Topper Sample Paper- 3 CLASS XI MATHEMATICS Time Allowed: 3 Hrs Maximum Marks: 100

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1 Topper Sample Paper- CLASS XI MATHEMATICS Time Allowed: Hrs Maximum Marks: 00. All questions are compulsory.. The question paper consist of 9 questions divided into three sections A, B and C. Section A comprises of 0 questions of one mark each, section B comprises of questions of four marks each and section C comprises of 07 questions of six marks each.. All questions in Section A are to be answered in one word, one sentence or as per the exact requirement of the question. 4. There is no overall choice. However, internal choice has been provided in some questions 5. Use of calculators is not permitted. You may ask for logarithmic tables, if required. Section A. Write the negation of the given statement, p: Intersection of two disjoint sets is not an empty set. f(.) - f(). If f(x) = x, the evaluate. (.) - (). Find the domain of the function f(x) = x 4 x 8x 4. If f(x) is a linear function of x.f: Z Z, f(x) = a x + b. Find a and b if { (,), (-, -7 ), (, 8) (-, - )} f. 5. With p: It is cloudy and q: Sun is shining and the usual meanings of the symbols:,,,, write the given statement symbolically. It is not true that it is cloudy if and only if sun is not shining. 6. Identify the quantifier in the given statement: P: There exists a whole number which is not natural. Write its truth value. 7. Show that the points (-,, 5 ), (,, ) and ( 7, 0, - ) are collinear. 8. What are the real numbers 'x' and 'y', if (x - iy) ( - 5i) is the conjugate of (- - i) 9. Find the focus of a parabola whose axis lies along the X- axis and the parabola passes through the point (5,0) 0. Write the equation, coordinates of foci, vertices and the eccentricity of the conic section given below. Section B. If the general term of a sequence is a linear polynomial, show that it is an AP.

2 OR Find the sum of the given sequence uptill the nth term: In a group of 400 people, 50 can speak Hindi and 00 can speak English. Everyone can speak atleast one language. How many people can speak both Hindi and English?. If Σn = 0, then find Σn. 4. An equilateral triangle is inscribed in the parabola y = 4ax, where one vertex is at the vertex of the parabola. Find the length of the side of the triangle. 5. Prove that (cos x cos x) cos x + (sin x + sin x) sin x =0 OR Simplify the expression: sin7x + sinx + sinx + sin5x 6. A pendulum 6 cm long oscillates through an angle of 0 degrees. Find the length of the path described by its extremity. 7. Let A = {a,b,c}, B = {c,d} and C = {d,e,f}. Find (i) A x (B C) (ii) (A x B) (A x C) (iii) A x (B C) (iv) (A x B) (A x C) 8. x If : R R;f(x).What is the range of f? x 9. What is the number of ways of choosing 4 cards from a pack of 5playing cards? In how many of these (i) four cards are of the same suit, (ii) four cards belong to four different suits, (iii) are face cards, (iv) two are red cards and two are black cards, 0. Evaluate (99) 5 Using Binomial theorem OR Find the ratio of the coefficient of x and x in the binomial expansion ( + a x) 9 a-ib. If x - iy =,find x y. c-id OR Let z = i and z = - + i, then find zz ire (ii)im z z z 0. Solve the equation x 4x 0 7

3 Section C. Plot the given linear inequations and shade the region that is common to the solution of all inequations x 0, y 0, 5x + y 500; x 70 and y 5. OR Solve the inequality given below and represent the solution on the number line. x 0 x The scores of two batsmen A and B, in ten innings during a certain season are given below: A B Find which batsman is more consistent in scoring 5. From the digits 0,,, 5 and 7, number greater than 5000 but of 4 digits are formed. What is the probability that the number formed is divisible by 5, if (i) the digits are repeated (ii) the digits are not repeated x 6. If x Q and cosx =- thenshow that sin. 7. (i) Find the derivative of the given function using first principle: f(x)=cos x- 6 x cos 4 (ii) Find the derivative of y sinx 8. If three lines whose equations are y = m x + c, y = m x + c and y = m x + c are concurrent, then find (i) the condition of concurrence of the three lines(ii) the point of concurrence OR A beam is supported at its ends by supports which are 4 cm apart. Since the load is concentrated at its centre, there is a deflection of 5 cm at the centre and the deflected beam is in the shape of a parabola. How far from the centre is the deflection cm?

