CBSE Board Paper Class-XII. Time allowed : 3 hours Maximum Marks : 100

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1 L.K.Gupta (Mathematic Classes) MOBILE: , CBSE Board Paper -011 Class-XII (SET-1) Time allowed : hours Maimum Marks : 100 General Instructions: (i) All questions are compulsory. (ii) The question paper consists of 9 questions divided into three sections A, B and C. Section A comprises of 10 questions of one mark each, Section B comprises of 1 questions of four marks each and Section C comprises of 7 questions of si marks each. (iii) All questions in Section A are to be answered in one word, one sentence or as per the eact requirement of the question. (iv) There is no overall choice. However, internal choice has been provided in 4 questions of four marks each and questions of si marks each. You have to attempt only one of the alternatives in all such questions. (v) Use of calculators is not permitted. SECTION A Question numbers 1 to 10 carry 1 mark each. 1. Let A {1,, }, B {4, 5, 6, 7} and let f {(1, 4), (, 5), (, 6)} be a function from A to B. State whether f is one-one or not. PIONEER EDUCATION (THE BEST WAY TO SUCCESS): S.C.O. 0, SECTOR 40 D, CHANDIGARH

2 L.K.Gupta (Mathematic Classes) MOBILE: , [Every Image has unique pre-image hence one one ]. 1 π 1 π. What is the principal value of cos cos + sin sin? π cos 1 cos + sin 1 π sin π 1 sin π + sin π π 1 sin π + sin π π π + π. Evaluate: 0 0 cos15 sin sin75 cos75. PIONEER EDUCATION (THE BEST WAY TO SUCCESS): S.C.O. 0, SECTOR 40 D, CHANDIGARH

3 L.K.Gupta (Mathematic Classes) MOBILE: , cos15 sin sin75 cos cos 15 0 cos75 0 sin 75 0 sin 15 0 cos ( ) [By Using : cos (A+B) cos A cos B Sin A Sin B ] cos (90 0 ) 0 4. If A 5, write 1 A in terms of A. A 5 A A 19 I Multiply by A 1 A 1 [A ] 19A 1 A 19A If a matri has 5 elements, write all possible orders it can have. 5 1 and 1 5. PIONEER EDUCATION (THE BEST WAY TO SUCCESS): S.C.O. 0, SECTOR 40 D, CHANDIGARH

4 L.K.Gupta (Mathematic Classes) MOBILE: , Evaluate : ( ) (a + b) ( + ) 4 1 a b a + b. + C a 4 1 ( a + b) 4 + C 4a 7. Evaluate : sin +c Write the direction-cosines of the line joining the points (1, 0,0 ) and (0, 1, 1). The direction ratios are proportional to 0 1, 1 0, 1 0 1, 1, 1 hence, their Direction Cosines are 1 1 1,, ( 1) + ( 1) + ( 1) ( 1) + ( 1) + ( 1) ( 1) + ( 1) + ( 1) 1 1 1,, 9. Write the projection of the vector ˆ i ˆ j on the vectorˆ i + ˆ j. Let a i ɵ j b i ɵ + j Projection of a a cos θ a.b 0 0 b PIONEER EDUCATION (THE BEST WAY TO SUCCESS): S.C.O. 0, SECTOR 40 D, CHANDIGARH

5 L.K.Gupta (Mathematic Classes) MOBILE: , Write the vector equation of a line given by 5 y z 6 5 y + 4 z i ɵ 4j ɵ +6 ɵ k+λ(i ɵ +7j ɵ + ɵ k) SECTION-B 11. Let f : R R be defined as f() Find the function g: R R such that gof fog IR. OR a + b, if a+b<6 A binary operation * on the set {0,1,,,4,5} is defined as: a b a + b 6, if a+b 6 Show that zero is the identity for this operation and each element 'a' of the set is invertible with 6 a, being the inverse of 'a'. gf -1 () y10+7 y y 7 10 f ( ) f ( ) 7 g() 10 PIONEER EDUCATION (THE BEST WAY TO SUCCESS): S.C.O. 0, SECTOR 40 D, CHANDIGARH

