Solved question paper for Math Mar-2017 (PSEB Class 12th)
Solved Question Paper
Math Mar-2017
PSEB • Class 12th • 1st • Mar-2017
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1 Relations and Functions:
Types of relations: Reflexive, symmetric, transitive and equivalence relations. One to one and onto functions, composite functions, inverse of a function. Binary operations.
2 Inverse Trigonometric Functions:
Definition, Range, Domain, Principal value branches. Graphs of inverse trigonometric functions. Elementary properties of inverse trigonometric functions.
3 Matrices:
Concept, notation, order, equality, types of matrices, zero matrix, transpose of a matrix, symmetric and skew symmetric matrices. Addition, multiplication and scalar multiplication of matrices, simple properties of addition, multiplication and scalar multiplication. Noncommutativity of multiplication of matrices and existence of non-zero matrices whose product is the zero matrix (restrict to square matrices of order 2). Concept of elementary row and column operations. Invertible matrices and proof of the uniqueness of inverse, if it exists; (Here all matrices will have real entries).
4 Determinants:
Determinant of a square matrix (up to 3×3matrices), properties of determinants, minors, cofactors and applications of determinants in finding the area of a triangle. Adjoint and inverse of a square matrix. Consistency, inconsistency and number of solutions of system of linear equation by examples, solving system of linear equations in two or three variables (having unique solution) using inverse of a matrix.
5 Continuity and Differentiability:
Continuity and Differentiability, derivative of composite functions, chain rule, derivative of inverse trigonometric functions, derivative of implicit function. Concepts of exponential and logarithmic functions. Derivatives of logex and ex. Logarithmic differentiation. Derivative of functions expressed in parametric forms. Second order derivatives. Rolle’s and Lagrange’s Mean Value Theorems (without proof) and their geometric interpretations.
6 Applications of Derivatives:
Applications of derivatives: rate of change, increasing/decreasing functions, tangents and normal, approximation, maxima and minima (first derivative test motivated geometrically and second derivative test given as a provable tool). Simple problems (that illustrate basic principles and understanding of the subject as well as real life situations).
7 Integrals:
Integration as inverse process of differentiation. Integration of a variety of functions by substitution, by partial fractions and by parts, only simple integrals of the type to be evaluated. Definite integrals as a limit of a sum. Fundamental Theorem of Calculus (without proof).Basic properties of definite integrals and evaluation of definite integrals.
8 Applications of the Integrals:
Applications in finding the area under simple curves, especially lines, areas of circles/parabolas/ellipses (in standard form only), area between the two above said curves (the region should be clearly identifiable).
9 Differential Equations:
Definition, order and degree, general and particular solutions of a differential equation. Formation of differential equation whose general solution is given. Solution of differential equations by method of separation of variables, homogeneous differential equations of first order and first degree.
10 Vectors:
Vectors and scalars, magnitude and direction of a vector. Direction cosines/ratios of vectors. Types of vectors (equal, unit, zero, parallel and collinear vectors), position vector of a point, negative of a vector, components of a vector, addition of vectors, multiplication of a vector by a scalar, position vector of a point dividing a line segment in a given ratio. Scalar (dot) product of vectors, projection of a vector on a line. Vector (cross) product of vectors, scalar triple product.
11 Three-dimensional Geometry:
Direction cosines/ ratios of a line joining two points. Cartesian and vector equation of a line, coplanar and skew lines, shortest distance between two lines. Cartesian and vector equation of a plane. Angle between (i) two lines, (ii) two planes, (iii) a line and a plane. Distance of a point from a plane.
12 Linear Programming:
Introduction, definition of related terminology such as constraints, objectives function, optimization, different types of linear programming (L.P.) problems, mathematical formulation of L.P problems, graphical method of solution for problems in two variables, feasible and infeasible regions, feasible and infeasible solutions, optimal feasible solutions (up to three non-trivial constrains)
13 Probability:
Multiplication theorem on probability. Conditional probability, independent events, total probability, Baye’s theorem, Random variable and its probability distribution, mean and variance of haphazard variable. Repeated independent (Bernoulli) trials and Binomial distribution.
