NEB Grade 11 Mathematics Model Paper Solution

Model question – 2077

Grade: 11

Full marks: 75

Time: 3 hours

Attempt all the questions

Group A (1 × 11 = 11)

Rewrite the correct option in your answer sheet

1. Which of the following is a statement?

(a) The fishes are beautiful

(b) Study mathematics.

(d) Water is essential for health

2. The value of $\sqrt { - 16\;} \times \sqrt { - 25}$ is:

(a) -20

(b) -20i

(d) 20

3. If ÐC = 60⁰, b = 5 cm and a = 4 cm of DABC, what is the value of c?

(a) 3.58 cm

(b) 4.58 cm

(d) 4.56

4. In a triangle ABC, B =120^o, a = 1, c = 1 then the other angles and sides are

(a) 35, 45, 2

(b) 10, 50, 3

(d) 30, 30, 3

5. The cosine of the angle between the vectors 𝑎 = i − 2j + 3𝑘 and 𝑏 = i + 3j+ 3𝑘 is

(a) 1/14

(b) 14

(d) 196

Note: Option does not match as per the question, so the correct answer is $\frac{4}{{\sqrt {266} }}$

6. The equation of parabola with the vertex at the origin and the directory y - 2 = 0 is.

(a) x² – 8y = 0

(b) y²+ 8y = 0

(d) y² - 8y = 0

7. A mathematical problem is given to three students Sumit, Sujan and Rakesh whose chance of solving it are 1/2 ,1/3, and 1/a respectively. The probability that the problem is solved is 3/4. The possible values of a are:

(a)

(b) 4

(c)

(d)

8. ${\lim }\limits_{\theta \to 0} \frac{{sin\theta }}{\theta }$ is equal to

(a) 0

(b) ∞

(d) 0

$9.The{\rm{ }}derivatives\frac{{(4{x^2} + 3)}}{{(3{x^2} - 2)}}$

(a) $\frac{{ - 34x}}{{{{\left( {3{x^2} - 2} \right)}^2}}}$

(b) $\frac{{ - 30{x^2}}}{{\left( {3{x^2} - 2} \right)}}$

(d) $\frac{{ - 32x}}{{{{\left( {3{x^2} - 2} \right)}^3}}}$

10. By Newton’s Raphson, the positive root of x³ -18 = 0 in (2, 3) is

(a)2. 666

(b) 2.621

(d) 2.622

11. Two forces acting at an angle of 45^o have a resultant equal to $\sqrt {10}$ N, if one of the forces be $\sqrt 4$ N, what is the other force.

(a) 1N

(b) 2N

(d) 4N

The total cost function of a producer is given as ${\bf{C}}{\rm{ }} = {\rm{ }}{\bf{500}}{\rm{ }} + {\rm{ }}{\bf{30Q}}{\rm{ }} + \frac{1}{3}{{\bf{Q}}^{\bf{3}}}$ . What is the marginal cost (MC) at Q = 4 is

(a) Rs.38

(b) Rs.34

(d) Rs.28

Group B

12. A function f(x) = x² is given. Answer the following question for the function f(x).

(i) What is the algebraic nature of the function?

· The Algebraic Nature of the Function is Quadratic.

(ii) Write the name of the locus of the curve.

· Parabola is the name for locus of the curve.

(iii) Write the vertex of the function.

· Vertex of function is (0,0).

(iv)Write any one property for sketching the curve.

· The Property of the given function are:

§ Function is Even.

§ Function is Symmetric about x-axis.

(v) Write the domain of the function.

· The function is defined for all the value of x that belongs to the real number. So it’s domain is (-∞,∞).

13. Compare the sum of n terms of the series: 1 + 2a +3a² +4a³ +……….and a+ 2a + 3a+ 4a …up to n terms.

Series 1st:

1 + 2a +3a² +4a³ +……….

It is an arithmetico-geometric series.

Sn= 1 + 2a +3a² +4a³ +……….+na^n-1

aSn= a + 2a²+3a³ +4a⁴ +……….+naⁿ

Subtracting above equations

(1-a)S_n = 1+a+a²+a³+…………+a^n-1-naⁿ

$(1 - a)Sn = \frac{{(1 - {a^n})}}{{(1 - a)}} - n{a^n}$

$Sn = \frac{{(1 - {a^n})}}{{{{(1 - a)}^2}}} - \frac{{(n{a^n})}}{{(1 - a)}}$

Series 2^nd: a+ 2a + 3a+ 4a … up to n terms

It is arithmetic series.

