Board Revision Maths Paper I

Mathematics Paper I

Out line of Paper I

• Logic of mathematics(6) – 3 marks each 2 out of 3

=

6/9

6/9

= =

4/8 3/6

• Matrices(7) – 4 marks each 1 out of 2 – 3 marks each 1 out of 2

7/14

• Vectors and three dimensional geometry(12) – 4 marks each 1 out of 2 – 3 marks each 2 out of 3 – 2 marks each 1 out of 2

12/21

= = =

4/8 6/9 2/4

=

4/8

• Linear Programming(7) – 4 marks each 1 out of 2

Out line of Paper I • Two dimensional coordinate geometry (12) – – – –

Pair of straight line Circle Parabola Ellipse and hyperbola

12/21 • Probability(6) – 3 marks each 2 out of 3

6/9

=

=

Out line of Paper I • In all 5 questions of 10 marks each • 10 = 6 + 4 • Knowledge based questions – 2+2+3 on Vector & geometry – 2+2 Lines, circle and conic sections

• Understanding base – – – –

3+3 logic 3+3 Probability 3 Vectors & geometry 3+3+3 (2D)

• Application base – – – –

3 logic 3+3+4+4 matrices 3 probability 3+4+4 vectors & geometry – 4 (2D) – 3+3 Linear programming

• Skill based – 4+4 Linear programing

Mathematical Logic • Symbolical representation – and (∧), or (∨), implication (⇒), equivalence (⇔), negation (¬ or ∼)

• Definitions: – Tautology: A statement which is always true for any truth values of component statements is called as tautology – Contradiction: A statement which is always false for any truth values of component statements is called as contradiction

Mathematical Logic – –

Converse: converse of implication p ⇒ q is q ⇒ p Inverse: Inverse of implication p ⇒ q is ∼p ⇒ ∼q Contra positive: Contra positive of p ⇒ q is ∼q ⇒ ∼p

–

•

Possible questions are— 1. Convert given statement into symbols 2. Convert given statements into indicated symbolic form 3. Verify using truth table 4. Verify or obtain truth values, if truth values of component statements are known

1. Represent given statement using Vein diagrams ( just use three facts)

X

Y U

All x are y

1. Represent given statement using Vein diagrams ( just use three facts)

X

Y U

No x are y

1. Represent given statement using Vein diagrams ( just use three facts)

X

Y U

Some x are y ⇔ some x are not y

Mathematical Logic 1. Write negation of the following statements ( to do so use – ∀ ∀ ∀ ∀

∼(p∧q) ⇔ (∼p∨ ∼q) ∼(p∨q) ⇔ (∼p∧ ∼q) ∼(p ⇒q) ⇔ (p ∨ ∼q) ∼(p⇔q) ⇔ ∼[(p ⇒q) ∧ (q ⇒p)] ⇔ [(p ∨ ∼q) ∨ (q ∨ ∼p)] 6. Note that negation of all is at least one and of some is no

2. Write the dual: means replace ∧ by ∨

Matrices • Types of matrices: – Row matrix – If a matrix has only one row (horizontal arrangement) then is called as row matrix its order is 1xn – Column matrix-If a matrix has only one column (vertical arrangement) then is called as column matrix its order is nx1 – Square matrix- A matrix is square matrix if number of rows equals number of column. – Diagonal matrix- A square matrix is called as diagonal matrics if aij = 0 if i ≠j

– Scalar matrix- A square matrix is called as scalar matrics if aij = 0 if i ≠j and aij = α if i = j (diagonal entries must be equal) – Unit matrix- A square matrix is called as unit matrics if aij = 0 if i ≠j and aij = 1 if i = j (diagonal entries must be unit) – Upper triangular matrix- A square matrix is called as upper triangular matrics if aij = 0 if i < j – Lower triangular matrix- A square matrix is called as lower triangular matrics if aij = 0 if i > j

– Symmetric matrix- A square matrix is called as symmetric matrics if aij = aji for all i & j – Skew symmetric matrix- A square matrix is called as skew symmetric matrics if aij = -aji for all i &j (diagonal entries must be zero) – Null matrix- A matrix of any order is called as null matrix if aij= o ∀i,j – Transpose of a matrix- A matrix AT or A’ is called as transpose of A if it is obtained by interchanging row by column and column by row. – Singular matrix- A square matrix is said to