Note:- The subtopics which are printed in the books published by Punjab School Education Board but are not mentioned in syllabus, should be considered as part of syllabus
Solved Questions
Solved-
(i) If a binary operation is defined by a'* b: ab then 2 * 2 is equal to:
(a) 4 (b) 2 (c) 9 (d) 8
Answer :
(a) 4
(ii) sin-1(1) is equal to :
(a) o (b) ({nover 6}) (c) ({nover 2}) (d) ({nover 3})
Answer :
(c) ({nover 2})
(iii) If order of matrix A is 4 * 3 and order of matrix B is 3 * 5 then order of matrix B'A' is:
a) 5 * 3 b) 4*5 c) 5 * 4 d) 3 * 2
Answer :
c) 5 * 4
(vi) If y: cos x then at X = ({nover 2}) , y2 is equal to:
a) -1 b) 1 c) 0 d) 1/2
Answer :
(c) 0
(v) ( intlimits_0^{x/2}) ({sin^{3/2} x over sin^{3/2} x + cos^{3/2}x}) dx is equal to :
a) 0 b) ({pi over 2} ) c) ({pi over 3}) d) ({pi over 4} )
Answer :
(d) ({pi over 4} )
(vi) If .({ sqrt{3}}) a .b = lA * Bl then angle between, vector a and vector b is :
a) ({pi over 2} ) b) ({pi over 6 } ) c) ({pi over 3}) d) ({pi over 4} )
Answer :
(b) ({pi over 6 } )
(vii) Direction ratio of line given by ({ x-1over 3} = { 6-2yover 10} = { 1-zover -7} ) are:
(a) < 3, 10, -7 > (c) < 3, -5, 7>
(b) < 3, 5, 7> (d) < 3, 5, -7>
Answer :
(d) < 3, 5, -7>
(viii) If P(A) = ({1 over 2}), P(B)= ({3 over 8}) and P(A (igcap) B) = ({1 over 5}) then P(B l A) is equal to:
a) ({ 2over5} ) b) ({8over 15 } ) c) ({2 over 3}) d) ({5over 8} )
Very Short Answer Mar-2017 • PSEB Class 12thb) ({8over 15 } )
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If A = (egin{bmatrix} 2 \[0.3em] -4 \[0.3em] 1 end{bmatrix}), B = [ 5 3 -1 ] then verify that (AB)' = B'A'
Short Answer Mar-2017 • PSEB Class 12thA = (egin{bmatrix} 2 \[0.3em] -4 \[0.3em] 1 end{bmatrix}), B = [ 5 3 -1 ]
AB = (egin{bmatrix} 10 & 6 & -2 \[0.3em] -20 & -12 &4 \[0.3em] 5 & 3 & -1 end{bmatrix})
((AB)^{-1}) = (egin{bmatrix} 10 & -20 & 5 \[0.3em] 6 & -12 &3 \[0.3em] -2 & 4 & -1 end{bmatrix})
(B^1 A^1) = (egin{bmatrix} 5 \[0.3em] 3 \[0.3em] -1 end{bmatrix})[ 2 -4 1 ]
(B^1 A^1) = (egin{bmatrix} 10 & -20 & 5 \[0.3em] 6 & -12 &3 \[0.3em] -2 & 4 & -1 end{bmatrix})
((AB)^{-1}) = (B^1 A^1)
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If Y = ({{sin^-1} }) ( ({2xover 1+x^2}) ) then find ({dyover dx})
Short Answer Mar-2017 • PSEB Class 12thY = ({{sin^-1} }) ( ({2xover 1+x^2}) )
put x = tanQ
y = ({sin^{-1}} ({ 2tanQ over 1+ tan^2Q}))
y = (sin^{-1} sin 2Q)
y = 2Q = (2 tan^{-1}Q)
({dy over dx} = {2 over 1+ x^2})
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Evaluate ({int {sin^4x cos^3x} }) dx
Short Answer Mar-2017 • PSEB Class 12thI = ({int {sin^4x cos^3x} }) dx
= ({int {sin^4x cos^2x cosx} }) dx
= ({int {sin^4x } (1 - sin ^2x) cosx }) dx
put sin x = y
cosx dx = dy
I = ({int y^4{(1 - y^2)} dy})
= ({int y^4 dy} - {int y^6 dy } )
= ({y^5 over 5} - {y^7 over 7 }+ c = {sin ^5 xover 5}- {sin^7xover7 } +c)