6 L.K.Gupta (Mathematic Classes) MOBILE: , OR S{0,1,,,4,5} If e is the identity element then always a* e ae*a Now we have to prove zero as e a + e if a + e < 6 a*e a + e 6 ifa + e 6 If e0 then a + 0 if a + 0 < 6 a*0 a if a a if a + 0 < 6 a*0 a 6 ifa Hence a {0,1,,,4,5} a<6. a*0 {a if a < 6 Hence proved a*0 a 0*a π Prove that : tan cos, L.H.S :- tan Putting cos θ 1 1 cos 1 cos tan + θ + θ 1 + cosθ + 1 cosθ 1 cos θ sin θ tan cos θ + sin θ PIONEER EDUCATION (THE BEST WAY TO SUCCESS): S.C.O. 0, SECTOR 40 D, CHANDIGARH

7 L.K.Gupta (Mathematic Classes) MOBILE: , cos θ sin θ tan cos θ + sin θ 1 1 tan θ tan 1 + tan θ 1 π tan tan θ 4 π θ 4 π 4 1 cos 1 π π θ 4 π π θ 8 4 π π π θ [ 1 1 cos θ [ cosθ θ cos θ cos ] Using properties of determinants, solve the following for : PIONEER EDUCATION (THE BEST WAY TO SUCCESS): S.C.O. 0, SECTOR 40 D, CHANDIGARH

8 L.K.Gupta (Mathematic Classes) MOBILE: , Let Applying C C C 1& C C C ( 1 1 ) 6 Applying R R R & R R R ( )( 6) 1 1 taking & 6commonfromR &R respectively) ( ) ApplyingR R R ( ) {( )( ) ( )} ( ) ( 4) 0 4 (epanding along C 1 ) 14. Find the relationship between 'a' and 'b' so that the function f defined by: a + 1, if f() b + if > continuous at. PIONEER EDUCATION (THE BEST WAY TO SUCCESS): S.C.O. 0, SECTOR 40 D, CHANDIGARH

9 L.K.Gupta (Mathematic Classes) MOBILE: , If y y e, show that Or dy log log(e) { } We have a + 1 if f b + if > (LHL at ) ( ) lim f ( ) lim a + 1 a + 1 ( R H L at ) + ( ) limf b + b( ) + b + Since f() is continuous at, Therefore b + a + 1 b a (a b ) (a b) a + b. or We have, y y e PIONEER EDUCATION (THE BEST WAY TO SUCCESS): S.C.O. 0, SECTOR 40 D, CHANDIGARH

10 L.K.Gupta (Mathematic Classes) MOBILE: , y log y y log ylog e e [ e e ] ylog y ylog + y y ( 1 + log) y 1 + log On differentiating both sides w. r. to, we get 1 ( 1 + log ).1. 0 dy + log 1 log 1 log ( + ) ( + ) y 4sinθ 15. Prove that y θ (+ cos θ) π is an increasing function in 0,. Or If the radius of a sphere is measured as 9 cm with an error of 0.0 cm, then find the approimate error in calculating its surface area. We have, f 4sin θ θ θ + cos θ ( ) ( ) ( )( ) ( + cos θ) + θ θ + θ cos 4cos 4sin f' θ 1 8cos θ + 4 f' θ 1 ( ) ( + cos θ) PIONEER EDUCATION (THE BEST WAY TO SUCCESS): S.C.O. 0, SECTOR 40 D, CHANDIGARH

11 L.K.Gupta (Mathematic Classes) MOBILE: , ( ) f' θ 4cos θ cos ( + cos θ) θ ( ) ( + cos θ) cos θ 4 cos θ π f' ( θ ) > 0 forall θ 0, π Hence, f() is increasing on 0,. or Surface area of Sphere 4πr Let r be the radius of the sphere and r be the error in measuring the radius. Then, r 9cm and r 0.0 cm Let S be the surface area of the sphere, Then, S 4πr ds dr 4π r ds dr r 9 4π 9 7 π Let S be the error in S due to error r in r Then, ds S dr r S 7 π π cm PIONEER EDUCATION (THE BEST WAY TO SUCCESS): S.C.O. 0, SECTOR 40 D, CHANDIGARH

