Mathematics. Section B. CBSE XII Mathematics 2012 Solution (SET 2) General Instructions:

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1 CBSE XII Mathematics 0 Solution (SET ) General Instructions: Mathematics (i) (ii) (iii) (iv) (v) All questions are compulsor. The question paper consists 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 7 questions of si marks each. All questions in Section A are to be answered in one word, one sentence or as per the eact requirement of the question. 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 onl one of the alternatives in all such questions. Use of calculators is not permitted. Section B Q. If abc,, are three vectors such that a, b,and c,and a bc 0, Find the value of a. bb. c c. a. a, b, c a, b and c. a b c 0 a b c a b c a a b c b a b c c a b c 0 a a a b a c b a b b b c c a c b c c 0 a a b c a a b b b c c a b c c 0 and a b c a b b c c. a 0 a b b c c a a b b c c. a 0 8 a b b c c. a 0 a b b c c. a a b b c c. a 69 Thus, the value of a b b c c. a is 69.

2 CBSE XII Mathematics 0 Solution (SET ) Q. Solve the following differential equation: d 0 d 0 d d... i This is a homogeneous equation of degree. On taking = v, we obtain d dv v... ii From equations (ii) and (i), we obtain dv ( v) ( v) v ( v) v dv v v dv v v v dv v dv v On integrating both sides, we obtain dv v log C v log C log C log C = 0 Thus, log C = 0 is the required solution.

3 CBSE XII Mathematics 0 Solution (SET ) Q. How man times must a man toss a fair coin, so that the probabilit of having at least one head is more than 80%? Let p denote the probabilit that the man gets head. Thus, p and q. Let us assume that the man tosses the coin n times. Let X denote the number of times for which the man gets head in n-trails. n r nr P X = r C p q ( r 0,,,..., n) r r nr n Cr p and q n n Cr Now, 80 P X (Given) 00 8 P X P X =... P X n 0 8 P X = 0 P X = P X =... P X = n P X = P X = 0 Sum of all probabilities is 0 8 P X = 0 0 P X = 0 P X = 0 n n 0 0 n0 C0 n C0 n n

4 CBSE XII Mathematics 0 Solution (SET ) n It is known that is true onl when n. Hence, the man must toss the coin at least times. Q4. If cos cos, find d. The given function is cos cos Taking logarithm on both the sides, we obtain log cos log cos Differentiating both sides, we obtain d d d d log cos log cos log cos log cos d d d log cos cos log cos cos cos cos d d log cos sin log cos sin cos cos d d log cos tan log cos tan d log cos tan tan log cos d tan log cos tan log cos OR If sin = sin(a + ), prove that d sin a sin a Given, sin sin a d sin sin a Differentiating both sides with respect to, we have d d d d cos sin a sin a d d cos sin a cos a d cos cos a sin a...

5 CBSE XII Mathematics 0 Solution (SET ) sin sina sin sin a () From () and (), we have d sin cos cosa sin a sin a d sin a cos sin cos a sin a d sin a sin a d sin a sin a d sin a sin a Q. Let A = R {}and B = R {}. Consider the function f : A B defined b f Show that is one-one and onto and hence find f. Given, A = R {} and B = R {}. f : A Bis a function defined b f. Let, A. Now, f f 6 6 f is one-one. Again, let be an arbitrar element of B. f.

6 CBSE XII Mathematics 0 Solution (SET ), which is a real number for all. Also, for an, because if we take, which is not possible. R {} for all R {}. Thus, for all R {} there eists R such that f f Ever element in B has a pre-image in A which is given b. f is onto. Hence, f is one-one and onto. To find f : Let f() = where R {} and R {}. f R R is defined b f f : for all R {}. Q6. cos Prove that tan,,. sin 4

7 CBSE XII Mathematics 0 Solution (SET ) cos sin 4 cos sin cos tan sin sin cos sin cos We have to prove that tan,,. tan cos sin cos sin cos tan tan sin sin cos cos sin cos tan tan sin cos sin cos sin cos cos tan tan sin cos sin cos cos tan tan tan sin tan tan tan cos tan tan 4 sin tan tan 4 cos sin 4 tan tan tan cos tan sin 4 Hence, the result is proved.