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Evaluate ({int {dx over {x^2-4x + 13}} })
Short Answer Mar-2017 • PSEB Class 12th({int {dx over {x^2-4x + 13}} })
I = ({int {dx over {(x-2)^2 - 4 + 13}} })
Put x = 2 = t
dx = dt
I = ({int {dt over {t^2 - (3)^2}}} )
= ({1 over 3 } tan^{-1} {t over 3} + c)
I = ({1 over 3 } tan^{-1} ({x-2 over 3}) + c)
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Find Particular solution of differential equation cos ( ({dy over dx}) ) = ({1 over 5}) , y (0) = 2
Short Answer Mar-2017 • PSEB Class 12th5dy = 1dx
5y = x+c
put y(0) = 2
5(2) = 0 + c
c = 10
5y = x + 10
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Find the integrating factor for the differential equation cot x ({ dy over dx}) + y = 2x + ({x^2})
Short Answer Mar-2017 • PSEB Class 12thcot x ({ dy over dx}) + y = 2x + ({x^2})
({ dy over dx}) + ({y over x} = 2 + x)
I.F = (e^{int pdx})
I.F = (e^{int {1 over x}dx} = e^{log x} = x)
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Find the angle between plane 3x + 4y - z = 8 and line ({x-1over 2}) = ({2-yover 7}) = ({3z + 6over 12})
Short Answer Mar-2017 • PSEB Class 12thThe equation of the line
({x-1over 2}) = ({2-yover 7}) = ({3z + 6over 12})
({x-1over 2}) = ({2-yover 7}) = (3({z + 6over 4}))
The Direction Ratio are : 2, -7, 4
The Equation of the plane is :
3x + 4y - z = 8
ie : 3x + 4y - z - 8 = 0
Direction Ratio of the normal to the plane are 3, 4, -1
Let Q be the angle between line & the plane
Since = ({(3)(2) + 4(7) + 4(-1) over sqrt{4+ 49 + 16 } sqrt{9 + 16 + 1}})
= (6 + -28 - 4 over sqrt{69} sqrt {26})
= (2 - 28 over sqrt{69} sqrt {26})
= (- 26 over sqrt{69} sqrt {26})
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Mar 2018
Find the angle between the plane 2x+3 y-5z= 10 and the line passing from the points (2, 3,-1) 2 Or (1, 2, 1)
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Mar 2018
Find the angle between the plane 2x+3 y-5z= 10 and the line passing from the points (2, 3,-1) Or (1, 2, 1)
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Probability of A,B and C of solving the problem are 1/3 , 1/2 and 1/4 respectively . If they all try to solve the problem then find the probability that exactly one of them will solve the problem
Short Answer Mar-2017 • PSEB Class 12thGiven that
P ((ar{A})) = (1 over 3 ) P (B) = (1 over 2) P (C) = (1 over 4 )
P(A) = 1 - P (A) = 1 - (1 over 3 ) = (2 over 3 )
P((ar{B})) = 1 - P (B) = 1 - (1 over 2 ) = (1 over 2 )
P((ar{C})) = 1 - (1 over 4 ) = (3 over 4 )
Req Prob
= P(A) P((ar{B})) P((ar{C})) + P((ar{A})) P(B) P((ar{C})) + P((ar{A})) P((ar{B})) P(C)
= (({1 over 3})({1 over 2})({3 over 4})+({2 over 3})({1over2})({3 over 4})+({2 over 3})({1over2})({1 over 4}))
= (11 over 24)
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Show that function f: R ( ightarrow) R, f(x) = ({2x+5over 8}) is invertible. Also find inverse of f.
