SOLUTIONS SAMPLE QUESTION PAPER - 6 Self Assessment Time : Hours Maximum Marks : 00 SECTION A. We have R {(x, y) : (x y) is odd natural number x A, y B}. R {(5, 4)}. f(x) 5 x.. It is defined when 5 x > 0 x < 5. Domain of f {x : x R, x < 5} lim x 0 sin Range (0, ). tan 8x 8 tan 8x x lim 8x 8 x x 0 sin x x x 4. cosec ( 40 ) cosec ( 60 4 + 0) cosec 0. 5. A (, ) x O (, 5) x 4 + 6 and y 5 B (x, y) y 0 y 0 + 7 Co-ordinates of B are (6, 7). 6. Contrapositive Statement : If x is not divisible by 4, then x is not an even number. MATHEMATICS Oswaal CBSE Class -, Examination Sample Question Paper
OSWAAL CBSE Sample Question Paper, Mathematics, Class XI 7. L.H.S. SECTION - B cos A cosb cosc + + a b c b + c a a + c b a + b c + + abc abc abc b + c a + a + c b + a + b c abc a + b + c abc R.H.S. Hence proved. Given that, cot a and sec b 5 tan a and tan b 5 tan b 5 9 9 tan b 4 tan (a + b) tan α+ tanβ tan αtanβ 4 4 + π [ β, π ] tan (a + b) 8. L.H.S. cot 7 cos 7 sin 7 cos 7.cos 7 sin 7. cos 7 cos 7 sin5 + cos5 sin5 + cos (45 0 ) sin (45 0)
Solutions + cos45 cos0 + sin 45 sin 0 sin 45.cos0 cos45 sin 0 + + ( + ) + + + ( + + )( + ) ( )( + ) 6 + + + + + ( ) 6 + + + 4 6 + + + 6 + + + 4 R.H.S. Hence proved. 9. We have kx 5k (k ).x (k ).x kx + 5k 0 ( k) 4(k ). 5k For real and unequal root 4k(4k 5) > 0 4k(4k 5) < 0 4k 0k(k ) 4k(k 5k + 5) 4k( 4k + 5) 4k(4k 5) > 0 0 < k < 5 4 ( + i 5)( i 5) ( + i ) ( i ) 9+ 5 i i i 7 i i 7 i 7 0 i
4 OSWAAL CBSE Sample Question Paper, Mathematics, Class XI 0. Veena visits to four cities A, B, C and D. Total arrangements in which Veena can visit 4! 4 (i) She visits A before B and B before C P(X) 4 4 6 (ii). We have a pack of 5 playing cards X {ABCD, DABC, ABDC, ADBC} Y Order of cities A first and B last Y {ACDB, ADCB} P(Y) 4 Total ways of selecting 4 cards 5 C 4 P(A). P(n) : 4 n + 5n is divisible by 9 at n, P() : 4 + 5 8 is divisible by 9 5 5 50 49 4 A Event of getting card from each suit P(A) 97 085 C C C C 5 5 50 49 4 4 5 5 50 49 So, P() is true Let P(k) is true. i.e., 4 k + 5k is divisible by 9 4 k +5k 9l, l N. 4 k 9l 5k +...(i) Now to show P(k +) is true. i.e., 4 k+ + 5(k + ) is divisible by 9. Taking 4 k+ + 5(k +) 4 k.4 + 5k + 5 Using equation (i), we get P(k + ) is true, when P(k) is true. So, P(n) is true for all n N. P(n) 4(9l 5k + ) + 5k + 4 6l 60k + 4 + 5k + 4 + + +... + 5 5 7 7 9 (n+ )(n+ ) Put n, 6l 45k + 8 9(4l 5k + ) 9m, (m 4l 5k + ) n ( n + )
Solutions 5 P() : 5 5 So, P() is true. Let P(k) is true. P(k) : + +... + 5 5 7 (k+ )(k+ ) k ( k + )...(i) Now prove that P(k + ) is true. i.e., + +... + + 5 5 7 (k+ )(k+ ) (k+ )(k+ 5) ( k + ) (k + 5) Take L.H.S. + +... + + 5 5 7 (k+ )(k+ ) (k+ )(k+ 5) Using equation (i), we get k + (k+ ) (k+ ) (k+ 5) k(k+ 5) + (k+ )(k+ 5) k + 5k+ (k+ )(k+ 5) (k+ ) ( k+ ) (k+ )(k+ 5) k + (k + 5) R.H.S. Hence, P(k + ) is true, when P(k) it true. P(n) is true for all n N.. f(x) For domain, as x 0 x x +, x R ½ x + > 0. ½ f(x) is defined for all x R. Domain of f R. For range, let f(x) y y x x + x y y ½ x y y y > 0 y < and y 0. Range : 0 y <. ½
6 OSWAAL CBSE Sample Question Paper, Mathematics, Class XI 4. f (x) sin (5x 8) fx ( + h) fx ( ) f '(x) lim h 0 h sin {5( x+ h) 8} sin(5x 8) lim h h 0 5x+ 5h 8+ 5x 8 5x+ 5h 8 5x+ 8 lim cos sin h h 0 5h 5h lim cos 5x 8 sin h 0 h + 5h sin 5h 5 lim cos (5x 8 + h 0 5h 5.cos (5x 8). 5 cos (5x 8) 5h sin lim 5h h 0 5. f(x), [ ] x x x R sin x lim x 0 x As, 0 (x [x]) <, " x R But when x Z x [x] 0 0 < (x [x]) <, " x R Z Domain of f R Z. For range, as 0 < < 0 < x [x] <, " x R Z x [ x] < < f(x) < < x [ x] Range (, ). 6. Let two positive integers be x and y. According to question, a x+ y...(i) and x, b, c, y are in G.P. Let r be common ratio. b xr, c xr, y xr r y x / y b x x /
Solutions 7 and y c x x / b b + c xy, c x xy xy + x x b + c xy (x + y) Using (i), and x + y a xy x x y + xy xy bc b + c abc 7. A (, 4, ), B (,, 4), and C (,, ) are co-ordinates of vertices of DABC, A (, 4, ) B (,, 4) C (,, ) AB ( ) + ( 4) + (4 ) + 49 + 4 54 BC ( ) + ( + ) + ( 4) BC 6 + 6 + 4 6 and AC ( ) + ( 4) + ( ) 9 + 9 8 BC + AC 6 + 8 54 AB DABC is a right triangle. A (, ), B (, ) and C (0, 0) are the vertices of DABC. P is concurrent point of altitude, so it is orthocentre. A (, ) P (x, y) E B (, ) D C (0, 0) Now, slope of BC slope of AD Equation of AD is y (x + ) x y + 5 0...(i) Now, slope of AC
8 OSWAAL CBSE Sample Question Paper, Mathematics, Class XI slope of BE 8. Equation of BE is y + ( ) x y + x x y 5...(ii) On solving equations (i) and (ii), we get x 4, and y. Co-ordinates of orthocentre P ( 4, ). Class f i x i u i xi - 05 0 u i f i u i f i u i 0 0 5 9 6 8 0 60 45 4 6 60 90 5 75 5 5 90 0 0 05 a 0 0 0 0 0 50 5 50 80 5 65 4 0 0 80 0 95 9 6 8 Total 0 76 Variance σ h N [N.Σ f i u i (Σ f i u i ) ] Here h 0, N 0. σ 0 0 76 () 0 [80 4] σ 76 9. We have word DAUGHTER which consists of vowels and 5 consonants. Number of words consisting vowels and consonants C 5 C 5! 5 4 0 600. SECTION - C cos 4x+ cosx+ cosx 0. L.H.S. sin 4x+ sin x+ sin x 4 4 cos x+ x cos x x + cos x 4x x 4x x sin + cos + sin x cos x.cos x+ cosx sin x.cos x+ sin x
Solutions 9 cos x( cos x+ ) sin x( cos x+ ) cos x R.H.S. Hence proved.. A Set of people who read English news-paper B Set of people who read Hindi news-paper According to question, Hindi B A n(u)5000 English n(a B)875 n(a) 50, n(b) 750 and n(a B) 875. n (A B) n(a) + n(b) n (A B) 50 + 750 875 Number of people who neither read Hindi nor English 400 875 5 n(u) n (A B) Number of people who read only English news-paper 5000 5 875. 50 875 75.. Equations of two straight roads are x y 4 0...(i) and x + 4y 5 0...(ii) Family is standing at the point P. On solving equations (i) and (ii), we get P, 7 7 N x 7, y 7 x + 4y 5 0 6x 7y + 8 0 P(x, y) x y 4 0 Because family is at P, want to reach 7 7
