MATHEMATICS Class: XII Mock Paper 1 Solutions

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1 MATHEMATICS Class: XII Mock Paper Solutions Section A. If A { a, b, c } and B {,, } and a function f : A B is given by f { (a, ), ( b, ), ( c, ) } Every element of set A is mapped to the unique element of set B.i.e each element in the set B has a unique pre image in B f is a one one function Range of f {,, } B f is an onto function f is a injective function. The function y cos can be inverted in the intervals where it is both one one and onto i.e in the intervals [ - π, - π ], [-π,0 ], [0, π ], [π, π]. (A + B ) (A + B) ( A + B) A (A + B) + B (A + B) A + AB + BA + B which may or may not be equal to A + AB + B [Since matri multiplication is not commutative] So the epression is not true in general. -4 ' 4. Let A A A A' ' ( A ) ( ) ( ) Transpose of A A' A A' ' ( A A' ) A is a skew symmetric matri ( Mark) 5. A 4 and A

2 5 8 A - 6A A - 6A A - 6A + 7I A - 6A + 7 I I sin sin 5 + c c 7. The unit vector in the direction of vector a r r $ a a r a $ $ i $ j $ + 6k i $ j $ + 6k$ a i $ j $ + 6k$ i $ j $ + 6k$ i $ j $ + 6k$ ˆ ˆ 6 i j + k ˆ

3 8. P(-,-,4) and Q(,0,-) position vector of P -i j+ 4k position vector of Q i+ 0 j k uuur PQ position vector of Q - position vector of P i+ 0 j k - i j+ 4k i+ j 6k ( mark) uuur Magnitude of PQ + + ( 6) 49 7 ( mark) 9. Given A(,-5,), B (-, 0, 8) and C ( 7, -0, -6) position vector of A i 5 j+ k position vector of B - i + 0 j+ 8k position vector of C 7 i 0 j 6k uuur AB position vector of B- position vector of A - i + 0 j+ 8k i 5 j+ k -4 i+ 5 j+ 7k uuur AC position vector of C -position vector of A 7 i 0 j 6k i 5 j+ k 4 i 5 j 7k uuur uuur AC AB uuur uuur AB and AC hasame magnitude but opposite directions The points A, B and C are collinear 7 0. Let f() sin 7 7 f(-) sin sin f() So f() is an odd function of π/ - π/ 7 sin. 0 Section B : Section B comprises of questions of four marks each.

4 .(i) (a, b) * (c, d) ( ac, ad + b) (c, d) *(a, b) ( ca, c b + d) ( ac, ad + b) ( ca, c b + d) So, * is not commutative (ii) Let (a,b ) ( c, d ), ( e, f ) A, Then (( a, b )* (c, d)) * ( e,f) (ac, ad + b) * (e,f) ( (ac) e, (ac) f + (ad+b)) ( ace, acf +ad +b) (a,b)* ((c,d)* (e,f)) (a,b) * ( ce, cf +d) ( a (ce), a ( cf+d) +b) ( ace, acf +ad +b) (( a, b )* (c, d)) * ( e,f) (a,b) ((c,d)* (e,f)) Hence, * is associative. (iii) Let (, y) A. Then (, y) is an identity element, if and only if (, y ) * (a, b) (a, b) (a,b) * (, y ), for every (a,b) A Consider (, y ) * (a, b) ( a, b + y) (a,b) * (, y ) ( a, ay +b ) ( a, b + y) (a, b) ( a, ay +b ) a a a b + y b ay +b b + y b ay +b y 0 ay y 0 Therefore, (, 0) is the identity element OR * [ Marks] From the table, the second row and second column are the same as the original set. 0*0 0, *0 0*, *0 0*, *0 0*, 4*0 0*4 4, 0*5 5*0 5 0 is the identity element of the operation *

5 Now, the element 0 appears in the cell *5 5* 0, *4 4 * 0, * 0, and 0* 0 0 Inverse element of 0 is 0, Inverse element of is 5, Inverse element of is 4, Inverse element of is, Inverse element of 4 is, Inverse element of 5 is..given:tan tan - - tan tan (>0) tantan tan tan + - tantan + - tantan ( )( + ) ( + ) ( ) ( ) ± e e 0. lim f() lim e + e e 0 lim f() lim lim + 0 lim f() lim f() So lim f() does not eist. 0 f() is not continuous at 0 e