12 L.K.Gupta (Mathematic Classes) MOBILE: , If tan logy, a 1 tan logy a d y dy show that (1 + ) + ( a) 0 1 atan logy a 1 dy d 1 y 1 1. [ [tan ] + ( 1 + ) ay ( 1 ) + dy a.dy d y dy ( 1 ). + + d y a ( ) ( ) dy 17. Evaluate: π/ 0 + sin 1 + cos π/ 0 + sin 1 + cos PIONEER EDUCATION (THE BEST WAY TO SUCCESS): S.C.O. 0, SECTOR 40 D, CHANDIGARH

13 L.K.Gupta (Mathematic Classes) MOBILE: , π/ + sin 0 cos π/ sin cos sec / + 0 cos π/ π/ sec + tan 0 0 π π tan tan π/ 1. tan π π tan π tan π 18. Solve the following differential equation : The given differential equation can be written as dy + y + y, 0 Clearly, it is a homogeneous differential equation. Putting y v and dy v + dv in it, we get dy y + y PIONEER EDUCATION (THE BEST WAY TO SUCCESS): S.C.O. 0, SECTOR 40 D, CHANDIGARH

14 L.K.Gupta (Mathematic Classes) MOBILE: , dv v + + v + v dv v dv 1+ v dv 1+ v 1 v v Integrating both sides, we get 1 1 dv 1 + v log v v log + logc [By separating the variables] v v C y y 1 C + + [ v y/] { y y } + + C 4 Hence, { } y y C gives the required solution 19. Solve the following differential equation :(y + ) dy ( y ) + dy dy y + PIONEER EDUCATION (THE BEST WAY TO SUCCESS): S.C.O. 0, SECTOR 40 D, CHANDIGARH

15 L.K.Gupta (Mathematic Classes) MOBILE: , dy 1 y..(i) This is a linear differential equation of the form dy Py Q + here 1 P Q Sp 1 log e I. F e e e Multiplying both sides of eq. (i) by I. F dy + y Integrating both sides w. r. t, ( ) ( ) ( ) y c y + c y + c y I.F ( ) ( ) Q I.F + c y + c which is the required solution. 0. Using vectors, find the area of the triangle with vertices A(l, 1, ), B(,, 5) and C(l, 5, 5). A (1, 1, ) B(,, 5) And C (1, 5, 5) AB (i ɵ +j ɵ +k) ɵ AC (4j ɵ +k) ɵ PIONEER EDUCATION (THE BEST WAY TO SUCCESS): S.C.O. 0, SECTOR 40 D, CHANDIGARH

16 L.K.Gupta (Mathematic Classes) MOBILE: , Area of triangle 1 AB AC AB AC ɵi ɵj kɵ i ɵ -j ɵ +4 ɵ k Find the shortest distance between the following lines whose vector equations are: r ( 1 t ) iˆ + ( t ) ˆj + ( t ) kˆ and r s + 1 iˆ + s 1 ˆj s + 1 kˆ ( ) ( ) ( ) r (1 t)i ɵ +(t )j ɵ +( t) k ɵ r (S+1)i ɵ +(S 1)j ɵ (S+1) k ɵ r (i ɵ i ɵ t+j ɵ t ɵ j+k t ɵ k) ɵ r Si ɵ +i ɵ +Sj ɵ j ɵ Sk ɵ k ɵ r (i ɵ -j ɵ +k)+t( i ɵ ɵ +j ɵ k) ɵ r (i ɵ -j ɵ k)+s(i ɵ ɵ +j ɵ k) Shortest distance between the lines r a 1+λb 1 and r a +µb is given by PIONEER EDUCATION (THE BEST WAY TO SUCCESS): S.C.O. 0, SECTOR 40 D, CHANDIGARH

17 L.K.Gupta (Mathematic Classes) MOBILE: , ( a a1).( b1 b) d b b 1 a1(i ɵ -j ɵ +k) ɵ a (i ɵ -j ɵ k) ɵ b1 ( i ɵ +j ɵ k) ɵ b (i ɵ +j ɵ k) ɵ b1 b ɵi ɵj kɵ i ɵ -4j ɵ k ɵ a a1 ɵ j 4k ɵ a a b b (j ɵ 4k). ɵ (i ɵ -4j ɵ k) ɵ ( 1 ). ( 1 ) b 1 b Shortest distance A random variable X has the following probability distribution : (i) K (ii) P (X<) (iii) P(X>6) (iv) P(0<X<) PIONEER EDUCATION (THE BEST WAY TO SUCCESS): S.C.O. 0, SECTOR 40 D, CHANDIGARH