Long Answer Mar-2017 • PSEB Class 12thf(x) = ({2x+5over 8}) , x (leftarrow) R
7(x1) = f(x2)
x1x2 (leftarrow) R
({2x_1 + 5 over 8} = {2x_2 + 5 over 8})
=> x1 = x2 f is 1 - 1
k = ({2x + 5 over 8} )
=> 8k = 2x+ 5
=> 8k - 5 = 2x
=> x = ({8k - 5 over 2} )
=> f is onto => f is invertible
Ley k (leftarrow) R
f(x) = k
f(k) = x
({2k + 5 over 8} ) = x
2k = 8x - 5
k = ({8x + 5 over 2} )
for x (leftarrow) R
f(x) = ({8x + 5 over 2} )
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Show that ({tan^{-1} 1over 3} + {tan^{-1} 1over 5} = {1over 2} {cos^{-1} 33over 65})
Long Answer Mar-2017 • PSEB Class 12th({tan^{-1} 1over 3} + {tan^{-1} 1over 5} = {1over 2} {cos^{-1} 33over 65})
= ( {tan^{-1}} ({1 over 3}) + {tan^{-1}} ({1 over 5}))
= ({tan^{-1}} [{{1 over 3}+{1 over 5} over 1 - {1over 3}+ {1 over 5}}])
= ({tan^{-1}} [{{8 over 15} over 1 - {1over 15}}])
= ({tan^{-1}} [{{8 over 15} over {14over 15}}])
= ({tan^{-1}} [{{8 over 14} }] = {tan^{-1}} [{{4 over 7} }])
Put ({1 over 2 }cos^{-1} ({33 over 65}) = Q)
= (cos^{-1} ({33 over 65}) = 2Q)
(cos 2Q = {33 over 65})
(1 + tan^2Q over 1 + tan^2Q) = (33 over 65)
(65 - 65 tan^2 Q = 33 + 33 tan^2Q )
((-65 - 33) tan^2 Q = 65 + 33 )
(- 98 tan^2 Q = -32 )
( tan^2 Q = {32 over 98})
( tan^2 Q = {16 over 49})
( tanQ = {4over 7})
(Q = tan^{-1} ({4over 7}))
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Express (egin{bmatrix} 6 & -4 & 5 \[0.3em] 1 & 4 & -2\[0.3em] 7 & 5 & 9 end{bmatrix})as sum of symmetric matrix and a skew-symmetric matrix,
Long Answer Mar-2017 • PSEB Class 12thA = (egin{bmatrix} 6 & -4 & 5 \[0.3em] 1 & 4 & -2\[0.3em] 7 & 5 & 9 end{bmatrix})
AT = (egin{bmatrix} 6 & 1 & 7 \[0.3em] -4 & 4 & 5\[0.3em] 5 & -2 & 9 end{bmatrix})
For symmetric matrix = (A + A^T over 2)
(A + A^T over 2) = (1 over 2) (egin{bmatrix} 6 & -4 & 5 \[0.3em] 1 & 4 & -2\[0.3em] 7 & 5 & 9 end{bmatrix}) + (egin{bmatrix} 6 & 1 & 7 \[0.3em] -4 & 4 & 5\[0.3em] 5 & -2 & 9 end{bmatrix})
= (1 over 2) (egin{bmatrix} 12 & -3 & 12 \[0.3em] -3 & 8 & 3\[0.3em] 12 & 3 & 18 end{bmatrix})
For skew-symmetric matrix = (A - A^T over 2)
(A - A^T over 2) = (1 over 2) (egin{bmatrix} 6 & -4 & 5 \[0.3em] 1 & 4 & -2\[0.3em] 7 & 5 & 9 end{bmatrix}) + (egin{bmatrix} 6 & 1 & 7 \[0.3em] -4 & 4 & 5\[0.3em] 5 & -2 & 9 end{bmatrix})
= (egin{bmatrix} 0 & -5 & -2 \[0.3em] 5 & 0 & -7\[0.3em] 2 & 7 & 0 end{bmatrix})
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Mar 2018
Express \\( \\begin{bmatrix} 2 & 5 & -1 \\\\ 3 & 1 & 5 \\\\ 7 & 6 & 9 \\end{bmatrix}\\) as sum of symmetric and skew-symmetric matrices.