0 OSWAAL CBSE Sample Question Paper, Mathematics, Class XI the path in least time. So they have to walk along PN. 6x 7y + 8 0...(iii) ½ Now, slope of path (iii) 6 7 slope of PN 45 0 Equation of PN y + 7 x 7 6 7 7y + 7 7 7x 6 7 0y + 9x + 7 9x + 0y 7 ½ 9x + 0y 05. So, he should follow the path given by 9x + 0y 05. Because of unavailability of road transport, they have less time to reach to station, that is why they have decided to take the shortest path.. The expansion is ( + x) n T r + n C r x r Coefficients of (r ) th, r th and (r + ) th terms are n C r, n C r and n C r+ According to question, n C r : n C r : n C r+ 45 : 0 : 0 n Cr n Cr 45 0 ( n)! r!( n r)! ( r )!( n r+ )! n! 45 0 n r + 0...(i) Also n Cr 0 n C r + 0 n! ( r+ )!( n r )! 4 r!( n r)! n! 7 4n r 7 0...(ii) On solving (i) and (ii), we get n 0 and r n 0 (5) 4 (50 + ) 4 Using Binomial theorem, we get 4 C 0 (50) 4 + 4 C 50 + 4 C 50 + 4 C 50 + 4 C 4 4 650000 + 8 5000 + 4 500 + 50 + 6 650000 + 000000 + 60000 + 600 + 6 7,,66
Solutions 4. Given in equations are Y x + y 60 x + y 0, x 0, y 0...(i)...(ii) 40 x + y 60 0 0 0 C(0, 0) B(5, 5) X' 0 0 0 0 40 A(0, 5) x + y 0 X Y' Take x + y 60 At x 0, y 0 and y 0, x 0 At (0, 0); 0 + 0 < 60, (True) i.e., (0, 0) is included. Now take x + y 0 At x 0, y 0 and y 0, x 0. At (0, 0); 0 + 0 < 0, (True) So, origin is also included. So, shaded region OABC is the solution region. 5. Given that, n+ n+ a n + + b n + (ab) ½ [a n + b n ] a n + + b n + a. b + a. b a n + + b n + a. b a. b 0 a + b n n a + b ab n+ n+ n+ n+ n+ n+ n+ n+ a. a + b. b a. b a. b 0 n+ n+ a b a b 0 n+ n+ a b 0 n + a b n +
OSWAAL CBSE Sample Question Paper, Mathematics, Class XI a b n + a b n + a b n + 0 n x + y + 6. Let AB be the line x + y 7 and image of p(, 8) be (x, y ) Middle point of PQ. (i.e.) M, 8 lies on AB P(, 8) A m B x + y 7 x + y + 8 + 7 Q (x, y ) x + + y + 4 4 x + y + 0...(i) Slope of AB Slope of PQ y 8 x AB PQ y 8 x y 8 (x ) x 9 y x...(ii) Putting Value of y in (i), we get x + (x ) + 0 0x + 0 0 x Putting x in (ii), we get y 4 The point Q, the image of P is (, 4) We have z+ i z+ 5+ 4i 5, where z is only complex number. z+ i z+ 5+ 4i 5
Solutions x + iy + i x + iy + 5+ 4i 5 ( x+ ) + iy ( ) ( x+ 5) + iy ( + 4) 5 As, z zz ( x+ ) + iy ( ) ( x+ ) + iy ( ) 5 ( x+ 5) + iy ( + 4) ( x+ 5) + iy ( + 4) [( x+ ) + iy ( )][( x+ ) iy ( ) ] 5 [( x+ 5) + iy ( + 4)] [( x+ 5) iy ( + 4)] (x + ) + (y ) 5 [(x + 5) + (y + 4) ] x + 4 + 4x + y + y 5 (x + 5 + 0x + y + 6 + 8y) 5x + 5y + 50x + 00y + 05 x y 4x + y 5 0 4x + 4y + 46x + 0y + 00 0 x + y + x + 0y + 50 0. qqq
SOLUTIONS SAMPLE QUESTION PAPER - 7 Self Assessment MATHEMATICS Oswaal CBSE Class -, Examination Sample Question Paper Time : Hours Maximum Marks : 00 SECTION-A. We have A {,,, 4} and B {,, 5, 7} R A relation is less than from A to B R {(, ), (, 5), (, 7), (, ), (, 5), (, 7), (, 5), (, 7), (4, 5), (4, 7)}.. 5x < x + x < 4 x <. (i) When x is an integer, then x {...,,, 0, } ½ (ii) When x is real number, then < x <. ½. Equation of line is y ( ) x + 4 x + 4 y + 6 0 x y + 0 0 is the required equation. 4. Centre of circle (, ), r 4 Equation of circle is (x + ) + (y ) 4 x + y + 4x 6y 0. + x+ x + x + x sin( + x) sin( x) cos sin 5. lim lim x 0 x x 0 x sin x lim cos x 0 x sin x cos lim x 0 x cos. 6. It is false that, the sum of and is 6. SECTION-B A 9A 7. L.H.S. cos A cos cos A cos A 9A cos A cos cos A cos A A cos cos A+ + A 9A 9A cos A+ cos A
Solutions 5 But cos ( A) cos A 5A A 5A A cos + cos cos cos 5A 5A c os cos 5 5 5 5 sin A A sin A A + ( ) 5A sin 5 A.sin R.H.S. Hence proved 8. A Set of students who take tea B Set of students who take coffee According to question, n(a) 50, n(b) 5 n(a B) 00 n(a B) 50 + 5 00 75 00 n(a B) 75. Number of students who neither take tea nor coffee n(u) n(a B) 600 75 5. ( x 4) f (x) x 4 For Domain x 4 0 i.e., x 4 Domain R {4} Range : When x 4; f(x) ( x 4) x 4 and when x < 4, f(x) ( x 4) x 4 Range of f {, }. 9. It is given that atan q + bsec q c (atan q c) bsec q a tan q + c actan q b sec q b ( + tan q) a tan q c + actan q 0 (b a ) tan q + actan q + b c 0. As a and b are its solutions tan a and tan b are their roots. ac tan a + tan b b a and b c tan a tan b b a tan α+ tanβ tan (a + b) tan αtanβ
6 OSWAAL CBSE Sample Question Paper, Mathematics, Class XI ac b a b c b a ac b a ( c a ) b a ac tan (a + b) a c ( a+ i) 0. As p + iq a i ( a i) p iq a+ i ( a+ i) ( a i) (p + iq) (p iq) a i ( a+ i) p + q ( a + ), 4a + Hence proved. z + i (given) Put rcos θ...(i) and r sin θ...(ii) Square and adding, we get r 4 r and tan q tan π q arg (z) π π π + i cos + i sin.. We have Using C and D, we get tan (a + β) ntan (a b) tan ( α+β) tan ( α β ) n tan ( α+β ) + tan ( α β) tan ( α+β) ta n ( α β) n + n sin ( α+β) sin ( α β) + cos ( ) cos ( ) α+β α β sin ( α+β) sin( α β) cos ( α+β) cos( α β) n + n sin ( α+β).cos ( α β ) + cos ( α+β).sin ( α β) sin ( α+β).cos ( α β) cos (( α+β).sin ( α β ) n + n