Let S_n= a+ 2a + 3a+ 4a … na

= $\frac{n}{2}\left[ {2a + \left( {n - 1} \right)a} \right]$

= $\frac{n}{2}\left[ {2a + na - a} \right]$

= $\frac{n}{2}\left[ {a + na} \right]$

= $\frac{{na}}{2}\left[ {1 + n} \right]$

14.a) In any triangle, prove that: $\left( {b + c} \right)Sin\left( {\frac{A}{2}} \right) = aSin\frac{A}{2} + B$

We Know

b+c =2R SinB + 2R SinC

=2R(SinB + SinC )

$= 2{\rm{R}}\left( {2{\rm{Sin}}\frac{{B + C}}{2}.Cos\frac{{B - C}}{2}} \right)$

$= 4R{\rm{Sin}}\frac{{B + C}}{2}.Cos\frac{{B - C}}{2}$ --------------(i)

And

a =2R SinA

=2R Sin (180⁰-(B+C))

=2R Sin (B+C))

$= 4R{\rm{Sin}}\frac{{B + C}}{2}.Cos\frac{{B + C}}{2}$ ---------------(ii)

Dividing above equations

$\frac{{b + c}}{a}\;\; = \frac{{4R{\rm{Sin}}\frac{{B + C}}{2}.Cos\frac{{B - C}}{2}}}{{4R{\rm{Sin}}\frac{{B + C}}{2}.Cos\frac{{B + C}}{2}}}$

$or,\;\frac{{b + c}}{a}\; = \frac{{Cos\frac{{B - C}}{2}}}{{Cos\frac{{B + C}}{2}}}$

$or,\;\frac{{b + c}}{a}\; = \frac{{Cos\frac{{B - \left[ {180^\circ - \left( {A + B} \right)} \right]C}}{2}}}{{Cos\left( {90^\circ - \frac{A}{2}} \right)}}$

$or,\;\frac{{b + c}}{a}\; = \frac{{Cos\frac{{A + 2B - 180^\circ }}{2}}}{{Sin\left( {\frac{A}{2}} \right)}}$

$or,\;\frac{{b + c}}{a}\; = \frac{{Cos\frac{{\frac{A}{2} + B - 90^\circ }}{2}}}{{Sin\left( {\frac{A}{2}} \right)}}$

$or,\;\frac{{b + c}}{a}\; = \frac{{Sin\frac{A}{2} + B}}{{Sin\left( {\frac{A}{2}} \right)}}$

$\therefore \left( {b + c} \right)Sin\left( {\frac{A}{2}} \right) = aSin\frac{A}{2} + B\;$

14. b) Express ${\rm{\vec r}}$ = (4, 7) as the linear combination of ${\rm{\vec a}}$ = (5, - 4) and ${\rm{\vec b}}$ = (-2, 5).

$Let{\rm{ \vec r = x\vec a + }}y{\rm{\vec b where x and y are scalars}}{\rm{.}}$

$Then,{\rm{ (4,7) = x(5, - 4) + y( - 2,5)}}$

$\begin{array}{l} {\rm{or, (4,7) = (5x, - 4x) + ( - 2y,5y)}}\\ {\rm{or(4,7) = (5x - 2y, - 4x + 5y )}}\\ {\rm{Equating Coressponding Elements,}} \end{array}$

$\begin{array}{l} {\rm{5x - 2y}} = 4{\rm{ - - - - - - - - - - (i)}}\\ {\rm{ - 4x + 5y}} = {\rm{ 7 - - - - - - - - - - (ii)}} \end{array}$

$\begin{array}{l} {\rm{Solving these Equations, We get:}}\\ {\rm{x = 2 }}\\ {\rm{and}}\\ {\rm{y = 3}} \end{array}$

${\rm{Thus, the Required combination is \vec r = 2\vec a + 3\vec b}}$

15. Calculate the appropriate measure of Skewness for the data below.

Ans: The given distribution is not an open ended. It will be better to use Karl person's Coefficient Method.

15. Calculate the appropriate measure of Skewness for the data below.

 
  
 
	 Class  
	 0-10 
	 10-20 
	 20-30 
	 30-40 
	 40-50 
	 50-60 
 
 
	 No of Workers 
	 10 
	 12 
	 25 
	 35 
	 40 
	 50

Class	0-10	10-20	20-30	30-40	40-50	50-60
No of Workers	10	12	25	35	40	50

Ans: The given distribution is not an open ended. It will be better to use Karl person's Coefficient Method.