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Using differentials find approximate value of (sqrt 360)
Long Answer Mar-2017 • PSEB Class 12thTake y = (sqrt x) , x = 361
dx = 8x = -1 so that x - 8x = 360
Now y + 8y = (sqrt{x + 3x})
y + 8y - y = (sqrt{x + 8x}) - (sqrt x)
8y = (sqrt{360} ) - (sqrt{361} )
(sqrt{360} = 8y + 19)
Now 3y is opp eqaul to dy and dy = (dyover dx) = (1 over 2 sqrt x)dx
dy = (1 over 2 * 19 ) dx = -(1 over38 )
(sqrt{360} ) = -(1 over38 ) + 19
(sqrt{360} ) = - 0.026 + 19
(sqrt{360} ) = 18.974
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Evaluate (intlimits_1^3) ((x^2+4) dx ) as limit of a sum
Long Answer Mar-2017 • PSEB Class 12th -
Using integration find the area of triangle whose sides are given by equations y= x + 1, y = 3x + 1, x=5
Long Answer Mar-2017 • PSEB Class 12thy= x + 1, --------(i)
y = 3x + 1, ----------(ii)
x=5 ----------(iii)
Subtrating (i) from (ii)
x = 0
putting x = 0 in (i) y = 1
Line 1 and 2 intersect in A(0,1)
From (i) & (iii) we get x = 5 , y = 5+ 1 = 6
Line 1 and 3 intersect in B(5, 6)
From (ii) and (iii)
x = 5 , y = 16
Line 2 and 3 intersect in C(5 , 16)
vertices of the ABC are A (0, 1); B(5, 6); C(5, 16);
Req Area = Area of triangle ABC
= Area of triangle OMC - Area of triangle OMB
= (int_0^5 (3x + 1)dx - int_0^5 (x + 1)dx)
= ({3x^2 over 2} + x |_0^5 - {x^2 over 2} + x |_0^5)
= ({3x^2 - x^2over 2} |_0^5)
= 24 sq units
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Find particular solution of differential equation (x^2dy -(3x^2+xy+ y^2)dx = 0, y(1) = 1)
Long Answer Mar-2017 • PSEB Class 12th(x^2dy -(3x^2+xy+ y^2)dx = 0)
(x^2dy = (3x^2+xy+ y^2)dx )
(({dy over dx}) = {3x^2 + xy + y^2 over x^2})
(({dy over dx}) = 3 +{ xy + y^2 over x^2})
({dy over dx} = { xy + y^2 over x^2} + 3)
Put y = Vx & ({dy over dx } = x {dv over dx} + v)
( x {dv over dx} + v) = ( { x(Vx) + (Vx)^2 over x^2} + 3)
( x {dv over dx} + v) = ({ x^2 (V + V^2)over x^2} + 3)
( x {dv over dx}) = (V^2 + 3)
({dv over V^2+ 3} = {dx over x})
({1 over sqrt 3} tan^{-1} {V over sqrt3} = log x + c)
({1 over sqrt 3} tan^{-1} {V over sqrt3x} = log x + c)
({1 over sqrt 3} tan^{-1} ({1 over sqrt3}) = c)
({1 over sqrt 3} { pi over 6} = c)
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Adjacent sides of a parallelogram are given by the vectors (2hat{i} - hat{j} + 2hat{k}) and (hat{i} + 5hat{j} - hat{k}) find a unit vector in a drection of it's diagonal. Also find the area of parellelogram.
Long Answer Mar-2017 • PSEB Class 12thLet ABCD be the parallelogram with
(overrightarrow{AB}) = (2hat{i} - hat{j} + 2hat{k})
(overrightarrow{AD}) = (hat{i} + 5hat{j} - hat{k})
Now = (overrightarrow{AB}) = (overrightarrow{AB}) + (overrightarrow{BC})
= (overrightarrow{AB}) + (overrightarrow{AD})
(overrightarrow{AC}) = ((2hat{i} - hat{j} + 2hat{k})) + ((hat{i} + 5hat{j} - hat{k}))
(overrightarrow{AC}) = ((3hat{i} + 4hat{j} + hat{k}))
Hence , a limit vector in the direction
(overrightarrow{AC}) = ({1 over |overrightarrow{AC} |} (overrightarrow{AC}))
= ({3hat{i} + 4hat{j} + hat{k}} over sqrt {9 + 16 + 1})
= (1 over sqrt {26})((3hat{i} + 4hat{j} + hat{k}))
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BagI contains 2 black and 8 red balls, bagII contains 7 black and 3 red balls and bag III contains 5 black and 5 red balls. One bag is chosen at random and a ball is drawn from it which is found to be red. Find the probability that the ball is drawn from bag II.
or
Two cards are drawn (without replacemant) from a well shuffled deck of 52 cards. Find probability distribution and mean of number of cards numbered 4.
