Mathematics Class XII

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1 Mathematics Class XII Time: hour Total Marks: 00. All questions are compulsory.. The question paper consist of 9 questions divided into three sections A, B, C and D. Section A comprises of 4 questions of one mark each, section B comprises of 8 questions of two marks each, section C comprises of questions of four marks each and section D comprises of 6 questions of six marks each.. Use of calculators is not permitted. SECTION A. This graph does not represent a function. Make the required changes in this graph, and draw the graph, so that it represents a function.. Find the value of for which (î + ĵ + ˆk ) is a unit vector.. Find the slope of the tangent to the curve y = x x + at the point where the curve cuts the y-axis. 4. This x matrix gives information about the number of men and women workers in three factories I, II and III who lost their jobs in the last months. What do you infer from the entry in the third row and second column of this matrix? Men workers Women workers Factory I 40 5 Factory II 5 40 Factory III 7 64

2 SECTION B 5. Evaluate: sin sin 6. Evaluate: x log dx x 7. For what value of 'a' the vectors i j 4k and ai 6j8k are collinear? 8. 5 Write A for A =. 9. Simplify cot for x <. x 0. If y = tan 5x, x <, then prove that dy. 6x 6 6 dx 4x 9x. Obtain the differential equation of the family of circles passing through the points (a,0) and (-a,0).. If P(A)=, P(B)=, P(A B)=, then find P(A / B). 5 5 SECTION C. Differentiate x 5 x 5 7 x 8 5x, wrt x OR '' ' If y = acos(log x) + bsin(log x), prove that x y xy y 0, 4. Let f(x) x, g(x) = x ; xn, Show that (i) f is not an onto function (ii) gof is an onto function

3 5. Find the distance between the parallel planes r. i j k 4 and r. 6i j 9k 0 6. A plane is at a distance of p units from the origin. It makes an intercept of a,b,c with the x, y and z axis repectively. Show that it satisfies the equation: a b c p 7. Four cards are drawn successively with replacement from a well shuffled deck of 5 cards. What is the probability that (i) All the four cards are spades? (ii) Only cards are spades? (iii) None is a spade? 5 8. Solve the equation:(tan x) (cot x) 8 9. Find the equation of a tangent to the curve given by a point, where t. 0. Evaluate: 4 sinx cosx dx 0 9 6sinx x asin t, y bcos t at. If cos sin A sin cos then prove that n cosn sinn A,nN sinn cosn

4 OR If is one of the cube roots of unity, evaluate the given determinant. Show that the function f defined by f(x) = -x + x, x R is continuous. OR Show that a logarithmic function is continuous at every point in its domain.. Evaluate: Evaluate : ( sin )cos d 5 cos 4sin dx x x OR SECTION - D 4. Show that the right circular cone of least curved surface and given volume has an altitude equal to times the radius of the base. OR 4 Find the points at which the function f given by f(x) = (x - ) (x ) 5. Obtain the inverse of the following matrix using elementary operations. 0 A = 6. Calculate the area x y (i) between the curves +,and the x-axis between x = 0 to x = a a b x y (ii) Triangle AOB is in the first quadrant of the ellipse +,where OA = a and OB = b. a b Find the area enclosed between the chord AB and the arc AB of the ellipse (iii) Find the ratio of the two areas found.

5 OR Find the smaller of the two areas in which the circle x y a is divided by the parabola y ax, a 0 7. Find the equation of a plane that is parallel to the x-axis and passes through the line common to two intersectiing planes r. i+j+k 0and r. i+j-k 4 8. Two trainee carpenters A and B earn Rs. 50 and Rs. 00 per day respectively. A can make 6 frames and 4 stools per day while B can make 0 frames and 4 stools per day. How many days shall each work, if it is desired to produce atleast 60 frames and stools at a minimum labour cost? Solve the problem graphically. 9. A random variable X has the following probability distribution : X P(X) 0 k k k k k k 7k +k Determine: (i) k (ii) P(X < ) (iii) P(X > 6) (iv) P( X < ) OR In answering a question on a MCQ test with 4 choices per question, a student knows the answer, guesses it or copies the answer. Let ½ be the probability that he knows the answer, ¼ be the probability that he guesses and ¼ be the probability that he copies it. Assuming that a student, who copies the answer, will be correct with the probability ¾, what is the probability that student knows the answer, given that he answered it correctly? Arjun does not know that answer to one of the questions in the test. The evaluation process has negative marking. Which value would Arjun violate if he resorts of unfair means? How would an act like the above hamper his character development in the coming years?

