4

$\displaystyle y={{x}^{3}}$ , so, $\displaystyle \frac{d}{{dx}}$ (y) = $\displaystyle \frac{d}{{dx}}$ (_x³) = 3 (x²) .

So 1+ $\displaystyle \frac{{dy}}{{dx}}~$ = 1+ 3x²= 1+ {3 x (1)²} = 1+ 3=4

So, option (d) is correct

21. Find the gradient of the curve y = 2x³ –3x² – 12x + 8 at x = 0

(a) – 12

(b) 12

(d) 4

y = 2x³ -3x² – 12x + 8. So, $\displaystyle \frac{{dy}}{{dx}}~$

= 2. (3x² )– 3. ( 2x ) – 12. 1. x⁰ + 0

= 6x² -6x -12

So, gradient of the curve at x=0, is 6. (0) – 6. (0) – 12

= 0 – 0 – 12 = – 12

So, option (a) is correct

22. y = x(x – 1) (x – 2). Evaluate $\displaystyle \frac{{dy}}{{dx}}$ _.

(a) 3x² – 6x + 2

(b) – 6x + 20

(d) 4x²+ 5

Let x=u, (x – 1) =v, (x – 2)=w. Then $\displaystyle \frac{d}{{dy}}$ (y) = $\displaystyle \frac{d}{{dy}}$ (uvw)

= { vw X $\displaystyle \frac{{dx}}{{dy}}$ (u)} + {uw X $\displaystyle \frac{{dx}}{{dy}}$ (v)} + {uv X $\displaystyle \frac{{dx}}{{dy}}$ (w)}

= {(x – 1) ((x – 2)} X 1} + {x (x – 2) X (1 -0) } + {x(x – 1) X (1-2)}

= {(x² – x – 2x + 2) X 1} + {(x² – 2x) X 1} + {(x² – x) X-1

= (x² – 3x + 2) + (x² – 2x) + x² – x

= x² – 3x + 2 + x² – 2x + x² – x = 3x² – 6x + 2

So, option (a) is correct

23. If y = x3, evaluate 1+ $\displaystyle \frac{{{{d}^{2}}y}}{{d{{x}^{2}}}}$

(a) – 5

(b) 5

(d) 4

y = x³. So $\displaystyle \frac{d}{{dx}}$ =3x^3-1 = 3x².

So $\displaystyle \frac{{{{d}^{2}}y}}{{d{{x}^{2}}}}$ = $\displaystyle \frac{d}{{dx}}$ (3x²) = 6x.

So, $\displaystyle \frac{{{{d}^{2}}y}}{{d{{x}^{2}}}}$ +1 = 6x +1

= 6 x (-1) +1 = – 6 +1= -5

So, option (a) is correct

24. If the total cost function is C = x³ – 2x² – 5x, find Marginal Cost (MC)

(a) 3x² + x + 2

(b) 3x² – 4x + 5

(d) None

We know MC = $\displaystyle \frac{d}{{dx}}$ (C) = $\displaystyle \frac{d}{{dx}}$ (x³ – 2x² – 5x)

= { 3(x^3-1 ) – 4(x^1-0 )+ 5 } = 3x² – 4x + 5

So, option (b) is correct

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