Possible types of questions • Matrices A, B & C will be given and – Check whether α A2+β A+γ I is null/scalar/upper or –

–

– – –

lower triangular/ diagonal/singular/ symmetric Problems of type verify associative property/Distributive law/ Laws of scalar multiplication If A is 3x3 then a missing term will be there which is to be identified as what is ‘k’ so that A is nonsingular Check whether AB =BA or not Verify IA.BI=IAI.IBI Find X such that AX =B

Possible types of questions  − 2 4   A =  3 2  1 6

 4 − 3 6 B= FindABwithout   − 3 0 5 computing BA showthatAB ≠ BA Here A is of order 3x2 B of 2x3 then AB is of order 3x3 and BA of 2x2 hence can’t be equal

Possible types of questions 9   3 A=    − 4 − 12

 4 3 B=   8 6 • Prove that (A+B)2 = A2 +AB+B2

For solving this problem do not show RHS = LHS instead you may try as (A+B)2 = A2 +AB+BA+B2 but required result implies that BA = 0 hence show BA = 0

Possible types of questions 1 2 A=    0 2

 2 1 B=   − 1 0 • Prove that (A+B)2 ≠ A2 +2AB+B2

 For solving this problem do not show RHS ≠ LHS instead you may do as, The inequality is because of AB ≠BA thus just show that AB ≠ BA

Possible types of questions 1 2  A=   0 − 2

 3 1 B=   − 1 0 • Prove that (A+B)(A-B) ≠ A2 - B2

 For solving this problem do not show RHS ≠ LHS instead you may do as, The inequality is because of AB ≠BA thus just show that AB ≠ BA

Possible types of questions  cosθ sinθ  If A(θ ) =    − sinθ cosθ   cos(nθ) sin(nθ)  n thenA ( θ ) =    − sin(nθ) cos(nθ) 

To prove above result you need to use principle of finite induction.Define P(n) ≡ An(θ)= A(nθ) prove P(1), assume P(n)

Possible types of questions

a 0   3 − 4   If A =   If A =   0 b  1 − 1   n   a 0 1 + 2 n − 4 n   n n thenA =  thenA =  n  1 − 2n  0 b   n To prove above result you need to use principle of finite induction as you did in previous problem.

Possible types of questions • To obtain inverse of matrix 1) by elementary row operations  R i ↔ R j interchanging two rows  Ri↔αR

j

multiply every element by non zero

element

 R i ↔ R i+ α R

j

• To obtain inverse of A consider A.A-1 = I • Go on performing elementary operations till A changes to I and at

Possible types of questions • To obtain inverse of matrix 1) by using matrix polynomial  For given A either a relation will be given or we will need to prove the same as A2+2A-3I=0 using this relation we can operate A-1 to get A + 2I-3A-1 hence A-1 = 1/3(A + 2I)

Possible types of questions • To solve system of equations 1) By using reduction method  The given system of equation is to be converted in matrix equation as AX=B for example: x-y + 4z = 4;2x + 6y –z = 3 & x + y –2x = -1 will 1 − 1 4   x  4  have the form— 2 6 − 1   y =  3         1 1 − 2  z  − 1

Possible types of questions  1 − 1 4   x  4  2 6 − 1   y =  3         1 1 − 2  z  − 1

• Now reduce the coefficient matrix to triangular matrix ( if fractions are not appearing then reduce it to unit matrix) get the values of x, y and z.

Points to remember for matrices • Copy the matrix by proper care. • Be careful about calculations which you are performing. • After finding inverse or values of x,y and z substitute and verify the same. • Take care of order of matrix

• You are good at calculations believe me you are!

Linear programmin g

•2 questions on find maxima/ minima of 3 marks each • 1 question on form LPP of 4 marks

Possible type of questions •

Define – – –

–

Convex set: A set of points is said to be convex set if the line joining any two points of the set entirely lies in the set Convex polygon: A bounded polygonal convex set is called convex polygon Corner points/ extreme points: The point of intersection of any two boundaries of the half planes determined by a system of linear inequalities is called an extreme point or corner point. Convex polygon theorem: z = f(x, y) be given linear function defined over a convex polygon X, maximum or minimum values will be at extreme points.