 
  
 
	 Class 
	 Mid Value (x) 
	 No. Of Workers 
	 d' = (x-25)/10 
	 fd' 
	 fd'² 
	 c.f. 
 
 
	 0-10 
	 5 
	 10 
	 -2 
	 -20 
	 40 
	 10 
 
 
	 10-20 
	 15 
	 12 
	 -1 
	 -12 
	 12 
	 22 
 
 
	 20-30 
	 25 
	 25 
	 0 
	 0 
	 0 
	 47 
 
 
	 30-40 
	 35 
	 35 
	 1 
	 35 
	 35 
	 82 
 
 
	 40-50 
	 45 
	 40 
	 2 
	 80 
	 160 
	 122 
 
 
	 50-60 
	 55 
	 50 
	 3 
	 150 
	 450 
	 172 
 
 
	  
	  
	 N=172 
	  
	 Σ fd'=233 
	 Σ fd'²=697

Class	Mid Value (x)	No. Of Workers	d' = (x-25)/10	fd'	fd'²	c.f.
0-10	5	10	-2	-20	40	10
10-20	15	12	-1	-12	12	22
20-30	25	25	0	0	0	47
30-40	35	35	1	35	35	82
40-50	45	40	2	80	160	122
50-60	55	50	3	150	450	172
		N=172		Σ fd'=233	Σ fd'²=697

Here, a=25 N=172 Σ fd'=233 Σ fd'²=697 h=10
We Know

$\bar X = a + \frac{{\Sigma f{d^'}}}{N} \times h{\rm{ = 25 + }}\frac{{{\rm{233}}}}{{172}} \times 10{\rm{ = 38}}{\rm{.55}}$

16. Define different types of discontinuity of a function. Also write the condition for increasing, decreasing and concavity of function.

First Part:

A discontinuous function may be of following types:

${{\rm{If\;}}\lim }\limits_{x \to a} f\left( x \right)\infty \;or - \infty \;{\rm{then\;the\;function\;is\;said\;to\;have\;indefinite\;discontinuity\;at\;x}} = {\rm{a}}.{\rm{\;}}\;$
${{\rm{If\;}}\lim }\limits_{x \to a} f\left( x \right) \ne f\left( x \right){\rm{then\;the\;function\;is\;said\;to\;have\;removable\;discontinuity\;at\;x}} = {\rm{a}}.$
If ${\lim }\limits_{x \to a} f\left( x \right)$ does not exists i.e., ${\lim }\limits_{n \to {a^ - }} f\left( x \right) \ne {\lim }\limits_{n \to {a^ + }} f\left( x \right)$ then the function is said to have an ordinary discontinuity.

Second Part:

Function	Condition
Increasing Function	f’(x)>0
Decreasing Function	f’(x)<0
Concavity i. Concave Upward ii. Concave Downward	f’’(x)>0 f’’(x)<0

17) Evaluate: $\smallint \frac{{{x^3}}}{{\sqrt {{x^2} + {a^2}} }}dx$

Ans:

$\int_a^b {f(x).dx \approx \frac{{b - a}}{{2n}}}$ $Let{\rm{ x = a sin}}\theta {\rm{ then, dx = a cos}}\theta .{\rm{d}}\theta$ and

${{\rm{a}}^2}{\rm{ - }}{{\rm{x}}^2} = {{\rm{a}}^2}{\rm{ - }}{{\rm{a}}^2}{\rm{si}}{{\rm{n}}^2}\theta {\rm{ = }}{{\rm{a}}^2}{\rm{co}}{{\rm{s}}^2}\theta$ . Thus,

$\int {\frac{{{x^2}dx}}{{\sqrt {{{\rm{a}}^2}{\rm{ - }}{{\rm{x}}^2}} }}}$ $= \int {\frac{{{{({\rm{a sin}}\theta )}^2}{\rm{ a cos}}\theta .{\rm{d}}\theta {\rm{ }}}}{{\sqrt {{{\rm{a}}^2}{\rm{co}}{{\rm{s}}^2}\theta } }}}$

$= {\rm{ }}\int {\frac{{{a^3}{\rm{si}}{{\rm{n}}^2}\theta {\rm{ cos}}\theta .{\rm{d}}\theta }}{{a{\rm{ cos}}\theta }}}$