Mathematics Class XII Solution Time: hour Total Marks: 00 SECTION A. If a vertical line intersects the graph of a relation in two or more points, then the relation is not a function.

6 Mathematics Class XII Solution Time: hour Total Marks: 00 SECTION A. If a vertical line intersects the graph of a relation in two or more points, then the relation is not a function. Graph should have no vertical lines. Magnitude of (î + ĵ + k )=, for it to be a unit vector..(i+j+k). y x x dy x dx At a point where the curve cuts y-axis, x = 0 dy dx x0 x x women lost their jobs in factory III in the last two months.

7 SECTION B 5. sin sin sin 6 sin 6 sin x 6. I log dx...() x x Let f(x) log and x x x f( x) log log f(x). x x f(x) is an odd function. x Thus, log dx 0 x 7. Let p = i j 4k and q = ai 6j 8k Two vectors p and q will be collinear if, i j k p q a 6 8 i(4 4) j( 6 4a) k( a) 0 0i (6 4a)j ( a)k 0i 0j 0k 6 4a 0 and +a = 0 a 4

8 8. 5 Given A =, then A and adja =, obtained from A by interchanging the main diagonal elements and multiplying by ( ) the non -diagonal elements. 5 adja 5 A A 9. Let sec x. Then,x sec and for x <, Given expression = cot ( cot ), cot (cot( )) sec as 0 < 0. y = tan 5x 6x y = tan x tan x dy dx 9x 4x

9 . x (y b) a b or x y by a...() dy x y dy dy x y b 0 b dx...() dx dx dy dx Substituting in (), dy (x y a ) xy 0 dx. P(A B) P(A / B)= P(B) = P(A B) P(B) [P(A) P(B) P(A B)] = P(B) 7 = 0

10 SECTION C. Let y = x 5 x 5 7 x 8 5x Taking log on both sides, we get log y = log x + log 5 x 5log 7 x log 8 5x dy y dx x 5 x 7 x 8 5x dy 5 5. y dx x 5 x 7 x 8 5x dy 5 y dx x 5 x 7 x 8 5x dy x 5 x 5 5 dx x 5 x 7 x 8 5x 5 7 x 8 5x 5 y = a cos (log x) + b sin (log x) OR dy a sin log x. b. coslog x. dx x x dy ' x xy a sinlog x b. coslog x dx Differentiating, x x '' ' xy y a cos log x. b. sin log x. '' ' x y xy a coslo x y xy y 0 '' ' g x b. sin log x y

11 4. f(x) x, xn, Domainof f {,,,...} Range {4,5,6,...} Codomain of f ={,,,...} f is not an onto function f(x)= x+ g(x) = x g[f(x)] x =x Domainof gof {,,,...} Range {,,,4,5,6,...} Codomain of gof ={,,,...} gof is an onto function 5. Distance between the parallel planes d k isgiven by n r. 6i j 9k 0 r. i j k r. i j k 4 and r. i j k the distance between the given parallel planes 4 is =

12 6. The equation of the plane in the intercept form is x y z a b c x y z 0 a b c Plane is at a distance of p units from the Origin a b c a b c p a b c a b c p =p 7. This is a case of bernoulli trials. p = P(Success) = P(getting a spade in a single draw) = 5 4 q P(Failure) p (i)all the four cards are spades = P(X = 4) = C4p q (ii)only cards are spades= P(X = ) = Cp q (iii)none is a spade = P(X = 0) = C p q

13 8. Here, (tan x) (cot x) (tan x cot x) tan x.cot x 8 5 tan x.cot x 8 5 tan x.cot x 4 8 ( 5) tan x.cot x 8 8 tan x. tan x 0 8 tan x (tan x) 0 8 6(tan x) 8tan x 0 tan x tan x tan x and 4 4 x so tan x 4 x tan 4 x