• Draw the graph for solution set of inequalities 3x + 5y ≥15, 5x + 2y ≤10, x ≥0, y ≥0 maximize Z = x + y Draw the figure, (0, 5) find corner points, (20/19, 45/19) (0, 3) substitute the value and (5, 0) compare the (2, 0) result for adjoining figure maxima is at (20/19,45/19 )

Possible type of questions • Using word problem form the LPP and solve graphically. You may start first to write LPP, check whether you want to maximize profit or production or to minimize expenses, choose proper variables and form the problem. • Note that objects, days can never be negative • Use tabular form for formation of LPP

Pair of straight line, circle, conic section

Important formulae • The equation ax2 + 2hxy + by2 = 0 represents 1.      Two real and different lines if h2 > a b 2.      Two coincident lines if h2 = a b • Consider by2 +2hxy + ax2 = 0 then dividing by ax2 and slope of lines passing through origin is y/x we get bm2 + 2hm + 1 = 0 then using relation between roots we have– • m1 + m2 = -2h/b and m1m2 = b/a then angle 2 between two lines will be m − m 2 h − ab −1 1 2 θ = tan −1 = tan 1+ m1.m2

a+ b

Important formulae • The equation ax2 + 2hxy + by2 = 0 represents Pair of two perpendicular lines if a+b = 0 • The difference of the slopes of lines given by ax2 + 2hxy + by2 = 0 is given by 2 2 h − ab m2 − m1 = b

Important formulae • The equation ax2 + 2hxy + by2 = 0 represents Pair of two lines and the equation of pair of lines which are bisectors is given by Pair of bisector is

x −y x.y = a− b h 2

2

Important formulae • The equation ax2 + 2hxy + by2+2gx +2fy+c = 0 represents Pair of two lines not passing through origin if

a h g h b f =0 g f c

Important formulae • The equation ax2 + 2hxy + by2+2gx +2fy+c = 0 represents Pair of two lines not passing through origin angle between these lines will be same as angle between lines represented by ax2 + 2hxy + by2=0

Important formulae • The equation ax2 + 2hxy + by2+2gx +2fy+c = 0 represents Pair of two parallel lines if

a h g = = h b f

Standard types of questions

• Homogeneous equation will be given and find the angle between them – Use formula and be careful about values of a,b,h.

• Homogeneous equation will be given find separate equations – Divide by term containing x2 then call y/x as m and find quadratic equation solve to find m1 and m2 and hence equation

• Homogeneous equation will be given with some relation between the slopes of lines as – Sum of the slopes is double the product or any such – Use relation between roots,sum = -2h/b and product = b/a to get the result

• Homogeneous equation will be given find difference between slope – Use standard result or relation between roots

Standard types of questions • A term will be unknown in equation ax2 + 2hxy + by2+2gx +2fy+c = 0 what must be the term so that the equation represents pair of straight lines – use determinant

• Find the equation of the pair of lines through origin inclined at 60o to line x + 2y = 1 – Let slope of required line be m, using relation between angle and slopes we get √3=(2m+1)/(2m) replace m by y/x to get solution

• Standard equation x2 + y2 = a2 • (x-h)2 + (y – k)2 = a2 • General equation x2 + y 2 +2 g x +2 f y + c =0 • Center (-g, -f) radius √g2 + f 2 + c • Diameter form (x-x1)(x-x2) + (y-y1)(y-y2)=0 • Parametric form x = a cos(t), y = a.sin(t) • Parametric form x-h = a cos(t), y-k = a.sin(t) • Equation of tangent to x2 + y2 +2g x +2 fy + c=0 at (x1 ,y1) is x.x1+yy1+ g(x+x1)+ f(y+y1)+c=0 • Condition of tangency: y = m x + c is tangent to std. Circle if c = ±a√m2+1 • Equation of director circle x2 + y2 = 2a2

•

The problems may be – 1. Find equation of circle with center at (2,3) – – –

and passing through (1,-1): Find radius using distance formula and get the answer and 2x + 3y+1 =0 is tangent : Find ⊥distance from tangent which equals radius and get - and a line x + 3y =7 cuts a chord AB of length 8: Find ⊥distance from chord, OM is ⊥ bisector, use Pythagoras theorem to find radius and get-

2. End points of diameter use formula 3. Three points given: Use general equation or consider (h,k) as center and equate radius to get (h, k) and using distance formula obtain radius. 1. These three points are (0,0), (a,0) and (0,b) then AB will be diameter and we get result using diameter form

1. To check whether given line is tangent to –

–

–

–

General circle then condition of tangency NOT applicable find point of intersection or ⊥distance from line of center is equal to radius Standard circle: use condition of tangency or you can find point of intersection, if is one, then tangent.