Long Answer Mar-2017 • PSEB Class 12th -
Solve the following sysJeur oflinear equations by matrix method :
3 x+y +z=10, 2x-y-z=0, x-y + 2z = 1
or
Using elementary transformation find the inverse of (egin{bmatrix} 3 & 2 & 1 \[0.3em] 2 & 4 & 3 \[0.3em] 2 & -1 & 2 end{bmatrix})
Long Answer Mar-2017 • PSEB Class 12th3 x+y +z=10,
2x-y-z=0,
x-y + 2z = 1
Let A = (egin{bmatrix} 3 & 1 & 1 \[0.3em] 2 & -1 & -1 \[0.3em] 1 & -1 & 2 end{bmatrix}); x = (egin{bmatrix} x \[0.3em] y \[0.3em] z end{bmatrix})
and B = (egin{bmatrix} 10 \[0.3em] 0 \[0.3em] 1 end{bmatrix})
|A| = 3(-2 -1) -1 (4 + 1) +1 (-2 + 1)
|A| = 3(-3) - 5 - 1 = - 9 - 6 = - 15 != 0
(A_{11}) = (egin{vmatrix} -1 & -1 \[0.3em] -1 & 2 \[0.3em] end{vmatrix}) = - 2 - 1 = -3
(A_{12}) = -(egin{vmatrix} 2 & -1 \[0.3em] 1 & 2 \[0.3em] end{vmatrix}) = - (4 + 1) = - 5
(A_{12}) = (egin{vmatrix} 2 & -1 \[0.3em] 1 & -1 \[0.3em] end{vmatrix}) = - 2 + 1 = - 1
(A_{21}) = (egin{vmatrix} 1 & 1 \[0.3em] -1 & 2 \[0.3em] end{vmatrix}) = - (2 + 1) = - 3
(A_{22}) = (egin{vmatrix} 3 & 1 \[0.3em] 1 & 2 \[0.3em] end{vmatrix}) = 6 - 1 = 5
(A_{23}) = - (egin{vmatrix} 3 & 1 \[0.3em] 1 & -1 \[0.3em] end{vmatrix}) = - (-3 - 1) = 4
(A_{31}) = (egin{vmatrix} 1 & 1 \[0.3em] -1 & -1 \[0.3em] end{vmatrix}) = - 1 + 1 = 0
(A_{32} ) = - (egin{vmatrix} 3 & 1 \[0.3em] 2 & -1 \[0.3em] end{vmatrix}) = - (- 3 - 2) = 5
(A_{33}) = (egin{vmatrix} 3 & 1 \[0.3em] 2 & -1 \[0.3em] end{vmatrix}) = - 3 - 2 = - 5
adj A = (egin{bmatrix} -3 & -5 & -1 \[0.3em] -3 & 5 & 4 \[0.3em] 0 & 5 & -5 end{bmatrix}^t) = (egin{bmatrix} 3 & -3 & 0 \[0.3em] -5 & 5 & 5 \[0.3em] -1 & 4 & -5 end{bmatrix})
(A^{-1}) = (adj A over |A|) = -(1 over 15)(egin{bmatrix} 3 & -3 & 0 \[0.3em] -5 & 5 & 5 \[0.3em] -1 & 4 & -5 end{bmatrix})
x = (A^{-1} B)
x = -(1 over 15)(egin{bmatrix} 3 & -3 & 0 \[0.3em] -5 & 5 & 5 \[0.3em] -1 & 4 & -5 end{bmatrix})(egin{bmatrix} 10 \[0.3em] 0 \[0.3em] 1 end{bmatrix})
= -(1 over 15) (egin{bmatrix} 30 + 0 + 0 \[0.3em] -50 + 0 + 5 \[0.3em] -10 + 0 - 5 end{bmatrix})
= (1 over 15)(egin{bmatrix} 30 \[0.3em] -45 \[0.3em] -15 end{bmatrix})= (egin{bmatrix} -2 \[0.3em] 3 \[0.3em] 1 end{bmatrix})
x = -2 ; y = 3 ; z = 1
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Mar 2018
Solve the following system oflincar equations by matrix mehord:
\r\n
\r\nx - 2y +3z = -5, 3 x +y +c= 8, 2x –y +2z = 1
\r\n
\r\n Or
\r\n
\r\n Using elementary transformations find inverse of \\( \\begin{matrix} 2 & 4 & 1 \\\\ 1 & 2 & 3 \\\\ 1 & -3 & 0 \\end{matrix}\\)
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A wire of length 25 cm is to be cut off into two pieces. One piece is to be made into a circle and other into a square. What should be the lengths of'two pieces so that combined area of,circle and square is minimum ?
