Differential Calculus MCQ

Competitive Exams, Entrance Exams are conducted on MCQ.  Click HERE to understand the Forms, Structure, Rules of MCQ, techniques of understanding, analysing and selection correct answer of MCQ,
Play the Video explaining some interesting aspects of selection of correct answer of MCQ.
For Complete resources on Mathematics MCQ
For  Complete resources on Calculus

1. If y = 5x5 then find value of  $\displaystyle \frac{{dy}}{{dx}}$

(a)    25x5

(b)    625x

(c)    25x4

(d)    5$\displaystyle {{x}^{5}}$

y = 5$\displaystyle {{x}^{5}}$. 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

(c)    0

(d)    1

$\displaystyle \frac{{dy}}{{dx}}=\frac{d}{{dx}}$(200) = 0. So, option (c) is correct

3. y = $\displaystyle {{x}^{2}}$ + x + 1 . Evaluate $\displaystyle \frac{{dy}}{{dx}}$

(a)    2x + 2

(b)    2x + 1

(c)    2x + 3

(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

(c)     x + 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 / ($\displaystyle \sqrt{x}$)

(c)     2$\displaystyle \sqrt{x}$

(d)     $\displaystyle \sqrt{x}$

y = $\displaystyle \sqrt{x}$. So, $\displaystyle \frac{d}{{dx}}$($\displaystyle \sqrt{x}$) = $\displaystyle \frac{d}{{dx}}$(x1/2) = $\displaystyle \frac{1}{2}${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

(c)    6

(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

(c)     26

(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

(c)       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+ 6x2 21

(c)     8x3 6x2 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)

(c)       4x + 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 .ex. Evaluate $\displaystyle \frac{{dy}}{{dx}}$ .

(a)    1

(b)    x2 ex + ex

(c)    x2 ex + 2ex

(d)    ex + xex

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

=  ex. 1 + x. ex = ex + xex

So, option (d) is correct

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

(a)    2x

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

(c)    2

(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, 2nd 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.

(c)    1/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

(c)     16x3

(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 (x3-1) = 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

(c)     – 1/ x2

(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

(c)    2x – 62

(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. x2 – 1 ) – ( 4. x 1 – 1 ) + 0

= 6x – 4. x= 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

(c)    0

(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

(c)    X + 3

(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

(c)     3

(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

(c)    8

(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

(c)       3x2 + 2

(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

(c)       3

(d)      4

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

(c)    4x2 – 3x + 2

(d)    None

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

= { 3(x3-1 ) – 4(x1-0 )+ 5 } = 3x2 – 4x + 5

So, option (b) is correct