To find equation of tangents from external point of circle: use slope point form and the fact that ⊥ distance from line of center is equal to radius using this relation you will get a quadratic equation whose roots are slopes of required tangents. To find angle between tangents from external point of circle: use slope point form and the fact that ⊥ distance from line of center is equal to radius using this relation you will get a quadratic equation whose roots are slopes of required tangents. Don’t find equations of tangent use relation between roots of QE and

–

To check whether given circles are orthogonal to each other : find and write values of g1, f1, c1 and g2, f2, c2 use the relation 2 g1 g2 + 2 f1f2 = c1 + c2

–

To find equation of director circle to general circle 1. Find circle with same center but radius double 2. Find locus of points from which tangents are perpendicular 3. Shift the origin at center of circle and use director circles std form, go back to original variables by substitution.

Conic section

Conic

eccentricity

Parabola

=1

Ellipse

(√a2-b2)/a<1

Hyperbola

(√a2+b2)/a >1

Conic

Std . Equation

Parabola

Y2 = 4ax

Ellipse

X2/a2 + y2/b2=1

Hyperbola

X2/a2 - y2/b2=1

Conic

Parametric Equation

Parabola

X=at2 ,Y = 2at

Ellipse

X = a cos(θ) y = b sin(θ)

Hyperbola

X = a sec(θ) y = b tan(θ)

Conic

Length of latus rectum

Parabola

4a

Ellipse

2b2/a

Hyperbola

2b2/a

Conic

focus/foci

Parabola

(a,0)

Ellipse

(ae,0);(-ae,0)

Hyperbola

(ae,0);(-ae,0)

Conic

Equation of directrix

Parabola

X = -a

Ellipse

X = ±a/e

Hyperbola

X = ±a/e

Conic

Eq. of tangent

Parabola

yy1 = 2a(x+x1)

Ellipse Hyperbola

xx1/a2 + yy1/b2=1 xx1/a2 yy1/b2=1

Conic

Condition Of tangency

Parabola

C = a/m

Ellipse

C = ±√a2m2+b2

Hyperbola

C = ±√a2m2 - b2

Conic

Point of contact

Parabola

(a/m2, 2a/m)

Ellipse

(-a2m2 /c,b2/c)

Hyperbola

(-a2m2 /c,-b2/c)

Conic

Locus of points from which tangents are ⊥

Parabola

X = -a

Ellipse

X2 + y2 = a2+b2

Hyperbola

X2 + y2 = a2- b2

Conic

Equation of chord joining points (x1,y1)(x2,y2)

Parabola

Y = mx –2a

Ellipse

(x-x1)(x1+x2)/a2 + (yy1)(y1+y2)/b2=1

Hyperbola

(x-x1)(x1+x2)/a2 + (yy1)(y1+y2)/b2=1

Conic

Equation of normal

Parabola

Y = m.x –2am – am3

Ellipse

a x secθ - b y cosecθ = a2 - b2

Hyperbola

a x cosθ - b y cotθ = a2 - b2

Important definitions • Set of all outcomes in a random experiment is called as sample space. • Any subset of sample space is called as an event • If A is an event then S\A = A’ is called as complementary event • A and B are said to be mutually exclusive events if A ∩ B = ∅ • A and B are mutually exclusive and exhaustive events if A ∩ B = ∅ and A ∪ B =S • Probability of event A = n(A) / n(S)

Important definitions • P(A) = 0 means event is impossible event • P(A) = 1 means event is certain event • 0 ≤ P(A) ≤ 1 • P(A ∪ B ) = P(A) + P(B) –P(A ∩ B) • P(A ∪ B ∪C ) = P(A) + P(B) +P(C) + P(A ∩ B ∩ C ) – [P(A∩B )+P(B∩ C )+P(C ∩A)] • P(A’) = 1 – P(A)

Tips and tricks • To determine number of elements in event space or sample space use following— – – – – –

Throwing of one die n(S) = 6 Throwing of k dies n(S) = 6k Choosing a card n(S) = 52 Event A or B n(Event) = n(A) + n(B) Event A & B n(Event) = n(A) x n(B)

Tips and tricks n

n! Pr = orderis important (n− r)!