14 9. Here, x asin t, y bcos t Differentiating () wrt t dx asin t cost dt and dy bcos t sint dt dy dy dt bcos t sin t b dx dx asin t cost a dt Slope of the tangent at t dy b cot 0 dx a cot t Hence, equation of tangent is given by y bcos 0 or y 0

15 0. Let I sin x cosx dx 9 6sinx 0 sin x cosx sin x cosx Put sin x cosx t cosx sin x dx dt For x, t 0 and For x 0, t 0 I 9 6 t 0 dt 5 6 t 0 5 4t dt dx a x log a x a a x dt c 5 4t log t 0 dx( mark) 40 log 9 log 9 40

16 . We shall prove by principle of mathematical induction n cosn sinn Here, let A sinn cosn n cosn sinn Pn : A, sinn cosn cos sin So, A sin cos P is true. Assuming result to be true for n = k i.e. P(k) to be true k cosk sink Pk : A sink cosk We have to prove P (k +)is true, k k P k : A A A k cos sin cosk sink A sin cos sink cosk coscosk sinsink cossink sincosk sincosk cossink sinsink coscosk cos k sin k sin k cos k cos k sin k sin k cos k P(k + )is true. Thus by principle of mathematical induction n cosn sinn A sinn cosn for all n N

17 C C C C OR But, is one of the cube roots of unity Since I column of the determinant is zero therefore, value of the given determinant is zero.. f(x) = x + x, xr Let g(x) = x+ x,xr x,being a polynomial function is continuous x,being a modulus function is continuous If a and b are two continuous functions, then a+b is a continuous function g(x) = x+ x,xr is a continuous function If it is continuous function, then g(x) is continuous. In the given problem, g x x + x...is continuous. but f x x + x...given f(x)iscontinuous.

18 Domainof log x is(0, ) f(x) log x,x (0, ) Let c (0, ) lim f(x) lim log x xc xc h0 lim log c h,c 0 h lim logc,c 0 h0 c h lim logc lim log h0 h0 c h logc lim log h0 c OR h log c h logc lim h0 h c c h log c h logc lim lim h0 h h0 c c log( x) But lim x0 x h log c lim h0 h c h log c h logc lim lim h0 h h0 c c h logc lim h0 c log c 0 logc f(c) logarithmic function is continuous at every point in its domain

19 . ( sin )cos Let I = d 5 cos 4sin Let t sin dt cosd ( t ) I dt 5 ( t ) 4t ( t ) ( t ) I dt dt t 4t 4 (t ) Using the method of Partial fractions ( t ) A B A,B 4 t (t ) (t ) 4 I t dt (t ) 4 I dt dt t (t ) 4 log sin C sin

20 Let I 6 OR dx x x dx x x Let x t x t dx 6t dt 6t dt I t ( t) 6 6t dt ( t) t dt ( t) t dt dt 6 6 ( t) ( t) t t t dt dt 6 6 ( t) ( t) dt 6 t t dt 6 ( t) t t 6 t 6log t 6 6 x x 6x 6log x C SECTION D 4.

21 Here, Volume V of the cone is S rl r h r...() Surface area Where h = height of the cone r = radius of the cone l = Slant height of the cone S r h r V r h r V h from equation () Let, s = s Substituting the value of r from equation (), we have, V V 9V S h Vh h h h Differentiating S with respect to h, we get ds V 9V dh h...() ds 0 for maxima/minima dh V 9V 0 h V 9V h 6V d S 54V 4 dh h h d S 6V 0 at h dh Therefore curved surface area is minimum at h Thus, r h h r 6 h r h V 6 Hence for least curved surface the altitude is times radius.

22 OR 4 f(x) = (x ) (x ) ' 4 f (x) =(x ) (x ) 4(x ) (x ) (x ) (x ) (x ) 4(x ) (x ) (x ) x 6 4x 4 (x ) (x ) 7x ' f (x) =0 (x ) (x ) 7x x,, 7 ' ' ' Let us examine the behaviour of f (x), slightly to the left and right of each of these three values of x (i) x = : When x< ; f (x) > 0 When x> ; f (x) > 0 x = is neither a point of local maxima nor minima It may be a point of inflexion (ii) x 7 ' When x< ; f (x) > 0 7 ' When x> ; f (x) < 0 7 x = is a point of local maxima f ( ) ( ) The local maximum value is 7 7 (iii)x When x < ; f (x) < 0 When x > ; f (x) > 0 ' ' x is a point of local minima 4 f() =( ) () =0 The local minimum value is 0