Solutions 7 sin( α+β+α β) sin( α+β α+β ) n + n sin( α) sin( β ) n + n (n + )sin b (n )sin a. Hence proved.. We have 6 red, 5 white and 5 blue balls, then 9 balls are to be selected with balls of each colour. Total ways 6 C 5 C 5 C 6 5 4 5 4 5 4 0 0 0 000. Let number of persons present in marriage anniversary party n Everybody shakes hand to each other. Total number of shake n C According to question, n C 66 n(n ) n n 0 (n ) (n + ) 0 n, n (not possible) Total persons present in the room. Shaking hand is the begining of friendship, so it is healthy thinking to establish good relationship.. As n P 5 4. n P n! n! ( n 5)! 4 ( n ) ½ ( n 5)! 4 ( n )( n 4)( n 5)! (n )(n 4) 4 ½ n 7n + 4 0 n 7n 0 0 ½ (n 0)(n+ ) 0 n 0, n (not possible) Given that, n C : n C : ( n)!!( n )! n!!( n )! ( n)!!( n )!!( n )! n! n( n )( n ) nn ( )( n ) ( n ) n 6 n n 6 n 5 ½
8 OSWAAL CBSE Sample Question Paper, Mathematics, Class XI 4. Two unbiased dices are thrown. S {(, ), (, ), (, ),..., (6, 4), (6, 5), (6, 6)} i.e., Total sample points 6 A Event of getting a doublet or a total of 0 A {(, ), (, ), (, ), (4, 4), (5, 5), (6, 6), (4, 6), (6, 4)} 8 P(A) 6 9 P (Neither a doublet nor a total 0 will appear) 9 ½ i.e., _ ( ) PA 7 9 ½ 5. In a Single throw of dice, n (s) 6 (a) Sample Space {(, 6) (4, 5) (5, 4) (6, )} 4 Probability 4 6 9 (b) Sample Space {(, ) (, ) (, ) (4, 4) (5, 5) (6, 6)} 6 Probability 6 6 6 (c) Probability 6 (d) Probability 9 + 6 4 + 6 6 6 6 6. Class Frequency (f) x u i xi 65 0 u i f i u i f i u i 0 40 5 9 9 7 40 50 7 45 4 4 8 50 60 55 60 70 5 65 a 0 0 0 0 70 80 8 75 8 8 80 90 85 4 6 90 00 95 9 6 8 Total 50 5 05 Here, a 65, h 0, Σf 50 Mean 65 + 5 0 50 65 6 Σfu i i Σ fu i i and Variance h N N
Solutions 9 05 5 00 50 50 550 5 00 50 50 Variance 0, and Mean 6 7. Equation of hyperbola is 6x 9y 44 x y 9 6 505 5 0 a 9 a and b 6 b 4 Length of transversal, a 6 Length of conjugate axis, b 8 c a + b 9 + 6 5 c 5 e 5 foci, (± c, 0) (± 5, 0) and vertices, (± a, 0) ( ±, 0) 8. f(x) x + x Using first principle, f '(x) fx ( + h) fx ( ) lim h h 0 ( x+ h) + (x+ ) lim h x+ h x h 0 (x+ h+ )( x ) ( x+ h )(x+ ) lim h ( x+ h )( x ) h 0 (x+ )( x ) (x+ )( x ) lim h 0 h ( x+ h )( x ) + h [x 4 x ] h(x 4 x ) lim h 0 hx ( + h )( x ) 7 ( x ) 7 f '(x) ( x ) 9. As roots of x x + p 0 are a and b. a + b and ab p...(i) and x x + q 0 have c and d roots. So, c + d, and cd q...(ii) But a, b, c and d are in G.P. b a c b d r (common ratio) c b ar, c ar and d ar...(iii)
0 OSWAAL CBSE Sample Question Paper, Mathematics, Class XI From (i), (ii) and (iii), we get a.ar p a r p and ar.ar q a r 5 q ½ 5 ar q ar p r4 q p Again, a + ar a( + r) and ar + ar ar ( + r) On dividing these, we get...(iv) r 4 q p (4) 6 (from iv) Applying componendo and dividendo, we get q+ p 6 + q p 6 7 5 Teacher can assess the ability of learning and understanding. On the basis of assesment he can guide and encourage them. ½ SECTION C 0. Take, 6 n ( + 5) n 6 n + n C (5) + n C 5 + n C 5 +... 6 n + 5n + 5 [ n C. + n C 5+...] 6 n 5n 5 [ n C + 5. n C +...] 5l ; l [ n C + 5 n C +...] 6 n 5n 5l 6 n 5n is divisible by 5.. Expansion /4 + + /4 n /4 /4 n 5 th n term from beginning ( 4 ) ( 4 ) and According to question, n C n C 4 4 4 n 4 4 4 5 th x term from end ( 4 ) ( 4 ) n n 4 4 n 4 4 ( n 4) 4 C C n C n C 4 4 4 / (6) ( x 4) 4 4 () ½...(i)...(ii) [Given, Ration is 6 : ] n 4 4 n 4 4 ( 6) (6) n 8 4 6
Solutions n 8 4 n 0 Expansion ( + x) n consecutive coeficients are n C r, n C r and n C r+ According to question n C r : n C r : n C r+ 6 : : 0 n! r!( n r)! 6 ( r )! ( n r+ ) n! r n r+ r n r + n r + 0...(i) and n! ( r+ )!( n r )! r!( n r) n! 0 ( r + ) ( n r) 0 n r 0r + 0 n r 0 0...(ii) On solving equations (i) and (ii), we get n and r.. P(x) : 0 n + is divisible by. Put x, P() 0 + is divisible by i.e., P() is true. Let P(k) is true P(k) : 0 k + l; (l N) 0 k 0 ( l )...(i) Now to prove P(k+) is true. i.e. 0 (k+) + is divisible by Take L.H.S. 0 k+ + 0 k+ + Using equation (i), we get 0 0 (l ) + 00 l 00 + 00 l 99 (00 l 9) m (... m 00 l 9) i.e., P(k + ) is true. Hence, P(n) is true for all n N.. A Set of players in Basketball team B Set of players in Hockey team C Set of players in Football team n(a), n(b) 6, n(c) 0 n(b A) 4, n(b C) 5 n(a C), and n(a B C) 8 n(a B C) n(a) + n (B) + n(c) n(a B) n(b C) n(a C) + n(a B C) + 6 + 0 4 5 + 8 67 + 8 4 75 4 n(a B C) 4 So, number of all players is 4.
OSWAAL CBSE Sample Question Paper, Mathematics, Class XI 4. We have lines y m x + c,...(i) y m x + c...(ii) and x 0. On drawing these equations graphically we get a D ABC. On solving equations, (i) and (ii), we get m x + c m x + c c c x m m m( c c) y + c m m mc mc + mc mc y ( m m ) ( mc mc ) y m m Y C y m x + c C B A y m x + c C X' X Co-ordinates of A (0, c ), c c mc mc B m m m m Y',, C(0, c ) mc mc ( c c) mc Ar (ABC) 0 c ( c c ) m m + + 0 c ( m m) m c ( c ) m m Ar (ABC) ( c ) c m m mc m Hence proved. Equation of line is, x y cosθ+ sin θ a b...(i) distance from ( ) a b,0 on (i)
Solutions p and distance of line (i) from ( a b,0 ) a b a cos θ+ 0 cos θ sin θ + a b...(ii) p Product of distance using equations (ii) and (iii) p p p p a b a cosθ+ 0 cos θ sin θ + a b a b + a b cos θ cosθ+ a a cos θ sin θ + a b a b cos θ a ( b cos θ+ a sin θ) ab ( a a cos θ+ b cos θ) ab ( b cos θ+ a sin θ) a...(iii) p p ( a sin θ+ b cos θ) b ( b cos θ+ a sin θ) Product of perpendicular distances is b. 5. Let a be first term and r be common ratio of G.P. So first 4 terms are a, ar, ar, ar. According to question, a a + 9 ar a + 9 a(r ) 9...(i) and a a 4 + 8 ar ar + 8 ar(r ) 8...(ii) From (i) and (ii), we get ar( r ) 8 ar ( ) 9 r From (i), a 9 4 Four terms of G.P. are, a ar ( ) 6 ar ( ) and ar ( ) 4 Four terms are, 6,, 4. b.