$= {\rm{ }}\int {{{\rm{a}}^2}{\rm{si}}{{\rm{n}}^2}\theta .{\rm{d}}\theta }$ $= {\rm{ }}{{\rm{a}}^2}\int {\frac{{1 - \cos 2\theta }}{2}} d\theta$

$= {\rm{ }}{{\rm{a}}^2}\int {\frac{1}{2}} d\theta {\rm{ - }}{{\rm{a}}^2}\int {\frac{{\cos 2\theta }}{2}}$

$= {\rm{ }}\frac{{{{\rm{a}}^2}}}{2}{\rm{ }}\theta {\rm{ - }}\frac{{{a^2}}}{2}.\frac{{{\rm{sin2}}\theta }}{2}{\rm{ + c}}$

${\rm{Also x = a sin}}\theta {\rm{,}}$ ${\rm{i}}{\rm{.e}}{\rm{. }}\theta {\rm{ = }}{\sin ^ - }\left( {\frac{x}{a}} \right).{\rm{ So,}}$

$\int {\frac{{{x^2}dx}}{{\sqrt {{{\rm{a}}^2}{\rm{ - }}{{\rm{x}}^2}} }}} \;{\rm{ = }}$ $\frac{{{{\rm{a}}^2}}}{2}{\rm{ }}{\sin ^ - }\left( {\frac{x}{a}} \right){\rm{ - }}\frac{{{a^2}}}{2}{\rm{ }}\frac{{2\sin \theta .\cos \theta }}{2}{\rm{ + c}}$

${\rm{ = }}\frac{{{{\rm{a}}^2}}}{2}{\rm{ }}{\sin ^ - }\left( {\frac{x}{a}} \right){\rm{ - }}\frac{{{a^2}}}{2}{\rm{ }}\sin \theta .\cos \theta {\rm{ + c}}$

${\rm{ = }}\frac{{{{\rm{a}}^2}}}{2}{\rm{ }}{\sin ^ - }\left( {\frac{x}{a}} \right){\rm{ - }}\frac{{{a^2}}}{2}.\frac{x}{a}.\sqrt {1 - {{\sin }^2}\theta } {\rm{ + c}}$

$= \frac{{{{\rm{a}}^2}}}{2}{\rm{ }}{\sin ^ - }\left( {\frac{x}{a}} \right){\rm{ - }}\frac{{{a^2}}}{2}.\frac{x}{a}.\sqrt {1 - \frac{{{x^2}}}{{{a^2}}}} {\rm{ + c}}$

${\rm{ = }}\frac{{{{\rm{a}}^2}}}{2}{\rm{ }}{\sin ^ - }\left( {\frac{x}{a}} \right){\rm{ - }}\frac{{{a^2}}}{2}.\frac{x}{a}.\sqrt {\frac{{{a^2} - {x^2}}}{{{a^2}}}} {\rm{ + c}}$

${\rm{ = }}\frac{{{{\rm{a}}^2}}}{2}{\rm{ }}{\sin ^ - }\left( {\frac{x}{a}} \right){\rm{ - }}\frac{{{a^2}}}{2}.\frac{x}{{{a^2}}}.\sqrt {{a^2} - {x^2}} {\rm{ + c}}$

${\rm{ = }}\frac{{{{\rm{a}}^2}}}{2}{\rm{ }}{\sin ^ - }\left( {\frac{x}{a}} \right){\rm{ - }}\frac{x}{2}.\sqrt {{a^2} - {x^2}} {\rm{ + c}}$

18. Define Trapezoidal rule. Evaluate using Trapezoidal rule for $\int {\frac{{dx}}{{1 + x}}}$ , n=4.

Ans:

Trapezoidal rule:

If a function f is continuous on the closed interval [a,b], then

$\int_a^b {f(x).dx \approx \frac{{b - a}}{{2n}}} [f({x_0}) + 2f({x_1}) + 2f({x_2}) + ......2f(x{}_{n - 1}) + 2f(x{}_n)]$

Where the closed interval [a, b] has been portioned into n sub intervals [x₀, x₁], [x₁, x₂], ……[x_n-1, x_n], each of the length $\frac{{b - a}}{n}$ .