23 5. Since A = IA A 0 0 Applying R R A 0 0 Applying R R R A Applying R R R A Applying R R 5R A Applying R R A Applying R R + R Applying R R R

24 A Hence A x y (i) between the curves +, and the x-axis between x = 0 to x = a a b a x ba b - dx a -x dx 0 a a 0 a b x a -x a x sin a a 0 0 a sin () 0 a sin (0) b a b a a a x b - dx ab 0 a 4

25 x y (ii) Area of triangle AOB is in the first quadrant of the ellipse +, where OA = a and OB = b. a b =the area enclosed between the chord AB and the arc AB of the ellipse. a x = Area of Ellipse ( In quadrant I)- Area of AOB= b - dx ab ab ab 0 a 4 ab 4 (iii) Ratio ab 4 ab 4

26 OR The circle is x y a C (O, a) The parabola is y ax, a 0 y 4 ax, a 0 4 Their point of intersection is given by :x ax a x ax a 0 x a x a 0 x a, a x a y a y a shade region is the smaller of the two areas in which the circle is divided by the parabola

27 a a / A= axdx a x dx 0 a a / a / x x a x a a x sin a a 0 a 4 / / x a a xa x a sin a a 4 / a / a a a a sin a a a a sin a a 4 / a a a sin a a a sin / / a a a a a 4 / a a a a 4 / a a a a sq. units

28 7. r. i + j + k 0and r. i + j - k 4 x y z 0 and x y z 4 0 The required plane passes through the line common to two intersectiing planes x y z k(x y z 4) 0 x( k) y( k) z( k) ( 4k) 0...() The required plane is parallel to the X-axis whose d.cs are,0,0.( k) 0.( k) 0.( k) 0 ( k) 0 k Substituting in (),we get x( ) y( ) z( x(0) y( ) z( ) 0 y z 6 0 y z 6 0 ) ( 4 ) 0

29 8. Let the two carpenters work for x days and y days respectively. Our problem is to minimize the objective function. C = 50x + 00y Subject to the constraints 6x + 0y 60 x +5y 0 4x + 4y x + y 8 And x 0, y 0 Feasible region is shaded. This region is unbounded. Corner points Objective function values C = 50x + 00y A(0, 0) 500 E(5, ) 50 D(0, 8) 600 The labour cost is the least, when carpenter A works for 5 days and carpenter B works for days.

30 9. 7 (i) P(X ) i0 i 0 k k k k k k 7k k 0k 9k 0k 9k 0 0k 0k k 0 0k(k ) (k ) 0 (k )(0k ) 0 k,k 0 k,also being a probability cannot be negative k 0 (ii)p(x ) P(0) P() P() 0 k k k 0 9 (ii)p(x 6) P(7) P(8) k 7k k (iii)p( X ) P() P() k k k 0

31 OR Let E, E, E and A be the events defined as follows: Let E be the event that the student knows the answer. Let E be the event that the student guesses the answer. Let E be the event that the student copies the answer. Let A be the event that the answer is correct. PE ;PE ;PE ; 4 4 Probability that he answers correctly given that he knew the answer is. That is, PA E If E has already occurred, then the student guesses. Since there are four choices out of which only one is correct, therefore, the probability that he answers correctly given that he has made a guess is 4. That is PA E. 4 It is given that, PA E 4 By Bayes Theorem, we have,

32 Required Probability=P E A PEP A E P E P A E P E P A E P E P A E Thus, the probability that student knows the answer, given that he answered it correctly is. Arjun is dishonest, as he copies from the other student(s). Copying once may become habit forming as he may continue resort to dishonest means in the coming years.

y mx 25m 25 4 circle. Then the perpendicular distance of tangent from the centre (0, 0) is the radius. Since tangent

y mx 25m 25 4 circle. Then the perpendicular distance of tangent from the centre (0, 0) is the radius. Since tangent Mathematics. The sides AB, BC and CA of ABC have, 4 and 5 interior points respectively on them as shown in the figure. The number of triangles that can be formed using these interior points is () 80 ()