4 OSWAAL CBSE Sample Question Paper, Mathematics, Class XI 6. We have linear inequations x y...(i) x + 4y...(ii) and x and y. Take x y At x 0, y and at y 0, x At (0, 0), 0 0 < (True) So origin is included. Now take x + 4y At x 0, y and at y 0, x 4 At (0, 0); 0 + 0 > (False) So origin is not included. x + 4y Y A B y C D X' 0 4 X x y Y' x x is a line parallel to y-axis and (0, 0) which is not included and y is a line parallel to x-axis and (0, 0) is not include. The shaded region is the solution region. qqq
SOLUTIONS SAMPLE QUESTION PAPER - 8 Self Assessment Time : Hours Maximum Marks : 00 SECTION A. Given that A {x : x < 70, x N} \ Roster form {,,, 4, 5, 6 7, 8}. R {(x, y) A B, x A, y B, x > y} \ R {(, ), (5, ), (5, 4)}. d sin x sin x cos x sin x dx 4. Foci of ellipse ( ± 5, 0) and e i.e., c 5 e c a 5 a 5 a a 0 a 00 and b 0 5 75 \ Equation of ellipse is 5. AD is median and G is centroid \ x y +. 00 75 + 4+x 8 x Similarly, y 5 and z. A (, 6, 4) Centroid B (4,, ) D 8 G(,, ) C(x, y, z ) Mid point \ Co-ordinates of centroid (, 5, ). 6. It is not the case that is greater than 7. MATHEMATICS Oswaal CBSE Class -, Examination Sample Question Paper
6 OSWAAL CBSE Sample Question Paper, Mathematics, Class XI SECTION B 7. L.H.S. cos 0 cos 0 cos 50 cos 70 [ cos 0 cos 50 ]cos 70 cos 0 [(cos 60 + cos 40 )cos 70 ], [ cos A cos B cos (A + B) + cos (A B)] cos70 + cos 40 cos 70 4 8 8 [cos 70 + cos 0 + cos 0 ] [cos 70 + cos (80 70) + cos 0 ] 8 [cos 70 cos 70 + ] R.H.S. Hence proved. 6 8. A Set of students who take tea B Set of students who take coffee \ According to question, n(a) 6, n(b) 8 and n ( A B) 8 But, na ( B) n( A B) 8 So, n(a B) n( ) n( A B) 40 8 Using n(a B) n(a) + n(b) n(a B) \ 6 + 8 n(a B) n(a B) 44 9. tan q + ( ) tan θ 0 It is quadratic equation in tan q, so, b 4ac ( ) + 4 + + 4 ( + ) \ tan q tan q Taking + ve sign, we get tan q ( ) ± ( + ) ( ) ± ( + ) π tan q np + π, n Z
Solutions 7 On taking ve sign, we get π π tan q tan tan 4 4 π q np, n Z. 4 0. The total number of ways in which three men can wear 4 coats is the number of arrangements of 4 different coats taken at a time, so three men can wear 4 coats in 4p ways. Similarly, 5 waist coats and 6 caps can be worn by three men in 5p and 6p ways respectively. Hence by fundamental principle of counting the required number of ways as desired 4P 5P 6P 7800. Value : Proper distribution avoid the chance of any quarrel. Given that, P (n, 4) : P (n, ) 9 : n! ( n 4)! 9 ( n 4)! ( n )! n 9 n 9. An Urn consists 5 blue and x red balls. Urn Blue 5 Red x \ Total balls 5 + x Probability of drawing two blue balls is 5 4 \ 5 4 5 C 5+ x C 5 4 5 4 (5 + x)(4 + x) x + 9x + 0 56 x + 9x 6 0 (x + )(x ) 0 x, x (not possible) \ x. lim x ( x ) π lim x sin π ( x ) x π xsin π( x ) [ ] π( x ) lim lim x x sin ( ) π π πx π x x lim x 0 sin x
8 OSWAAL CBSE Sample Question Paper, Mathematics, Class XI Now lim x f (x) lim x (ax + ) [( f (x) ax + ) for x < ] a. + a + and lim f (x) lim (x + a) + a [ f (x) x + a for x > ] + + x x Thus, lim f (x) lim f (x) + + x x Hence, lim x Note that existance of. Given that, f (x) exists and is equal to + a. lim x f (x) does not depend upon it. x+ iy a + ib p + iq (x iy) \ (x + iy) (x iy) x + y (x + y ) a ib p iq a + ib a ib p + iq p iq a p a p + b + q + b + q Hence proved Let x + iy 6 + 0i Squaring on both sides, x + y + ixy 6 + 0i x y 6 and xy 0...(i) \ (x + y ) (x y ) + (xy) 56 + 900 56 x + y 4...(ii) On solving equations (i) and (ii), we get x 9 and y 5 x ± and y ± 5 As xy is positive. So, Similarly take 6 + 0i ± ( + 5i) u + iv 6 0i u v + iuv 6 0 i u v 6, uv 0...(iii) (u + v ) ( 6) + ( 0) 56 u + v 4...(iv)
Solutions 9 On solving (iii) and (iv), u 9 and v 5 u ± and v ± 5 As uv is negative. \ 6 0 i ± ( 5i) Hence 6 + 0i + 6 0i ± ( + 5i) ± ( 5i) 4. We have two equations, xsec q + ycosec q a ± 6. and xcos q ysin q acos Q...(ii) p and p' be ^ distances from O(0, 0) to line (i) and (ii) respectively....(i) p O (0, 0) p' M x sec + y cosec a N x cos y sin a cos So, OM p \ p a + cos θ sin θ p a.sin θ.cos θ (sin θ+ cos θ) p a sin q cos q...(i) Now ON p' 0 0 acos θ cos θ+ sin θ (p') a cos q...(ii) From (i) and (ii), 5. There are 5 playing cards. 4p + p' 4a sin qcos q + a cos q 4p + p a a sin q + a cos q a (sin q + cos q) a Hence proved. Total ways of drawing two cards 5 C A Two cards drawn are red. B Two cards drawn are kings \ P(A) 6 5 C C 6 5 5 6 5 5 5 5
0 OSWAAL CBSE Sample Question Paper, Mathematics, Class XI P(B) and P(A B) 4 C 5 C 4 5 5 C 5 C 5 5 5 5 5 5 \ P(A B) P(A) + P(B) P(A B) 5 6 + 5 5 5 5 5 5 \ P(A B) 55 [5 + 6 ] 5 5 0 5 5 55 6. C.I. f C.F. C.I. x i x i Med f i x i Med 6 0 5 5 5.5 0.5 8 0 00 5 6 0.5 5.5 5 90 6 0 5.5 0.5 8 0 0 5 4 7 0.5 5.5 5 70 6 40 6 6 5.0 40.5 8 0 0 4 45 75 40.5 45.5 4 5 60 46 50 6 9 45.5 50.5 48 0 60 5 55 9 00 50.5 55.5 5 5 5 Total 00 75 N 00, l 5 5, f 6, cf 7 50 7 \ Median 5 5 + 5 6 5 5 5 + N 50 Median 8 Σ fi xi Med \ Mean deviation from median 75 Σ f 00 7 5 M.D. (from median) 7 5 7. We have a greatest integer function f (x) [x], is less than or equal to x f(0) [0] 0 f( 5) [ 5] 0 f() [] f [ 5] [ 5] f [ ] [ ] f [ 7] [ 7] i
Solutions Y X' 0 X Signum function f : R R is f (x) Y' ; x > 0 0; x 0 ; x < 0 Y x > 0 X' x < 0 0 X 8. It is given that PA PB PA PB Y' P (x, y, z) A (, 4, 5) B (,, 4) (x ) + (y 4) + (z + 5) (x + ) + (y ) + (z 4) x + 9 6x + y + 6 8y + z + 5 + 0z x + 4 + 4x + y + y + z + 6 8z 6x 4x 8y + y + 0z + 8z + 50 0 0x 6y +8z + 9 0 0x + 6y 8z 9 0 is the required equation 9. L.H.S. + + + cos 8θ But \ + cos A cos A 4 cos 4 + + θ + + (+ cos 8 θ) + + cos 4θ + 4 cos θ ( + cos θ) cos q R.H.S.