Second Part

Since n=4, h = $\frac{{b - a}}{n}$ = $\frac{{1 - 0}}{4}$ =0.25 and the five points to be considered are x₀= 0, x₁=0.25, x₂= 0.5, x₃= 0.75, x₄= 1. Evaluating the values of the functions at these points:

End Points	x₀= 0	x₁=0.25	x₂= 0.5	x₃= 0.75	x₄= 1
y= $\frac{1}{{1 + x}}$	1	0.8	0.66666	0.57143	0.50000

Now, From trapezoidal rule,

$\int {\frac{{dx}}{{1 + x}}}$ = $\frac{h}{2}$ [y₀+ y₁ +2y₂ + 2y₃ + y₄]

= $\frac{{o.25}}{2}$ [1+ 2*0.8 + 2*0.66666 +2*0.57143+0.50000]

=0.1250*5.57618

=0.69702

19. State sine law and use it to prove Lami’s theorem.

Ans:

Sine Law:

In any triangle ABC,

$\frac{a}{{\sin A}} = \frac{b}{{\sin B}} = \frac{c}{{\sin C}}$

Lami’s Theorem:

If three forces acting at a point, be in equilibrium, each force is proportional to the sine angle between the other two.

Let P, Q and R be three forces acting at equilibrium at point O, represented as OA, OB, and OD respectively. Complete the Parallelogram OACB in which diagonal OC represents the resultant of forces P and Q represented by OX will be valanced by force R. That is, the force represented by OC is equal and opposite to force R. So, CO represents the force R. Since AC and OB are equal and parallel, so AC represent the force Q.

In triangle OAC, Using Sine Law

$\frac{{OA}}{{\sin OCA}} = \frac{{AC}}{{\sin COA}} = \frac{{CO}}{{\sin OAC}}$

$OR,\frac{P}{{\sin OCA}} = \frac{Q}{{\sin COA}} = \frac{R}{{\sin OAC}}$

Also,

sin OCA =sin COB = sin (180-QOR) = sin QOR

sin COA = sin (180-ROP) = sin ROP

sin OAC = sin (180-POQ) = sin POQ

Thus, $\frac{P}{{\sin QOR}} = \frac{Q}{{\sin ROP}} = \frac{R}{{\sin POQ}}$

A decline in the price of good X by Rs. 5 causes an increase in its demand by 20 units to 50 units. The new price is X is 15. (i) Calculate elasticity of demand. (ii) The elasticity of demand is negative, what does it mean?

Ans:

Given:

ΔP= -Rs5, Q₁= 20units,Q₂=50units, P₂ = Rs. 15

Then,

ΔQ = Q₂-Q₁ =50-20 =30 units

And P₁=P₂-ΔP = 15- (-5) =Rs.20

i. Elasticity of demand:

$\frac{{{\bf{\Delta }}Q\;}}{{{\bf{\Delta }}P}} \times \frac{{{P_1}}}{{{Q_1}}}$

$= \frac{{30\;}}{{ - 5}} \times \frac{{20}}{{20}} = - 6$

ii. The elasticity of demand is negative means there is inverse relationship between price and quantity demand i.e., demand will increase when the prices decreases and demand will decrease and demand will decrease when price increases.

20. a The factors of the expression ${\omega ^3}$ -1 are $\omega$ -1 and ${\omega ^2} + \omega + 1$ . if ${\omega ^3}$ -1 =0.

(i)Find the possible value of $\omega$ and write the real and imaginary roots of $\omega$ .

Ans:

Here:

${\omega ^3} - 1 = 0$

$or,(\omega - 1)({\omega ^2} + \omega + 1) = 0$

$Either,(\omega - 1) = 0{\rm{ }} \to {\rm{ }}i.e.{\rm{ }}\omega = 1$

$OR,({\omega ^2} + \omega + 1) = 0$

$i.e.{\rm{ }}\omega = \frac{{ - 1 \pm \sqrt {1 - 4} }}{2}{\rm{ = }}\frac{{ - 1 \pm \sqrt { - 3} }}{2} = \frac{{ - 1 \pm \sqrt 3 i}}{2}$

$Thus\,the\;possible{\rm{ Value of }}\omega {\rm{ are 1, }}\frac{{ - 1 + \sqrt 3 i}}{2},\frac{{ - 1 - \sqrt 3 i}}{2}$ .

(ii) Prove that $\left| {\begin{array}{ccccccccccccccc} 1&{{\omega ^n}}&{{\omega ^{2n}}}\\ {{\omega ^{2n}}}&1&{{\omega ^n}}\\ {{\omega ^n}}&{{\omega ^{2n}}}&1 \end{array}} \right|$ = 0, where n is integer.