18 L.K.Gupta (Mathematic Classes) MOBILE: , Since the sum of all the probabilities in a probability distribution is always unity. Therefore, P (0)+p(1)+. +P(7)1 O+k+k+k+k+k +k +7k +k1 10k +9k 10 (10k 1)(k+1)0 10k 10 [ k 0 k+1 0] k 1 10 (ii)p(<) P(0)+P(1)+P() 0+k+k k (iii)p( > 6) P (7) 7k +k (iv)p (O<<) P (1)+P() k+k PIONEER EDUCATION (THE BEST WAY TO SUCCESS): S.C.O. 0, SECTOR 40 D, CHANDIGARH

19 L.K.Gupta (Mathematic Classes) MOBILE: , k SECTION-C. Using matrices, solve the following system of equations : 4 + y + z 60 + y + z y + z y + z 60 + y + z y + z 70 Or y 45 6 z 70 Or AX B A , y,b 45 6 z 70 4 A 1 6 4( 6 6) ( 18) + ( 1) ( 15) + ( 10) So, the given system of equations has a unique solution by X 1 A B PIONEER EDUCATION (THE BEST WAY TO SUCCESS): S.C.O. 0, SECTOR 40 D, CHANDIGARH

20 L.K.Gupta (Mathematic Classes) MOBILE: , The co- factors of elements in A are ( ) ( ) ( ) C C C ( ) ( ) ( ) C C C ( ) ( ) ( ) C C 1 0 C Adj A t Adj 1 A t A /5 1/5 /5 /5 1/5 t A /5 0 /5 1 X A B X X y z , y 8, z 8 is the required solution PIONEER EDUCATION (THE BEST WAY TO SUCCESS): S.C.O. 0, SECTOR 40 D, CHANDIGARH

21 L.K.Gupta (Mathematic Classes) MOBILE: , Show that the right-circular cone of least curved surface and given volume has an altitude equal to times the radius of the base. Let r and h be the radius and height of the cone respectively. Volume 1 V π r h πk (Constant) l k r h k or h r Surface S π rl π r h + r k Put h from (1). r k S π r. + π r k + r r 6 r r 4 π r + k r 6 5 6r 6 r r + k.1 ds 6 π r + k dr r PIONEER EDUCATION (THE BEST WAY TO SUCCESS): S.C.O. 0, SECTOR 40 D, CHANDIGARH

22 L.K.Gupta (Mathematic Classes) MOBILE: , ( ) 6 6 r r + k 6 r k r r + k r r + k 6 6 ds 0 at k r 6 dr () ds dr changes sign from ve to +ve as r increases through the point 6 k r S is least at this point. From (1) k h r 4 h r 4 r 6 or h r Or A window has the shape of a rectangle surmounted by an equilateral triangle. If the perimeter of the window is 1 m, find the dimensions of the rectangle that will produce the largest area of the window. + y 1m 1 y --(1) A y + 4 PIONEER EDUCATION (THE BEST WAY TO SUCCESS): S.C.O. 0, SECTOR 40 D, CHANDIGARH

23 L.K.Gupta (Mathematic Classes) MOBILE: , ( 6) da ( 6) d A 1 ( 6) 0 < So, area will be maimum for all values of and y. Put da 0 [ for critical points] ( 6) Put in (1) (6 ) 6 y (6 ) 1 6 6( ) 18 6 (6 ) Evaluate: π/ π /6 + 1 tan Or PIONEER EDUCATION (THE BEST WAY TO SUCCESS): S.C.O. 0, SECTOR 40 D, CHANDIGARH

24 L.K.Gupta (Mathematic Classes) MOBILE: , Evaluate: π/ π /6 + 1 tan Let I π/ π /6 + 1 tan π/ cos (1) cos sin π /6 + π/ π/6 π cos π π cos + sin π/ sin () sin cos π /6 + Add 1 and I π/ π/6 I 1 [ ] / π π/6 1 π π 6 1 π π 6 PIONEER EDUCATION (THE BEST WAY TO SUCCESS): S.C.O. 0, SECTOR 40 D, CHANDIGARH

25 L.K.Gupta (Mathematic Classes) MOBILE: , π or ( )( ) ( + ) 1/ ( 1/ + 1) log c log c 6. Sketch the graph of y + and evaluate the area under the curve y + above -ais and between 6 to 0. PIONEER EDUCATION (THE BEST WAY TO SUCCESS): S.C.O. 0, SECTOR 40 D, CHANDIGARH

L.K.Gupta (Mathematic Classes) www.poineermathematics.com.