OSWAAL CBSE Sample Question Paper, Mathematics, Class XI 0. Four lines (given) forms a parallelogram. On solving (i) and (ii), Solving (ii) and (iii), we get SECTION 'C' x+ y 0...(i) y+ x 0...(ii) x+ y...(iii) y+ x...(iv) x 0 and y 0, i.e., A (0, 0) ( y) + y y y and x \ B, Solving equations (iii) and (iv), we get ( x) + x x x and y i.e., C, D, + + + C, P A (0, 0) Now, solving (i) and (iv), ( x) + x x B, x
Solutions y Hence, D, \ slope of AC fig. and slope of BD + \ Product of slopes of AC and BD ( ) Hence, diagonals of ABCD are ^ to each other. 6 6 ( ) ( ). + + We have an expansion x 6 6 5 6 4 6 + C + C + C ( ) ( ) ( ) ( ) + 6 C 6 6 6 4( ) + C5( ) + C6 + ( ) 6 6 C( ) C4( ) + ( ) ( ) ( ) 6 6 C5( ) C6 6 6 4 6 6 + C + C4 + C6 6 5 6 5 8+ 4+ + 6 5 4 C + C ( ) 6 ( ) 6 ( ) + [8 + 60 + 0 + ] 99 98. x r r \ T r+ ( ) Cr ( x ) x For term involving x 5 r 5 r r r r r r ( ). C. x.. x r r r r r ( ). C... x r 7 (which is not an integer) \ This expansion does not include any term containing x 5.. (i) R.H.S. By sine rule, we know that \ a ksin A, b ksin B, c ksin C a sin b c a A b sin R.H.S. r B C sinc k (say) sin B sin C sin A
4 OSWAAL CBSE Sample Question Paper, Mathematics, Class XI But A + B + C p A p (B + C) (sin B+ sin C)(sin B sin C) R.H.S sin [ π ( B+ C)] B+ C B C B+ C B C sin cos cos sin sin ( B+ C) ( B+ C) B+ C B C B C sin cos sin.cos sin ( B+ C) sin ( B+ C) sin ( B C) sin ( B+ C) sin( B C ) L.H.S. Hence proved sin( B+ C) (ii) L.H.S. bcos B + ccos C Using sine rule, we get b ksin B and c ksin C \ ksin Bcos B + ksin C cos C k (sin Bcos B + sin Ccos C) k [sin B + sin C] k [sin (B + C).cos (B C)] ksin (B + C)cos (B C) ksin (p A)cos (B C) [ B + C p A] ksin Acos (B C) acos (B C) R.H.S. Hence proved.. An aged person writes letters to 4 of his friends and ask to copy of letter mailed to 4 people and with same instruction. So, sequence of letters (numbers) is 4, 4, 4, 4 4,... which is G.P. with a 4 and r 4. So, number of mailed letter up to 8th mail. sum of 8 terms of this G.P. 8 4(4 ) (4 ) 4 (56 56 ) 4 (6556 ) 4 6555 8780 Cost of mailing to each letter Rs. (0 5). So, total cost 8780 0 5 Rs. 4690.
Solutions 5 Importance : (i) The ancient means was so popular, effective and people wrote and share all emotional feelings and wait for the response. (ii) Modern means works faster and we don t feel distances among relationship. 4. x + y...(i) x + y 6, x 0, y 0...(ii) Take x + y At x 0, y x and at y 0, At (0, 0), 0 + 0 (True) So origin is included. Y x + y 6 B (0, 6) 4 X', 0 O (0, 0) A (, 0) X x + y 6 (0, ) Y' Now take x + y 6 At x 0, y 6 and at y 0, x At (0, 0), 0 + 0 < 6 (True) So here origin is also included x 0, y 0 mean first quadrant region. The shaded region OAB is the solution region. 5. P(n) 5 n is divisible by 4. Put n. P() : 5 4, it is divisible by 4 i.e., P() is true. Let P(k) is true. \ P(k) 5 k 4l ; (l N)...(i) Now to prove that P(k + ) is true. i.e., 5 (k + ) is divisible by 4. Take L.H.S. 5 (k+) 5 k. 5 5.(4l + ) [using equation (i)] 5 4l + 5 4(5l + ) 4m ; (m 5l + N) i.e., P(k + ) is true. Hence, P(n) is true when P(k) is true for all n N.
6 OSWAAL CBSE Sample Question Paper, Mathematics, Class XI Let P(n) : + +... + n > P() : Put n n () > () > which is true. So, P() is true. Assume that P(k) is also true. P(k) : + + +... + k > k, k N...(i) We show that P(k + ) is also true. Adding (k + ) on both sides of (i), we get + + +... + k + (k + ) > k + (k + ) k + k + k + (k + k + 6k + ) [(k + ) + k + ] > (k + ) P(k + ) is also true, whenever P(k) is true. Hence, P(n) is true by principle of mathematical induction. 6. There are Mathematics, 4 Hindi and 5 Physics books. So, Total books \ Total arrangement of these books in a shelf! ways When all books of same subject are together, then these can be arranged in!, 4! and 5! ways respectively. \ Total ways of such arragements!! 4! 5! 6 6 4 0,0,680. The students, who have such habits of well and proper arrangements of things can be good in behaviour, and are disciplined also. qqq
SOLUTIONS SAMPLE QUESTION PAPER - 9 Self Assessment Time : Hours Maximum Marks : 00 SECTION A. A {a, b, c} i.e., n(a) \ Number of subsets 8.. Equation of line is x 4y + 9 0 \ Equation of ^ line is 4x + y + l 0. 6 6 C r C r + (Given) r + r + 6 r 7 \ 7 C 4 7 6 5 4 4 4. Equation of circle is x + y 4x + 6y 0 \ C (, ) R d 5. log a x dx.log a e 6. Parabola is x x + y 0 x y + ( ) + 5 \ Focus 0, and Directrix 8y 0 ½ + ½ 8 SECTION B 7. A {,,, 4, 5}, B {, }, C {5}. \ (B C) { } A (B C) {,,, 4, 5} { } {,,, 4, 5} L.H.S. and (A B) {,,, 4, 5} {, } {, 4, 5} and (A C) {,,, 4, 5} {5} MATHEMATICS Oswaal CBSE Class -, Examination Sample Question Paper
8 OSWAAL CBSE Sample Question Paper, Mathematics, Class XI {,,, 4} (A B) (A C) {,,, 4, 5} R.H.S. \ L.H.S. R.H.S. Hence verified. 8. A horse is tied to a point at O, OA OB, (Rope) horse moves along a circular path keeping rope tight. O A 7 B \ Length of rope 76 m Arc AB AOB 76 80 7 π Length of rope 76 80 7 80 m. 7 9. Modulus function : A function f : R R is defined as f(x) x x R x : x 0 f(x) x x: x < 0 i.e., Absolute/modulus value of any function is always positive. Graph : Y X' f(x) x f(x) x 0 X Given, f (x) ax + b f( ) 5 a + b 5...(i) and f () a + b...(ii) On solving (i) and (ii), we get a, b 0. cos q + sin q cos q, (Given) sin q cos q cos q sin q ( ) cos q Y'
Solutions 9 sin θ On rationalising the denominator sin θ ( + ) ( )( + ) cos q sin θ+ sin θ cos q cos q sin q sin θ If tan a m m + and tan b m + \ tan (a + b) tan (a + b) tan α+ tanβ tan α.tanβ m + m m + + m m+ m+ m(m+ ) + ( m+ ) ( m )(m ) + + ( m+ ) (m+ ) m ( m+ )(m+ ) m + m+ m+ m + m+ m+ m m + m+ tan (a + b) m + m+ π tan (a + b) tan 4 Hence proved. a + b 4 π. P(n) : + 4 + 7 +...+ (n ) ( ) n n Put n P() : ( + ) So P() is true. Let P(k) is true ½ i.e., P(k) : + 4 + 7 +...+ (k ) ( ) k k...(i) Now to prove P(k +) is true. i.e., + 4 + 7 +... + (k ) + [ (k + ) ] ( )[( ) ) k k + + ½ Taking L.H.S. + 4 + 7 +... + (k ) + (k + )
40 OSWAAL CBSE Sample Question Paper, Mathematics, Class XI Using equation (i), we get ( k k ) ( ) k k+ 6 k+ + + [ ] k + 5 k+ [ ] ( k + k+ k+ ). Class f x u i [ k( k+ ) + ( k+ )] ½ [(k+ )( k+ )] ( ) [( ) ] k+ k+ ½ R.H.S. Hence proved. xi 45.5 6 u i f i u i 0.5 6.5 4 5.5 4 6 8 6.5 4.5 0 9.5 0 0 4.5 48.5 4 45.5 (a) 0 0 0 0 48.5 54.5 7 5.5 7 7 54.5 60.5 45 57.5 4 80 90 Total 00 99 Here, a 45 5, h 6, N 00 Variance h Σ fu i i Σ fu i i 99 6 00 00 6 [ ( 99) ] 6 [ 0 980] 6 5 Variance 48 6.. An urn contains red and black ball, a ball is drawn, if gets red toss a coin. If gets black throw a dice, then S {RH, RT, B, B, B, B4, B5, B6} A Event of getting an even number i.e., (B, B, B4) + + 6 6 6 \ P(A) + + 6 6 N + N f i u i