6 4. y a+ a+ a+,where a is a constant. y a a+ a+ [Mark] + dy d a a+ a+ a a+ a+ [Mark] + + d y d a a+ a+ a a+ a a+ [Mark] dy a a+ a+ a a+ a+. [Mark] + + dy a + a+ a+. a + a+. a+ 4 π π 5.f() sin in the interval -, Let h() sin, g() goh() f() sin π π h()sin is a continuous function in the interval -, π π g() is a continuous function in the interval -, π π goh() sin is also a continuous function in the interval -, π π h()sin is a differentiable function in the interval -, π π g() is not a differentiable function in the interval -, π π goh() sin is also not a differentiable function in the interval -, Conditions of Lagrange's theorem are not satisfied Lagrange's theorem is not applicable for the given function

7 ( + ) 0 0 ( + + ) ( + ) + [ Marks] + 4 OR

8 LetI π/ 0 π/ 0 - sin tan (sin ) - sin cos tan (sin) Let sin t cos dt π 0 t 0, t I t tan t dt t tan t dt 0 0 Integrating by parts, we have t t tan t dt ( t ) t 0 ( + t ) tan 0 dt 4 0 ( + t ) π t + dt π dt 4 0 ( t ) + π [t 4 + [tan t 0 0 π π π

9 π π 7.Let sin α and y sin β, such that α, β, sin α + sin β a sinα-sinβ cos α + cos β a sinα-sinβ α + β α β α + β α β cos cos a cos sin α β α β cos a sin α β cot a α β cot a α β cot a [ Marks] sin sin y cot a [ Marks] Differentaiting w.r.t, we have dy y dy -y - 8. (-)dy + y (-) y dy (-) + y (-) y dy y + - y / / dy dt Let y t y dt t + - dt t + - This is a linear differential equation, whose integrating factor is log() - log / IFe e e ( ) [ Mark]

10 Solution of Differential equation is / / t ( ) ( ) + c / / / y ( ) ( ) + c 5/ 5 / / / ( ) ( ) y ( ) + C / ( ) C / / 5 / 8 / y ( ) ( ) ( ) + C 5 0 / / y ( ) ( ) + C ( ) [ Mark] 5 0 OR dy 8. sec y + tany Let tan y t sec y dy dt dt + t This is a linear differential equation with integrating factor : IF e e [ Mark] Solution of the differential equation is given by t e e + C [ Mark] To solve e Let z dz e z ze z e z dz C + ze z e z C + ze dz e C + t ( ) + Ce tany ( ) + Ce 8 /

11 r r r 9. Here, a + b + c 0 r r r a + b c r r r ( a + b) ( c ) [ mark] r r r r r a + b + a.b c [ mark] r r r r r a + b + a b cos θ c, [ mark] r r where θ is the angle between a and b () + (5) + ()(5) cos θ (7) (Mark) cos θ 49 5 cos θ 0 o θ cos 60 [ Mark] a i λ j+ k andb 4 i 5 j+ k vectors are perpendicular if a.b 0 [ mark] a.b i λ j+ k. 4 i 5 j+ k [ mark] 4 + λ λ λ [ mark] 0 + 5λ 0 λ [ mark] OR

12 + 5 y + z 6 0. The given line is 4 9 Given point is (, 4, -) uur The distance of a point whose position vector is a from a line whose vector equation is r d v (a a ) v ( Mark) a i + 4j 9k ( i + 4 j k) ( 5i j+ 6k) i + 4j 9k i + 4j 9k (7 i + 7 j 7k) i + 4j 9k (Mark) ( Mark) i $ j k i + 4j $ 9k (7 i + 7 j 7k) 4 9 5i $ 56$ j k $ (Mark) i 8 j k 98 7 units (Mark) 98 98

13 { { }}.S (,y,z):,y,z,,,4,5,6 S contains cases ' [/ mark] Let E : an odd number appears atleast once E : an odd number appears none of the times ' i.e E : an even number appears all three times ' ' { { }} E (,y,z):,y,z,4,6 E contains 7 cases Now, P(E) - P(E ) ' [ mark] [ mark] [/ mark] 7 7 [ mark] sinθ. A -sinθ sinθ sinθ ( + sin θ) sin θ( sin θ + sin θ ) + ( + sin θ) ( + sin θ) [ mark] 0 sin θ ( + sin θ) () ( + sin θ) () [ mark] ( + sin θ) 4 [ Marks] A 4 A,4 [ mark] So value of A is in interval [,4]