26 L.K.Gupta (Mathematic Classes) MOBILE: , The graph of y ( + ) meets X ais at point P (, 0) To evaluate the area under the curve y + above X ais & between 6 & 0 The graph will be The Required Area ( ABC + CDO) is 0 ( ) ( ) {( ) ( ) } ( ) ( ) ] { } {( )} ( ) PIONEER EDUCATION (THE BEST WAY TO SUCCESS): S.C.O. 0, SECTOR 40 D, CHANDIGARH

27 L.K.Gupta (Mathematic Classes) MOBILE: , [9 6] [] + [ 9] + [9] [ 7] 9 + 9sq. units 7. Find the distance of the point ( 1, 5, 10), from the point of intersection of the line r (i ˆ ˆj + k) ˆ + λ.(i ˆ + 4j ˆ + k) ˆ and the plane r.(i ˆ ˆj + k) ˆ 5. The coordinates of any point on the line is ( + λ, 1+ 4 λ, + λ ) (i) If it lies on the plane r ɵ ɵ ( i ɵ j + k) 5 them ( + λ ) 1 + ( 1 + 4λ)( 1) + ( + λ ) ( 1) 5 + λ + 1 4λ + + λ 5 λ 0 r ɵ ɵ ( i ɵ ɵj + k ) + λ i ɵ + 4j ɵ + k ( ) Putting λ 0in (i) we get (, 1, ) as the coordinates of point of intersection of given line plane Required distance Distance between point (, 1, ) & ( 1, 5, 10) ( 1) ( 1 5) ( 10) units PIONEER EDUCATION (THE BEST WAY TO SUCCESS): S.C.O. 0, SECTOR 40 D, CHANDIGARH

28 L.K.Gupta (Mathematic Classes) MOBILE: , Given three identical boes I, II and III each containing two coins. Ii bo I, both coins are gold coins, in bo II, both are silver coins and i: bo III, there is one gold and one silver coin. A person chooses a boat random and takes out a coin. If the coin is of gold, what is the probability that the other coin in the bo is also of gold? Consider the following events E1 Bo I is chosen E Bo II is chosen EBo III is chosen AThe coin drawn is of gold We have, 1 P( E ) P 1 ( E ) P( E ) P(A/E1) Probability of drawing a gold coin from bo I P( A/E1 ) 1 P(A/E) Probability of drawing a gold coin from bo II ( ) P A/E 0 P(A/E) Probability of drawing a gold coin from bo III 1 P( A/E ) Now, Probability that the other coin in the bo is of gold Probability that gold coin is drawn from the bo I PIONEER EDUCATION (THE BEST WAY TO SUCCESS): S.C.O. 0, SECTOR 40 D, CHANDIGARH

29 L.K.Gupta (Mathematic Classes) MOBILE: , (PE1/A) P( E ) P( A/E 1 1 ) ( ) ( ) + ( ) ( ) + ( ) ( ) P E P A/E P E P A/E P E P A/E A merchant plans to sell two types of personal computers a desktop model and a portable model that will cost Rs. 5,000 and Rs. 40,000respectively. He estimates that the total monthly demand of computers will not eceed 50 units. Determine the number of units of each type of computers which the merchant should stock to get maimum profit if he does not want to invest more than Rs. 70 lakhs and his profit on the desktop model is Rs. 4,500 and on the portable model is Rs. 5,000. Make an L.P.P. and solve it graphically. Sol: Regular problem, still can send doubts in doubt forum. PIONEER EDUCATION (THE BEST WAY TO SUCCESS): S.C.O. 0, SECTOR 40 D, CHANDIGARH

MATHEMATICS. Time allowed : 3 hours Maximum Marks : 100

MATHEMATICS. Time allowed : 3 hours Maximum Marks : 100 MATHEMATICS Time allowed : hours Maimum Marks : General Instructions:. All questions are compulsory.. The question paper consists of 9 questions divided into three sections, A, B and C. Section A comprises