20.(b) Verify that: |x+y| ≤|x|+|y| Where x= 2 and y= -3.

Here, x=2 and y= -3
|x+y|=|2-3|=|-1|=1

|x|+|y|= |2|+|-3|=2+3=5
∴|x+y| ≤|x|+|y|

21. (a) The single equation of pair of lines is 2x² +3xy +y² +5x +2y -3 = 0

(i) Find the equation of pair straight lines represented by the single equation.

Given:

2x² +3xy +y² +5x +2y -3 = 0 -----------------(i)

Or, y² +(3X+2)y+(2X²+5x-3) = 0

Which is quadratic in y so,

$y = \frac{{ - (3x + 2) \pm \sqrt {{{(3x + 2)}^2} - 4(2{x^2} + 5x - 3)} }}{2}$

$or,y = \frac{{ - (3x + 2) \pm \sqrt {9{x^2} + 12x + 4 - 8{x^2} - 20x + 12} }}{2}$

$or,y = \frac{{ - (3x + 2) \pm \sqrt {{x^2} - 8x + 16} }}{2}$

$or,y = \frac{{ - (3x + 2) \pm \sqrt {{{(x - 4)}^2}} }}{2}$

Taking -ve sign:

$\begin{array}{l} or,y = \frac{{ - (3x + 2) - (x - 4)}}{2}\\ or,y = \frac{{ - 4x + 2}}{2}\\ or,y = - 2x + 1\\ or,2x + y - 1 = 0 \end{array}$

Taking +ve Sign:

$\begin{array}{l} or,y = \frac{{ - (3x + 2) + (x - 4)}}{2}\\ or,y = \frac{{ - 2x - 6}}{2}\\ or,y = - x - 3\\ or,x + y + 3 = 0 \end{array}$

Thus Separate Equations are :

2x+y-1=0...............................(2)

x+y+3=0...............................(3)

(ii) Are the pair of lines represented by the given equation passes through origin? Write with reason.

Ans:

The pair of line represented by the given equations does not pass through the origin because the given equations is not homogenous.

(iii) Find the point of intersection of the pair of lines.

2x+y-1=0

-x+y+3=0_

x-4=0

or, x=4.

Again, From (3)

4+y+3=0

Or, y=-7

Thus, point of intersection is (4,7).

21. (b) If three vectors $\vec a,{\rm{ }}\vec b{\rm{ and }}\vec c$ are mutually perpendicular unit vectors in space then write a relation between them.

Ans:

If three vectors $\vec a,{\rm{ }}\vec b{\rm{ and }}\vec c$ are mutually perpendicular unit vectors in the space then

$\vec a.\vec b = \vec b.\vec a{\rm{ = 0 }}$
$\vec b.\vec c = \vec c.\vec b\;{\rm{ = 0}}$
$\vec c.\vec a = \vec a.\vec c = 0$
$\vec a.\vec a = \vec b.\vec b = \vec c.\vec c = 1$
$\left| {\vec a + \vec b + \vec c} \right| = \sqrt 3$

22. (i) Distinguish between derivative and anti-derivative of a function. Write their physical meanings and illustrate with example in your context. Find, the differential coefficient of log (Sin x) with respect to x.

Ans:

Derivative

Antiderivative

It gives the slope of the function f(x).

It gives the area under the curve of the function f(x).

The Derivative of Function is denoted by $\frac{d}{{dx}}[f(x)]$

The anti-derivative of a function, denoted by ∫f(x)dx

The derivative is defined as eh instantaneous rate of change of the function at the given point. For example, the instantaneous velocity v(t) is the derivative of the position function s(t).

$v(t) = {\lim }\limits_{\Delta t \to 0} \frac{{\Delta s}}{{\Delta t}} = {\lim }\limits_{\Delta t \to 0} \frac{{f(t + \Delta t) - f(t)}}{{\Delta t}}$

That is, v(t) = s'(t). Furthermore, the acceleration a(t) is the derivative of the velocity v(t), i.e. a(t)=v'(t).Now suppose we are given an acceleration function a(t), but not the velocity function v(t) or the position function s(t). since a(t)=v'(t) determining the velocity function requires us to find an antiderivative of the acceleration function. Then, since v(t)=s'(t) determining the position function requires us to find the antiderivative of the velocity function.

Let y=log (sin x)

Differentiating both sides with respect to x.