4 9. Prove by using the principle of mathematical induction that x n y n is divisible by (x + y). 4

5 TOPPER SAMPLE PAPER CLASS XI MATHEMATICS (Solutions). p: Intersection of two disjoint sets is not an empty set. p: Intersection of two disjoint sets is an empty set [ Mark]. f(x) x f() ;f(.).. f(.) f(). 0.. [ Mark] (. ) f(x) = x 4 x 8x For f(x) to be defined x -8x+ must be non zero i.e x -8x+0 ( x-)(x-6) 0 ie x and y6 So domain of f = R-{,6} [ mark] 4. f(x) ax b (,) f f() a. b a b (,8) f f() a. b 8 a b 8 solving the two equations, we get a = 5, b=- a = 5, b=- also satisfy the other two ordered pairs f( ) 5. (, ) f( ) 5. 7(, 7)...[ Ma rk] 5. (p q) 6. P: There exists a whole number which is not natural. Quantifier in the given statement is There exists Statement P is true since 0 is a whole number which is not natural. 7. Let the given points be A, B and C, such that A(-,, 5 ), B (,, ) and C( 7, 0, - ). 5

6 AB( )( )( 5) 4 BC(7 )(0 )( ) 56 4 AC( 7)( 0)(5 ) 6 4 AB BC AC Therefore, the points A, B, C are collinear points. 8.(x iy)( 5i) i i i( 5i) i 5i 9i 5i (x iy) ( 5i)( 5i)( 5i) (9 5i) 5i 9i 5 4i 6 7i 6 7i x ;y [ Mark] The parabola is y² = 4ax The point (5, 0) lies on it (0²) = 4a (5) 00 a 5 [ Mark] 0 0. Given conic sec tion represents a hyperbola x y a b c Foci( c,0);vertices( a,0);eccentricity [ Mark] a SECTION B. Since, the general term of the AP is a linear polynomial So let a n be kn+ p Then a = k()+p a = k() +p a = k()+p... a n- = (n-)k+p Then a - a = k()+p - [k() +p]= k - k =k a - a = k() +p - [k()+p] = k - k = k a n a n- = kn+ p - (n-)k- p = k Hence a - a = a - a = a n a n- =k [Marks] Since the difference between two consecutive terms is same so the sequence is an A.P OR 6

7 a n = n (n+) = n + n [ Mark] n n n n S k + k k k k k k n(n )(n ) n(n ) [mark] 6 n(n )(n ) n(n )(n ) [mark]. Let H denote the set of people who can speak Hindi, and E denote the set of people who can speak English. Given everyone can speak atleast one language Therefore, n(h U E) = 400 and n(h) = 50, n(e) = 00 n(h U E) = n(h) + n(e) - n(h E) n(h E) = n(h) + n(e) - n(h U E) n(h E) = = 50 [Marks]. Σn=0 n(n ) = 0 n(n+) = 40 x 0 = 40 so n= 0 Σn n(n )(n ) 40 4 = = 6 6 = 870 [Marks] 4. Let the two vertices of the triangle be Q and R 7

8 Points Q and R will have the same x - coordinate = k(say) Now in the right triangle PRT, right angled at T. k k k tan0 RT RT RT k R(k,) Now R lies on the parabola : y = 4 ax k 4 a(k) k 4a k a [Marks] k a Length of side of the triangle = (RT)=.. 8 a 5. Consider LHS(cos x - cos x) cos x + (sin x + sin x) sin x x+x x-x x+x x-x = sin sin cos x sin cos sin x [ Mark] sinx cos x sinx sinx sinx cos x [ Mark] sinx cos xsin x sinx cos xsinx=0 OR sin 7x + sin x + sinx + sin 5x = sin 7x + sin x + sinx + sin 5x 7x x 7x x 8x 6x sin 7x + sin x sin cos sin cos sin4x cosx...(i) x 5x x 5x 8x x sinx + sin 5x sin cos sin cos sin4x cos( x) sin4x cos x...(ii) From(i)and (ii) sin7x + sinx + sinx + sin5x sin4x cosx sin4x cos x sin4x cosx cos x x x x x 4x x sin4x cos cos sin4x cos cos 4 sin4x cosx cos x [Marks] 6.Length of pendulum 6 cm long 8