Long Answer Mar-2017 • PSEB Class 12th -
Find the image of the point (5, - 3, l) in the plane 2x - 2y - 3z = 10
Long Answer Mar-2017 • PSEB Class 12thThe equation of the plane is 2x - 2y - 3z = 10 -------------(i)
From P (5,3,-1) draw PM plane and perpendicular is to P' s.t . M is the mid-point of PP' Then P' ((lpha eta gamma)) is Image of P
Divertion ratio of PM are 2, -2 , -3
The equation of PM are:
(x- 5 over 2) = (y + 3 over -2) = (z - 3 over 1)
Any pt on it is
M((lpha + 5 , -eta -3, gamma - 3))
M lies on (i)
2(r + 5) - 2(-r -3) - 3( r - 3 ) = 10
2r + 10 + 2r + 6 -3r + 9 = 10
r = -15
M is (-10, 12, -18)
since M is mid point of PP'
({lpha + 5 over 2 }= -10 {eta - 3 over 2 } = 12)
(lpha)= -20 - 5 (eta ) = 24 + 3
(lpha) = -25 (eta ) = 27
(gamma + 1 over 2) = -18
(gamma) = -37
P' (-25, 27, -37) which is the image of P in the plane
Or
Find the shortest distance between the lines :
({x+1 over 4}= {y-3 over -6} = {z+1 over 1} and {x+3 over 3}={y-5 over 2}={z-7 over 6})
Answer :
The equation of two line :
({x+1 over 4}= {y-3 over -6} = {z+1 over 1} ) or ({x-(-1) over 4}= {y-3 over -6} = {z-(-1) over 1} )
And
( {x+3 over 3}={y-5 over 2}={z-7 over 6}) or ( {x- (-3) over 3}={y-5 over 2}={z-7 over 6})
(x_1 = -1 , y_1 = -3, z_1 , -1; )
(a_1 = 4 , b_1 = -6 , c_1 = 1)
and
(x_2 = -3 , y_2 = 5, z_2 = 7; )
(a_2 = 5 , b_1 = -2 , c_1 = 6)
(x_2 - x_1 = -3 - (-1 ) = -3 + 1 = -2)
(y_2 - y_1 = 5 + 3 = 8)
(z_2 - z_1 = 7 + 1 = 8)
(egin{vmatrix} x_2- x_1 & y_2 -y_1 & z_2 - z_1 \[0.3em] a_1 & b_1 & c_1 \[0.3em] a_2 & b_2 & c_2 end{vmatrix})
= (egin{vmatrix} -2 & 8 & 8 \[0.3em] 4 & -6 & 1 \[0.3em] 5 & 2 & 6 end{vmatrix})
= (-2(-36 -2) -8 (24 -5) +8(8 + 30))
= (-2(-38) -8 (19) +8(38))
= 76 - 152 + 304
= 228
(sqrt {(b_1 c_2 - b_2 c_1)^2+ (c_1 a_2 - c_2 a_1)^2 + (a_1 b_2 - a_2 b_1)^2})
= (sqrt {((-6) (6) - (2) (11))^2+ ((1) (5) - (6) (4))^2 + ((4) (2) - (5) (-6))^2})
= (sqrt {(36 - 2)^2+ (5 -24)^2 + (8 + 30)^2})
= (sqrt {(36 - 2)^2+ (5 -24)^2 + (8 + 30)^2})
= (sqrt {(34)^2+ (-19)^2 + (38)^2})
= (sqrt {(1156 + 361 + 1444)})
= (sqrt {2961})
S.D = (|{228 over sqrt {2961}}|)
= ({228 over 3sqrt {329}})
S.D = ({76 over sqrt {329}})
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Maximize Z = 12x + 24y subject to the constrains x+y >= 5, 5x+7y <= 35, x-y>=0, x,y, >-0 graphically.