Can be used when to form numbers, arrange different books, n! n Cr = orderisnotimportant r!(n− r)! Can be used when to form group without order, types of books

Vectors and Geometry

Important results • Vector: directed segment • Two vectors are equal if have same direction and magnitude • Triangle law of vector addition • Parallelogram law of vector addition • Dot product and their properties • Cross product and their properties • Scalar multiplication, generates parallel vector • Scalar triple product • Two vectors are perpendicular if their dot OA product is zero

Important results • A vector is unit vector if its magnitude is one • To determine unit vector along AB if point A & B are given use – Find position vector of A and B, find b – a thus you have vector AB now divide it by its magnitude to get the answer.

• To determine vector perpendicular to given two vectors – To find this take cross product of the two

Important results • To check whether given three points A, B and C are collinear you can use— – Cross product of AB and AC must be zero – Using section formula we can prove colinearity by showing that c = αa + βb – Prove that AB = α.AC

• To check whether A,B,C and D are coplanar you can use – AB.(AC x AD) = 0 – AB = α.AC + β.AD

Important results • Section formula: If P divides AB in ratio m:n then position vector of P is (m.b+n.a)/(m+n): – to prove the result use AP.n=PB.m (note the direction). Use OA + AP = OP substitute AP = m.PB/n and now use position vectors for relation OA + m.PB/n =OP – If P divides externally then AP.n=BP.m (note the direction).

• If r is any vector coplanar to a and b (a and b are non zero) then r is uniquely expresses as linear combination of a and b

Important results • Applications in geometry – Area of ∆ABC is ½ I AB x AC I = ½ I AB x BC I – Area of parallelogram ABCD is = IAB X AC I – Volume of parallelepiped =AB . (AC X AD)

• Application in trigonometry: – Rules of T- ratios of sum and difference of angles can be proved by vectors as α−β αβ

Consider cross product of OA and OB by definition of cross product and by analytical method – take modulus and you have the result, use the same for +

Important results

• Sine Rule: in triangle ABC

a/sin A = b/sin B =c/ sin C use AB + BC + CA = 0 and consider cross product with AB to get one equality and then with AC to get other equate one side of these equality and get the result by dividing by proper fraction

• Cosine rule : Use AB + BC + CA = 0 hence AB + BC = AC and equate magnitude of both sides considering IABI = c, IBCI = a and ICAI = b we get, cos(A) = (b2 + c2 – a2)/2bc lly other results

Important results

• Application to Physics:

– Work done : F.s – Angular momentum: moment of momentum means if M is momentum and P is any point on line of action then moment about O is cross product of OP and M – Torque: moment of force means torque – Projection of a vector a along a vector b means projba = a.b / b = a.eb    a.  b – Resolution of a along b means ( 2 )b b

Important results

• Direction angles: If L is any line then its angle with X+ ,Y+,Z + are called as direction angles and are denoted by α, β and γ • Direction cosine: If α, β and γ are direction angles then cosα, cosβ and cosγ are called as direction cosine denoted by l, m and n • l2 + m2 + n2 = 1 to prove this result let P(x,y,z) be any point, let IOPI = r then using definition of dot product OP.i = r cos α = (xi+yj+zk).i = x lly y = rcos β and z = r cos γ square these results and add to get the required result.

Important results

• Direction ratio: If l, m and n are direction cosine then a,b and c are called as direction ratio if a/l=b/m = c/n • Relation between dr’ s and dc’s: If a,b and c are direction ratio and l, m and n are direction cosine then a/l=b/m = c/n = k say then using l2 + m2 + n2 = 1 we get required result as • l = a /√a2+ b2 + c2 • m = b /√a2+ b2 + c2 • n = c /√a2+ b2 + c2