9 Angle of oscillation = 0 degrees 80 degrees = radians so, 0 degrees= radians 8 θ= radians( Mark) 8 so using this formula =rand substituting the values of r=6, θ= radians ( Mark) 8 we get, =6 x π = x (.4) = 6.8 cm ( Marks) 8 7.A = {a,b,c} B = {c,d} C ={d,e,f} (i) (B C) = {d} A x (B C) ={(a,d), (b,d), (c,d)}. (ii) A x B ={(a,c), (a,d), (b,c), (b,d), (c,c), (c,d)} A x C={(a,d), (a,e), (a,f), (b,d), (b,e), (b,f), (c,d), (c,e), (c,f)} (A x B ) ( A x C) = {(a,d),(b,d),(c,d)} (iii) (B C) = {c, d, e,f} A x (B C) = {(a,c), (a,d), (a,e), (a,f), (b,c), (b,d), (b,e), (b,f), (c,c), (c,d), (c,e), (c,f)}. (iv) (A x B ) ( A x C) ={(a,c), (a,d), (a,e), (a,f), (b,c), (b,d), (b,e), (b,f), (c,c), (c,d), (c,e), (c,f)}. 9

10 8. f(x) x x Let y = f(x) x x 0 x Denominator Numerator y Now, y y x x yx y x x(y ) y x x x x y y y 0 y y y Now, x 0 Case : y 0; y 0 y 0; y or y,but y i.e y [0,) Case : y 0; y 0 y 0; y or y Not possible [ Mark] Range of f(x) = [0,) 0

11 9. The number of ways of choosing 4 cards from a pack of 5 playing cards 5! !48! = C4 (i) The number of ways of choosing four cards of any one suit! !9!...4 Now, there are 4 suits to choose from, so = C4 The number of ways of choosing four cards of one suit=4 75=860 (ii) four cards belong to four different suits, ie, one card from each suit The number of ways of choosing one card from each suit = C C C C 4 (iii) are face cards There are face cards The number of ways of choosing four face cards from = C4 =495 (iv) two are red cards and two are black cards, There are 6 red cards and 6 black cards, The number of ways of choosing red and black from 6 red and 6 black 6 6 6! 6! = C C 5 5!4!!4! = To be able to use binomial theorem, let us express 99 as a binomial : 99= C0 00 C 00 C 00 C C4 00 C [Marks] [ Marks] OR

12 Given:( + ax) 9 Generalterm in the exp ansionof( + ax) r 9 9r tr Cr ax r r 9 Coefficient of x Cr a 9r Coefficient of x C a C a Coefficient of x 9 C a C a Coefficient of x C a. C.9!!6!. 9 Coefficient of x C a a. C!7!9!a 7a 7a a-ib. x - iy = c-id a-ib x - iy = c-id Now, x - iy x - iy But x - iy x y x - iy = x y x y...(i) a-ib x - iy = x - iy = c-id a b a b x y a-ib a-ib a-ib a b a b...(ii) c-id c-id c id c d c d From (i) and (ii), we have x y a c b d c d c d OR We have, z = i and z = - + i z z z i To find Re

13 z z z i i 4 i 4i = i i 4i 4i i =- = - i i i 6 i 8i 4 i 4 6 i 4 i i [ Mark] z z i i Re =Re Re [ Mark] z (ii)im z z i i z 4 i i i z Im Im z z 0 7. x 4x 0 x 8x 0 0 D [ Mark] The equation has complex roots b b 4ac b D 8 56 x= a a 8 56i 8 4i [ Mark] i 8 4i, i 4 4i, [Marks]

14 Section C. System of inequations x 0, y 0, 5x + y 500; x 70 and y 5. Converting inequations to equations 500 5x 5x + y = 500 y = x y x 70 is x =70 and y 5 is y =5. Plotting these lines and determining the area of each line we get [ Marks for correct plotting of lines, three for correct shading] OR x 0 x 6 5 x 0 x 6 0 Multiplying both sides by LCM of 0 and, i.e 0, we get 0 0 x 0 x 6 0 (x 0) 0 x 6 9x 60 0x x 9x 0 x x 0 x(,0] 4

15 Thus, all real numbers less than or equal to 0 are the solution of the given inequality. The solution set can be graphed on real line as shown [4 marks for correct solution, marks for correct plotting on real number line] 4. Coefficient of variation is used for measuring dispersion. In that case the batsman with the smaller dispersion will be more consistent. (/ mark) Cricketer A Cricketer B For cricketer A: x di= x - 50 di y fi = x Total Tota l Mean = 50 + = 46 fi ( marks) S.D = ( Mark) Therefore C.V = (/ mark) For Cricketer B: Mean = 50 + = 50 S.D = 5

16 ( Mark) Therefore C.V = = 49 (/ mark) Since the C.V for cricketer B is smaller, he is more consistent in scoring. (/ mark) 5. There are 4 places to be filled ThH T U 4 The number has to be > 5000, so in place 4 only 5 or 7 out of 0,,,5,7 can come (i)when repetition of digits is allowed The number of choices for place 4 = The number of choices for place = 5 The number of choices for place = 5 The number of choices for place = 5 Total number of choices = x 5 x 5 x 5 = 50 Now, for the number to be divisible by 5, there should be 0 or 5 in the units place The number of choices for place 4 = The number of choices for place = 5 The number of choices for place = 5 The number of choices for place = Total number of choices = x 5 x 5 x = P(number divisible by 5 is formed when digits are repeated ) =