14 Section C : Section C comprises of 07 questions of si marks each. Given A AIA A 0 0 R R + R R A ( Mark) 0 0

15 R R R R R R, A R R A R R R A A ( mark) A 0 [ mark]

16 Let radius of the cone be R and height h. Let r be the radius of the cylinder and be its height. Consider triangles PAN and PBS.!PAN ~!PBS PA AN PB BS h r h R R ( h ) r h The volume of the cylinder V πr ( ) ( ) R h π h πr h h π R h + h ( Marks) h

17 dv π R h + 4h h dv 0 π R h + 4h 0 h + h 4h 0 ( h)( h) 0 h h,, h but < h, so h [ marks] ( ) d V πr 6 4h h [ mark] d V πr ( 6 4h) πr < 0 h [ mark] h h h h/ is a point of local maima V ( ) πr h h h/ h 4πR h 4πh tan α R tan 7 7 Q α h [ mark] 5.

[ mark] The points where the two parabolas meet in the first quadrant are obtained by solving the two equations y.() and y..() Substituting from () into (), we get X ( ) 4 4 0, i.

18 [ mark] The points where the two parabolas meet in the first quadrant are obtained by solving the two equations y.() and y..() Substituting from () into (), we get X ( ) 4 4 0, i.e ( - ) 0 0, So y 0, The points where the two parabolas meet in the first quadrant are (0,0) and (0, ). The area gets divided into parts as shown in three different colours. / Area I (In Blue) - squnits / 0 0 / Area II (In Red) squnits / 0 0 Area III (In Green) sq.units 0 0 Area I Area II Area III The curves y and y divide the square bounded by 0, y0, and y into three parts that are equal in area. [Each area mark, conclusion mark] 6 Let E be the event that a patient used Drug A P(E ) ½ Let E be the event that a patient used Drug B P(E ) ½ Let E be the event that a patient had a heart attack Required probability :P(E /E) [ mark]

19 P(E /E ) P(E /E ) [ marks] 00 P(E /E )P(E ) P(E /E) [ mark] P(E /E )P(E ) + P(E /E )P(E ) [ marks] OR S : the bulb is manufactured by machine X S : the bulb is manufactured by machine Y S : the bulb is manufactured by machine Z Required probability :P(S /E) [ mark] P(S ) /6 P (S ) / P(S ) ½ P(E S ) 00, P(E S ) 00 P(S )(P(E S ) P(S E) [ mark] P(S )(P(E S )) + P(S )(P(E S )) + P(S )(P(E S )) [ marks

20 8. [ mark] [/ marks] mark] [

21 [/ marks] ) ( ) ( ) ( ) ( ) )( ( 4 [ mark] OR

22 [ Marks] 9.Let the two tailors work for days and y days respectively, The problem is to minimise the objective function C y Subject to the constraints 6 + 0y 60 +5y 0 4+4y + y 8 And 0,y 0

23 [ Marks] Feasible region is shown shaded.. [ marks] This region is unbounded. corner points objective functi A( 0,0) 500 E(5,) 50. D(0,8) 600 The red line in the graph shows the line y 50 or + 4y 7 We see that the region + 4y > 7 has no point in common with the feasible region. Hence, the function has minimum value at E (5,). Hence, The labour cost is the least, when tailor A works for 5 days and Tailor B works for day [ mark]

MATHEMATICS (SET -3) Labour cost Z 300x 400y (to be minimized) The constraints are: SECTION - A 1. f (x) is continuous at x 3 f (3) lim f (x)

MATHEMATICS (SET -3) Labour cost Z 300x 400y (to be minimized) The constraints are: SECTION - A 1. f (x) is continuous at x 3 f (3) lim f (x) 8 Class th (SET -) BD PPER -7 M T H E M T I C S () SECTION -. f () is continuous at f () lim f () ( ) 6 k lim ( )( 6) k lim ( ) k. adj I 8 I 8 I 8I 8. P : z 5 5 P : 5 5z z 8 Distance between P & P sin