Solutions 4 P(A) 4. 4 4. According to definition of ellipse PF + PF a (constant) Y C P (x, y) X' B F ( c, 0) F (c, 0) A X ( x c) + y + ( x+ c) + y a D Y' Squaring both sides, we get ( x+ c) + y a ( x c) + y x + c + cx + y 4a + x + c + cx + y 4a ( x c) + y 4cx 4a 4 a ( x c) + y (cx a ) a ( x c) + y Again squaring both sides, we get c x + a 4 a cx a (x + c cx + y ) c x + a 4 a cx a x + a c a cx + a y (a c )x + a y a 4 a c (a c )x + a y a (a c ) x y + a a c x a y + ( a c b ) b It is standard equation of ellipse. 5. We have 600 litres of a % acidic solution. So total acid 600 00 litre. Let x litres solution of 0% is to be added. Total mixture (600 + x) litres. So, according to question, 5 600 x 0 (600 + x ) < + < 8(600 + x ) 00 00 00 00 Take 5 700 + 0x (600 + x ) < 00 00 9000 + 5x < 700 + 0x 5x > 800 x > 0
4 OSWAAL CBSE Sample Question Paper, Mathematics, Class XI Now take 600 x 0 + < 8 (600 + x) 00 00 00 700 + 0x < 0800 + 8x x < 600 x < 00 i.e., 0 < x < 00 So we have to mix 0% acid solution between 0 litres to 00 litres. \ Linear inequations are x + y > 4...(i) x + y <...(ii) and x y < 6...(iii) ½ Take x + y 4 At x 0, y 4 and at y 0, x At (0, 0), 0 + 0 > 4 (False) ½ i.e., origin is not included. Now take x + y At x 0, y and At y 0, x At (0, 0), 0 + 0 < (True) So origin is included. Y 4 A (, ) X' 0 C (, 0) x + y X B x + y 4 x y 6 Y' Now take x y 6 At x 0, y and at y 0, x At (0, 0), 0 0 < 6 (True) So origin is included. The shaded region ABC is the solution region excluding the lines. All the points within this region will satisfy these inequations.
Solutions 4 6. Given a + b + c 0 b+ c c+ a a+ b and,, are in A.P. a b c b+ c c+ a a+ b +, +, + are also in A.P. a b c a+ b+ c a+ b+ c a+ b+ c,, a b c,, a b c 7. f (x) are in A.P. are also in A.P. f'(x) 00 99 x x x + +... + + x 00 99 99 98 00x 99x x + +... + + 00 99 f'(x) x 99 + x 98 +... + x + \ f'() + + +... 00 times 00 and f'(0) 0 + 0 +... + 0 + f'() 00 f'(0) Hence proved. Give that, f(x) cos x fx ( + h) fx ( ) f '(x) lim h 0 h cos( x+ h) cos x lim h 0 h lim h 0 lim h 0 cos( x+ h) cos x h cos( x+ h) + cos x x+ h h sin sin h cos( x+ h) + cos x.. lim h 0 x+ h sin cos( x+ h) + cos x sin x cos x h sin lim h 0 h
44 OSWAAL CBSE Sample Question Paper, Mathematics, Class XI 8. There is a girder (large metal beam) is in the shape of parabola. Y P (0, ) Q (0, y) y A 50 O 0 50 B (50, 0) X So equation of girder is x 4a(y 0)...(i) B (50, 0) lies on (i) \ 500 4a (0 0) 4a 50 x 50(y 0) Also Q (0, y) lies on it \ 400 50 (y 0) 400 y 0 50 y 0 6 8 4 metre \ Height of girder 0 m from centre is 8 4 metre. Equation of ellipse 6x + y 6 x y + 6 a a ± and b 6 b ± 4 (b > a) Y-axis is axis of ellipse. \ c b a 6 5 \ F (0, ± 5 ), V (0 + 4) c ± 5 Length of major axis b 4 8 Length of minor axis a and Latusractum e c b 5 4 a b 4. 9. There are 5 men and 4 women; seated in a row such that woman at even place. So we have to place all men like M M M M 4 M 5 at (boxes) 4 women can sit in 4! ways and men in 5! ways. \ Total ways of arrangements 4! 5! 4 0 880.
Solutions 45 SECTION-C 0. Quadratic equation is x + (p + ) x + p + 0 Let a and b be their roots such that b a \ a + a (p + ) a (p + )...(i) and Product of roots c/a a.a (p + ) a p +...(ii) On solving equations (i) and (ii), (p + ) p + (4p + + 4p) 9p + 8 8p + + 8p 9p + 8 p 8p + 6 0 (p 4) 0 p 4 z x + iy and w iz z i As w iz z i i( x + iy) ( x + iy) i ( + y) ix x+ iy ( ) ( + y) + x x + ( y ) + y + y + x x + y + y 4y 0 y 0 i.e., z x + i.0. z is purely real. π π. Let x t, then if x, then \ t 0 and x t + p lim tan x π x π lim ( t + π) tan t 0 x t lim t 0 ( t) tan t sin t lim t 0 t lim 0 cos t t [as cos t when t 0 and sin t t when t 0 ]
46 OSWAAL CBSE Sample Question Paper, Mathematics, Class XI In DABC, b+ c b+ c k c+ a k a+ b 5k c+ a a+ b k, (say) 5...(i) (a + b + c) 40k a + b + c 0k...(ii) Using equation (i) and (ii), a 8k, b 7k and c 5k cos A cos B and cos C b + c a bc 49k + 5 k 64k 7k 5k 0k 70k 7 b + c a ac 64k + 5 k 49k 8k 5k 40k 80k 64k + 49 k 5k 8k 7k cos A : cos B : cos C : : 44 7 56 8k 44 k 56 6 : 56 : 88 cos A : cos B : cos C 6 : 56 : 88 cos A : cos B : cos C : 7 : cos A cos 7 B cosc. Total students in a hostel 400 A Set of students who take tea B Set of students who take milk \ According to question n(a) 50, n(b) 00 n( A B) 50 n( A B) 50 n(a B) 400 50 50
Solutions 47 We know that n(a B) n(a) + n(b) n(a B) 50 50 + 00 n(a B) n(a B) 450 50 n(a B) 00 \ Number of students who takes both drinks 00. It seems that mostly students like tea, it shows the traditional drink items are changing in our country.. General equation of circle having (h, k) centre and radius r is (x h) + (y k) r...(i) Passing (, ) and (, 4) ( h) + ( k) r...(ii) and ( h) + (4 k) r...(iii) From (ii) and (iii), ( h) + ( k) ( h) + (4 k) 4 + h 4h + 4 + k + 4k 9 + h 6h + 6 + k 8k h + k 7...(iv) Also (h, k) lies on x + y h + k h + k...(v) On solving equations (iv) and (v), we get From equation (ii), we get + + r \ From (i), equation of circle is x+ + y k and h r 74 4 5 49 + 74 4 4 4 x + y + x y + + 9 74 4 4 4 x + y + x y 6 0 It is required equation of circle. The mid, points D, E and F of DABC are respectively (,, ), (, 0, ) and (,, 4). Using mid point formula, and On adding all these, we get x x x + x + x + x x + x x + x x + x 6
48 OSWAAL CBSE Sample Question Paper, Mathematics, Class XI x + x + x 6 (x + x + x ) + + 6 6 x + x + x...(i) A(x, y, z ) F (,, 4) (, 0, ) E G ( Centroid (x, x, y y,, z) z) B ( x, y, z) x i.e., x + x + x Now, and y y y + y + y + y D (,, ) C ( x, y, z) (x coordinate of centroid) y + y y + y 4 0 y + y 0 (y + y + y ) 6 y + y + y...(ii) Similarly, z z + z + z 4 z + z 8 z + z 6 z + z z and + z (z + z + z ) z + z + z...(iii) \ From equations (i), (ii) and (iii), we get co-ordinates of centroid. i.e., G (,, ). 4. Binomial expansion is (x 7y) \ Its general term i.e., T r+ C r. (x) r. ( 7y) r T r+ ( ) r. C r x r.7 r. y r ( ) r. C r x r y r.7 r For coeff. of x 5 y 7, we need to put r 7 \ Coefficients of x 5 y 7 ( ) 7 C 7 (7) 7 0 9 8 5 4 7 7 7 7 7 7 7 65,,46,056.