17 (ii) When repetition of digits is not allowed The number of choices for place 4 = The number of choices for next place = 4 The number of choices for next place = The number of choices for next place = Total number of choices = x 4 x x = 48 For the number to be divisible by 5 there should be either 0 or 5 in the units place, giving rise to cases. Case I : there is 0 in the units place The number of choices for place 4 = Total number of choices for remaining places = x x =6 Total number of choices = x x x = Case II : there is 5 in the units place Then there is 7 in place 4 The number of choices for place 4 = Total number of choices for remaining places = x x =6 Total number of choices=6 From case I and case II : Total number of choices = 6 + = 8 P( a number divisible by 5 is formed when repetition of digits is not 8 allowed)=

18 6. x Q III quadrant and cos x =- cos cos x cos x cos x x - cos cos x cos x sec Now, x Q n x n n x n x n n 4 Case I:When n is even k(say) x k k 4 x Q Case I:When n is odd k (say) x k k 4 x k k 4 x 7 k k 4 x Q 4 x x sin cos x InQ sin x InQ4sin x So sin 8

19 7.(i) f(x)=cos x- 6 f(x+ x)=cos x+ x - 6 f(x+ x)-f(x)=cos x+ x - cos x x+ x - x - x+ x - x sin sin x+ x - 8 x sin sin x+ x - x+ x - 8 x 8 x sin sin sin sin f(x+ x)-f(x) x x x x sin x lim sin x+ - lim x0 6 x0 x sin x - 6 9

20 x cos 4 x (ii)y cos sinx 4 sinx d d sin x(x) x(sin x) dy cos dx dx dx 4 sin x dy x.sinx x.cos x cos dx 4 sin x dy x.sin x x.cos x cos dx 4 sin x x x.sin.cos dy cos dx 4 x sin..0 dy 4 cos dx x 4 0

21 8. The three lines whose equations are y = m x + c (), y = m x + c () and y = m x + c...()aregiven The point of intersection of () and ( ) can be obtained by solving y = m x + c, y = m x + c m x + c = m x + c m x + c - m x - c 0 m - m x c - c c - c m - m x y = m c - c + c m c - m c m- m m- m c - c mc - mc The point of intersection =, m- m m- m If this point also lies on line (), then the three lines are concurrent and it is the point of concurrence c - c m - m m c - m c m - m We substitute, into y = m c - m c c - c m c m- m m- m m x + c(),to verify Som c - m c = m c - c + c m - m m c - m c m c - m c + c m -c m ` m (c c ) + m (c c ) + m (c c ) = 0 is the required condition for concurrence. ( Marks) c - c mc - mc (ii) Point of concurrency is, ( Mark) m- m m- m OR Let the vertex be at the origin and the axis vertical along y axis Therefore general equation is of the form x = 4ay ( mark) Since deflection in centre is 5 cm, So (7,5) is a point on parabola

22 Therefore it must satisfy the equation of parabola i.e 49 = 0 a i.e a = 49/0 ( mark) So the equation of parabola becomes x = 49/5y = 9.8 y ( mark) Let the deflection of cm be t cm away from the origin. Let AB be the deflection of beam,so the coordinates of point B will be (t,) ( mark) Now, since parabola passes through ( t, ), it must satisfy the equation of parabola Therefore t =9.8 = 9.4 t=5.4 cm Therefore distance of the deflection from the is 5.4 cm ( mark) ( mark) n n 9. Let P(n) : x - y is divisible by x + y P() : x - y is divisible by x + y P() : x - y =(x -y) (x+y) is divisible by x + y which is true k k Now we assume P(k) : x - y is divisible by x + y k+ k+ To Prove :P(k ) : x - y is divisible by x + y k+ k+ k+ k+ k k k k x - y x - y x x - x y + x y - y y k k k k k x x - y y x - y x - y is divisible by x + y from P(k) k k x x - y is divisible by x + y from P(k) x - y is divisible by x + y k y x - y is divisible by x + y k k k x x - y y x - y is divisible by x + y [Marks]

Grade XI Mathematics

Grade XI Mathematics Exam Preparation Booklet Chapter Wise - Important Questions and Solutions #GrowWithGreen Questions Sets Q1. For two disjoint sets A and B, if n [P ( A B )] = 32 and n [P ( A B )] =