8 CBSE XII Mathematics 0 Solution (SET ) OR Prove that sin sin cos Let sin and sin 7 8 sin and sin 7 We have to prove that sin sin cos. cos sin 8 cos 7 cos 7 Similarl, cos sin cos 4 cos We know that, cos ( + ) = cos cos sin sin 4 8 cos cos 8 6 cos 8 6 cos sin sin cos 7 8 Hence, the result is proved.

9 CBSE XII Mathematics 0 Solution (SET ) Q7. Find the point on the curve at which the equation of tangent is =. The equation of the given curve is. Let P, be the point on the given curve. Since, lies on, () Differentiating both sides with respect to, we have d d, Given, the line = is tangent to the given curve. Slope of the tangent to the curve at, = Slope of the line = d, 4 When =, we have When 9 Using, we have 9 Using So, the points are (, 9) and (, 9). Now, (, 9) lies on = if 9 = which is true. And, (, 9) lies on = if 9 = which is not true. Thus, the required point is (, 9). OR Using differentials, find the approimated value of 49..

10 CBSE XII Mathematics 0 Solution (SET ) Let f For 49, For = 49, we have Let = = 0. Now, d d d d We have d d 0. (For 49) 4 d 8 8 d (approimatel) Thus, the approimate value of 49. is Q8. Evaluate: sin sin sin

11 CBSE XII Mathematics 0 Solution (SET ) I sin sin sin sin sin sin sin sin sin sin cos cos sin cos sin cos sin cos sin cos sin cos sin 4 4 cos sin 4 4 sin cos sin 6 sin cos cos 6 cos 4 C cos cos 6 cos 4 = C cos 6 cos 4 6cos C 48 OR Evaluate: I = Let A B C Multipling both sides b ( ) ( + ), we get = A ( + ) + (B + C) ( ) = A ( + ) + B ( - ) + C ( ) = (A B) + (B C) + (A + C) () Comparing the coefficients of, and constant term on both sides of (), we get 0 = A B A = B ()

12 CBSE XII Mathematics 0 Solution (SET ) 0 = B C B = C () A + C = A + A = A = A = So, B = A = and C = B = A =, B =, C = So, I = log log Hence, log log tan C log log tan C Q9. Using properties of determinants, prove the following: a b c a b ( b c)( c a)( a b c) a b c Let a b c. a b c Appling C C C and C C C, we have:

13 CBSE XII Mathematics 0 Solution (SET ) 0 0 a c b c c a c b c c 0 0 a c b c c a c a ac c b c b bc c c 0 0 c a b c c Appling C C + C, we have: a ac c b bc c c 0 0 c a b c 0 c b a bc ac b bc c c 0 0 b c c a a b 0 c a b c b bc c c 0 0 a b b c c a a b c 0 c Epanding along C, we have: 0 a bb cc aa b c c a bb cc aa b c Hence, the given result is proved. b bc c c Q0. If = cos (log ) + 4 sin(log ), show that d d 0 = cos (log ) + 4 sin (log )... ()

14 CBSE XII Mathematics 0 Solution (SET ) 4 cos log d sin log d sin log 4 cos log d sin log cos log 4 cos log 4sin log d d cos log 4sin log d d cos log d d 4sin log d d cos lo g 4sin log d d cos log 4sin log 0 d d 0 cos log 4sin log Q. Find the equation of the line passing through the point (,, ) and perpendicular to the lines z and z Let the direction ratios of the required line be <a, b, c>. The equations of the given lines are: z... z and... Direction ratios of line () and () are <,, > and <,, > respectivel. Since the required line is perpendicular to line () and line (), a b c 0 a a b b c c 0 a b c 0... and a b c 0 a b c Solving () and (4), we have

15 CBSE XII Mathematics 0 Solution (SET ) a b c 0 a b c 0 a b c k a 4 k, b 4 k, c 8k The direction ratios of the required line are < 4k, 4k, 8k > or <, 7, 4>. Equation of line passing through (,, ) and having direction ratio s <, 7, 4 > is z z z 7 4 a b c Thus, the equation of the required line is z. 7 4 Q. Find the particular solution of the following differential equation: d e ; 0 when = 0 d e d e Integrating both the sides, d e e d e Let e = t e d = dt e d = dt dt t log t = log ( + ) + c log ( e ) = log ( + ) + c...() When = 0, = 0 Putting = = 0 in equation (), we get c = 0 log ( + ) + log ( e ) = 0 ( + ) ( e ) = So, ( + ) ( e d ) = is the particular solution of e.