Solutions 49 We have (x + a) n T r+ n C r (x) n r a r rd, 4 th and 5 th terms respectively are n C (x) n a, n C (x) n a and n C 4 (x) n 4 a 4 \ According to question, Now take n n n C (x) n a : n C (x) n a : n C 4 (x) n 4 a 4 n a n 84 : 80 : 560 Cx C ( x ) 80 a 84 nn ( )( n ). a 6 nn ( ) x 70 n a 70 x (n )a 0x...(i) Also take n n n 4 4 4. a n a Cx C x 560 80 nn ( )( n )( n ) 6 a 4 nn ( )( n ) x On solving equations (i) and (ii), n a 4 x 0 n a x 8 n 8 n...(ii) 0n 0 8n 6 n 4 n 7 From (i) we get 5a 0x a x...(iii) Now using n C (x) n a 84 7 C (x) 7 (x) 84 7 6 4 84 x 7 84 7 6 x 7 () 7 x \ a x a Hence x, n 7 and a. 5. There are questions in test and it consists of two parts say : Part I consists of 5 questions and part II consists of 7 questions.
50 OSWAAL CBSE Sample Question Paper, Mathematics, Class XI A student must attempt 8 questions, selecting atleast from each. So total ways of selecting 8, having atleast questions from each part { in I and 5 in II) or (4 in I and 4 in II) or (5 in I and in II) 5 C 7 C 5 + 7 C 4 5 C 4 + 5 C 5 7 C 5 4 7 6 7 6 5 5 5 7 6 5 + + + 5 0 + 5 5 + 5 0 + 75 + 5 40 Teacher can judge the consistency of students through this test, because it is already mentioned that students are unique. 6. We have two A.P. s say A.P. I. First term a and common diff. d ½ and A.P. II. First term a and common diff. d \ According to question, Sum of n terms of I A.P. 4 5 n Sum of n terms of II A.P. n + 5 ½ n [ a + ( n ) d ] 4 5 n n [ a + ( n ) d ] n + 5 [ a + ( n ) d] 4 5 n [ a + ( n ) d] n + 5 For ratio of 8 th term of two A.P. s, we have to put n 4 n 5 [a + 4 d] 4 5 5 \ [a + 4 d] 5 + 5 [ a + 7 d] 4 75 [ a + 7 d ] 45 + 5 th 8 term of I A.P. th 8 term of II A.P. 6 50 qqq
SOLUTIONS SAMPLE QUESTION PAPER - 0 Self Assessment Time : Hours Maximum Marks : 00 SECTION-A. A f { } P(A) element.. tan π tan π π+ tan π. π π π π tan tan tan 4 6 4 6 π π + tan tan 4 6 5 x + x 5 6 +. 0 4x x 0 5x 40 x > 8 x [8, ]. 4. (x + ) 8 T r+ 8 C r x 8 r r \ Coefficient of x 5 8 C 8 7 6 5. A(, 6), B(4, 8) C(8, ) and D(x, 4) Slope of AB 6 Slope of CD x 8 56 7 5. MATHEMATICS Oswaal CBSE Class -, Examination Sample Question Paper
5 OSWAAL CBSE Sample Question Paper, Mathematics, Class XI AB ^ CD x 8 x 8 4 x 4 6. If two integers a and b are such that (a b) is always positive integer, then a > b. SECTION-B 7. L.H.S. : Let x (A B)' x (A B) x A and x B ½ x A' and x B ½ x A' B' Thus x (A B)' x (A' B') (A B)' (A' B')...(i) Now taking R.H.S. Let y A' B' y A and y B' ½ y A' and y B y (A B) y (A B)' ½ Thus y A' B' y (A B)' (A' B') (A B)'...(ii) From equations (i) and (ii), (A B)' A' B' Hence proved. 8. f(x) 6 x, x R Domain : f(x) is defined when 6 x > 0 ½ x 6 0 (x 4)(x + 4) 0 ½ 4 x 4 \ Domain of f [ 4, 4] Range : f(x) 0, "x ( 4, 4) Let f(x) y y 6 x x 6 y x 6 x 6 y It is defined, if 6 y 0 y 6 0 4 y 4 But y 0 \ 0 y 4 \ Range [0, 4]. 9. L.H.S cos 6x cos x, [cos x cos x ] [4 cos x cos x] [6 cos 6 x + 9cos x 4cos 4 x] cos 6 x + 8cos x 48cos 4 x R.H.S. Hence proved.