Long Answer Mar-2017 • PSEB Class 12th -
Or
One kind of cake requires 300 gm of flour and 15 gm of fat and another kind of cakerequires 150 gm of flour and 30 gm of fat. Find the maximum number of cakes that can be made from 7. 5 kg of flour and 600 gm of fat, From a linear programing problem and solve it graphically
Long Answer Mar-2017 • PSEB Class 12thLet x and y be the number of cakes of first and second type that can be made. Clearly x>= 0 , y >= 0
Let Z be the no of cakes
Kind Number of cakes Flour required
(in grams)
Fat required
(in grams)
I
II
x
y
300x
150y
15x
30y
Total x+y 300x + 150y 15x + 30y Mathematical formulation of the L.P.P is as follow:
Maximise : Z = x + y
subject to the constrains
300x + 150y <= 7500 i.e 2x + y <= 50
15x + 30y <= 600 i.e x + 2y <= 40
x,y >= 0
Consider a set of rectangular cartesian axes OXY in the plane.
It is clear that any point which satisfied x>= 0 and y >= 0 lies in the first quadrant
Now we draw the graph of 2x + y = 50
For x = 0 , y = 50
For y = 0 , 2x = 50 or x = 25;
lines meet OX in A(25,0) and OY in L(0,50)
Again we draw the graph of x + 2y = 40
For x = 0 , 2y = 40 or y = 20
For y = 0 , x = 40
Lines meet OX in B(40, 0) and OY in M(0, 20)
Since feasible region satisfied of all the constraints
OACM is the feasible region
The corner point are O(0,0), A(25,0), C(20,10), M(0,20)
At O(0,0), Z = 0 + 0 = 0
A(25,0) Z = 25 + 0 = 25
C(20,10) Z = 20 + 10 = 30
M(0,20) Z = 0 + 20 = 20
Minimum value = 30 at (20,10)
Maximum number of cake is 20 of one kind and 10 of the second kind.
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01 Where can I find PSEB 12th Class Math previous year question papers for 1st semester on BRpaper?
This Math subject page is meant to help students find PSEB Class 12 previous year Math question papers in one place. You can use it to access subject-wise paper links for revision and to review how board-level Math questions are presented.
02 Are the PSEB 12th Math papers on BRpaper arranged subject-wise?
Yes, this page is focused on Math for PSEB 12th Class Question Papers, so it helps students navigate previous year papers by subject. That makes it easier to revise Maths separately instead of searching across all subjects.
03 How should I use previous year Math papers for PSEB board exam revision?
Use old Math papers to practice question selection, step-based solving, and time management. They can also help you spot repeated topic areas, understand common question styles, and check which chapters need more revision before the exam.
04 What is the difference between previous year Math papers and sample or model papers?
Previous year papers show questions asked in earlier board exams, while sample or model papers are practice papers used to understand the current paper style. For best revision, students often use previous year Math papers for real exam trends and the latest official model paper for updated pattern guidance.
05 Can PSEB 12th Math previous year papers help me understand the exam pattern and marks distribution?
They can help you understand the general paper pattern, question types, and level of presentation commonly seen in Math exams. For exact sections or marks distribution for your session, it is better to also check the latest official sample or model paper because patterns can change.
06 How many previous year PSEB 12th Math papers should I practice?
Students often try to solve multiple recent papers so they can compare question style, difficulty, and recurring topics. A practical approach is to start with the latest available papers first, then use older ones for extra revision and self-assessment.