Solutions 5 sin A sin B cos C sin A sin B cos C ka kb. a a + b c ab a + b c a a a + b c b c DABC is isoceles. 0. A man has a board of length 9 cm. Let length of smallest piece x cm \ Length of second piece (x + ) cm and length of third piece x cm ½ \ According to question, x + x + + x < 9 + 4x < 9 4x < 88 x < cm. Also x > x + + 5 x > 8. Hence, 8 < x < \ Smallest piece of board is more than or equal to 8 cm and less than or equal to cm. ½ i. z π π cos + i sin i + i ( i ) ( i ) ( + i ) ( i ) ½ [ i i + i ] ( i ) Let and Squaring and adding eqns. (i) & (ii), + [( ) + i ( + )] z i + + r 4 rcos q... (i) rsin q...(ii) ½ ( ) ( + ) + 4 4 [+ + + ] 4 r
54 OSWAAL CBSE Sample Question Paper, Mathematics, Class XI r and tan q tan q + + π π tan + tan 4 6 π π tan.tan 4 6 π π tan q tan + 4 6 π π q + 5 π 4 6 \ z This is required polar form. As m 5π 5π cos + i sin + i i [Given] m + i + i i + i m ( + i) i m + i + i m + i i m It is possible when least value of m is 4. i 4.. We have a word INDEPENDENCE consists letters with N, D, E 4 and others are different. (i) There are 5 vowels in it with 4E and I, they should occur together i.e., EEEEI as a single word and remaining 7 letters are separate. 8! 5! \ Total such arrangements!! 4! (ii) Word begin with I and end with P i.e., I NDEENDENCE P 8 7 6 5 4! 5 4! 4! 6800. ½ ½ ½
Solutions 55 So, we have to arrange only 0 letters. \ Total such arrangements 0! 4!!! 0 9 8 7 6 5 4! 4! 70 5 60 5 ½ 600. We need to use digits 0,,,,, 4, 4, to obtain a number more than 000000. It is 7 digit number and we also have 7 digits so all digits should be used once. 7! Total 7 digits numbers using these digits 7 6 5 4!!!! 40 But it includes numbers that also start with zero. 6! So, Numbers that start with zero 6 5 4! 60!!! \ Total 7 digits numbers more than 000000 40 60 60.. P (n) : ( + x) n > + nx, x N and x >. Put n P () : ( + x) > + x (True) So, P() in true. Let P(k) is true i.e., ( + x) k > + kx...(i) ½ Now to prove P(k + ) is true. i.e., ( + x) k + > + (k + )x...(ii) ½ Taking L.H.S. ( + x) k + ( + x) k ( + x) Using in equation (i), ( + x) k + > ( + kx) ( + x) > kx + kx + x + As x > 0 kx > 0 ( + x) k + > (kx + x + ) ( + x) k + > [ + (k + ) x] i.e., P (k + ) is true. Hence P(n) is true for all n N. 4. Let a and b are two numbers then A.M., A a+ b and G.M., G ab Let a and b are roots of quadratic equation, then x (a + b)x + ab 0 But a + b A, and ab G x Ax + G 0 Using quadratic formula, we get \ x A± 4A 4G A± A G x A + ( A+ G)( A G) ½ ½
56 OSWAAL CBSE Sample Question Paper, Mathematics, Class XI i.e., a A+ ( A+ G)( A G) and b A ( A+ G)( A G) Hence proved. 5. There are 5 students, 0 are to be chosen, students decided that either all of them will join (or) none of them will join. So, when all these students are joining then we need to select 7 students out of. i.e., in C 7 ways and when these decided not to go, then select 0 students out of. i.e., in C 0 ways So, total ways of selection of 0 students C 7 + C 0 6. Equation of line is 4x y 0...(i) Y 4x y 0 P (x, y) X' Q (4, ) 5 X Y' It is passing through origin, we need to find distance of line (i) from Q(4, ) along a line, makes angle 5 with + ve x-axis \ Slope of PQ tan 5 tan (80 45 ) tan 45 \ Equation of PQ, y (x 4) x + y 5...(ii) On solving (i) and (ii), we get x, y 4 i.e., P (, 4) \ PQ (4 ) ( 4) + PQ units. i.e., Distance of line (i) from (4, ) units. 7. The vertices of triangle PQR are P(a,, 6), Q( 4, b, 0) and R (8, 4, ac) and centroid is at origin. A (a,, 6) 9+ 9 G (0, 0, 0) \ Q ( 4, b, 0) a 4+ 8 D 0 R (8, 4, c)
Solutions 57 8. a + 4 0 a + b + 4 0 b + 6 0 b 6/ 6 0 + c and 0 4 + c 0 c 5 5 5 ( x + ) lim x 0 ( x + ) () lim x 0 ( x + ) x 0 form 0 Let x + y y sin( π x) lim x π π( π x) ; 0 form 0 Apply R.H.L., put x p + h, h 0 lim y 5 5 y y n n x a n lim na x a x a 5 () 5 5. sin( π π h) lim π( π π h) h 0 sin( h) lim π ( h) h 0 sin h lim h π h 0, 9. z i and z + i z + z + \ z z + i i+ + i+ i i+ i \ z + z + z z + i π x 0 4 i 4( + i) ( + i)( i) 4( + i) i ½ ½ sin x lim x ( + i) ( + i) + ().
58 OSWAAL CBSE Sample Question Paper, Mathematics, Class XI SECTION-C 0. Let, A Set of students in Chemistry class. i.e., n(a) 40 and B Set of students in Physics class. i.e., n(b) 60 (i) When two classes meet at the same hour than there is no common student in these classes i.e., n(a B) 0 \ n(a B) n(a) + n(b) n(a B) 40 + 60 0 n(a B) 00 (ii) When classes meet at different hour, then there are 0 common students. i.e., n(a B) 0 \ n(a B) 40 + 60 0 00 0 n(a B) 80. Parents encourage their wards to these streams because of (i) Jobs oriented streams. (ii) Develope a scientific skill. (iii) Technical education.. f(x) 4 x +, x R x For domain : 4 x 0 x 4 and x > 0 (x )(x + ) > 0 x < and x > x < x > x < 4 4 i.e., domain of f (, ) (, 4]. A {,,, 4, 5, 6} R {(x, y) : y x + ; x, y A}. i.e., R {(, ), (, ), (, 4), (4, 5), (5, 6)} \ Domain {,,, 4, 5} Co-domain {,,, 4, 5, 6} and Range {,, 4, 5, 6} ½ A 4 5 6 R :A A Arrow diagram. There are 0 points in a plane, so straight lines can be formed using two at a time. \ Total straight lines 0 C 0 9 45 But 4 points are collinear. \ Lines using these points 4 C 4 6 But these 4 points can make a single line. B 4 5 6 ½
Solutions 59 So, Total different lines 45 6 + 40 Now triangle can be formed using points at a time. So, Number of triangles 0 C 0 9 8 0. Using 4 points taking at a time. No. of triangles 4 C 4 But collinear points can t make only tirangle. So, Total no. of triangles 0 4 6. m. Expansion of x x m C 0 x m m C x m m m + Cx x x \ Coefficients of first three terms of this expansion are m C 0,. m C and 9. m C ½ According to question, m C 0 + ( ). m C + 9. m C 559 ½ m + 9 mm ( ) 559 9m 9m + 6m 559 9m 5m + ( 559) 0 9m 5m 558 0 m 5m 7 0 m 6m + m 7 0 m(m ) + (m ) 0 (m )(m + ) 0 m (Not possible) ( m N) and m \ Expansion is now x x \ T r+ ( ) r C r (x) r x ( ) r C r x r r. r For coefficients of x, put r r \ Term containing, x ( ) C x 0 x 45 x 5940x a and b are distinct integers (given), then a n (a b + b) n Using binomial theorem, a n [(a b) n + n C (a b) n. b + n C (a b) n b + C (a b) n. b +...+ n C n b n ] a n b n [(a b) n + n C (a b) n (a b) b + n C (a b) n (a b) b +...] a n b n (a b) [(a b) n + n C (a b) n b + n C (a b) n b +...] Let l (a b) n + n C (a b) n b + n C (a b) n b +...] \ a n b n l (a b) \ (a b) is a factor of a n b n. r +...
60 OSWAAL CBSE Sample Question Paper, Mathematics, Class XI 4. AOB is a beam supported at the ends A and B and deflection is cm at centre and beam is in parabolic shape. Y A m B P 6, 00 B E A M cm Q (x, ) x X \ Equation of beam is x 4ay But B 6, 00 lies on it i.e., 6 4a 00 4a 00 x 00 y...(i) Let PQ cm deflection in beam at x cm distance from centre. \ Q x, 00 lies on (i). \ x 00 00 x 4 x 4 x 6 metres \ cm deflection is at 6 metres distance from centre. 5. There are 5 playing cards. One card is to be drawn. (i) P (Diamond card) 5. 4 (ii) A Black card \ n(a) 6 \ P(A) 6 5. (iii) B Not an ace \ n(b) 5 4 48 \ P(B) 48 5 (iv) C Not a diamond card \ n(c) 5 9 cards \ P(C) 9 5 4 (v) D Not a black card i.e., red cards \ n(d) 6 \ P (D) 6 5 (vi) E A face card
Solutions 6 \ n(e) cards (4; king, queen and jack each) \ P(E) 5 6. There are two firms belonging to the same industry say A and B. In firm A, No. of earners 586 Mean wages of earners 55 and Variance 00 \ Wages (total) paid by firm 586 55 0,78,58 and S.D. Variance 00 0 In firm B, No. of wage earners 648 Mean wages 55 and Variance \ Total wages paid by firm 648 55 4,0,944. S.D. Variance \ Firm B pays larger amount of wages to their workers. Now, to compare variability between two firms we need to find coefficients of variation. C.V. (Firm A) S.D. 00 Mean 0 00 55 0 90 C.V. (Firm B) 00 55 0 094 Hence, firm B shows greater variability between firms. It means firm A looks more consistent than firm B about their wages. qqq