ANNUAL EXAMINATION - ANSWER KEY II PUC - MATHEMATICS PART - A

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1 . LCM of and cosec ( ) -. π a a A a a. A A A A sin d ANNUAL EXAMINATION - ANSWER KEY -7 + d + + C II PUC - MATHEMATICS PART - A 7. or more vectors are said to be collinear vectors if they are parallel to the same line, irrespective of their magnitudes and directions. 8. Let the direction cosines be cos α, cosα, cosα cos α + cos α + cos α cos α ± Direction cosines are ±, ±, ± 9. The common region determined by all the constrains including non negative constraints of a linear programming problem is called the feasible region.. P(A B) P(A) P(B) f() + -,, f() f(-) PART - B - but f(-) f() f is not one one Only the +ve real numbers have pre images hence f is not onto.. put cos θ θ cos - Sin - ( cos θ sin θ) sin - (sin θ) θ cos -. use the formula tan - tan -

2 tan tan ( ) ( + ) tan + tan - ± >. K ± ± 8 K + 8 K, 8 5. a + (by + sin y) d a d by + sin y 6. f is continuous in [ -, ] f is differentiable in ( -, ) and f(-) f() There eists a real number c (-, ) such that f (c) c + c - (-, ) Rolle s theorem is verified with C - dv 7. V % of. d Change in volume 8. put t tan dv V..9 m d dt sec d f ( ) + 8 f '( ) + f '( c) c + sec d dt 5 t 5 t dt + c tan + c 5 5

3 9. / / d tan tan + π 6. order degree. m : n : internally uuur ( iˆ + ˆj + kˆ ) + ( iˆ + ˆj kˆ ) iˆ ˆj kˆ OR R,, m : n : eternally uuur ( iˆ + ˆj + kˆ ) ( iˆ + ˆj kˆ ) OR iˆ + kˆ R ( -,, ) r. A a b r iˆ ˆj kˆ a b iˆ + 5 ˆj 5kˆ 7 a b ( 5) 5. a i j 5 k, b i j + 6k Vector equation is a + λ ( b a). (i j 5k) + λ(k) Cartesian form X p() y y z z y y z z y + z + 5

4 PART - C 5. Refleive: Let a R. Then a a is true ( a, a) R R is refleive Symmetric: Let a, b R and ( a, b) R a b b a may not be true Eg: (, ) R But is not true (, ) R R is not symmetric. Transitive: Let a, b, c R and ( a, b), ( b, c) R a b and b c a c is true ( a, c) R R is transitive. 6. Put tan θ θ tan - + tan θ secθ tan tan tan cosθ tanθ tan θ sinθ cosθ θ sin cosθ θ tan tan tan tan sinθ θ θ sin cos θ tan. 7. Let A and B are symmetric matrices of same order then A T A and B T B Consider AB BA. To prove AB is symmetric Now (AB) T B T A T B.A AB AB is symmetric Consider AB is symmetric. To prove AB BA AB (AB) T B T A T B A 8. Let y (log) cos log y cos log(log ) cos + log (log ) ( sin ) y d log cos d log cos (log ) sin log (log ) 9. Let u sin, V e cos du d dv d cos sin cos, e ( sin )

5 du sin cos cos cos cos dv e sin e. Given + y 6 6 y Let P y ( 6 y) y 6y y dp 8y y d p 6y y dp At maimum, y ( 5 y) When d p y or y 5 y, when y 5, d P is ve At y 5 P is maimum Numbers are 5 and 5. A B ( + ) ( + ) + + A( + ) + B( + ) When -, B - When -, A d d log ( + ) log ( + ) + c log ( + ) ( + ). Let I e sin I sin + c d e e cos d I e sin cos e e ( sin ) d e sin e cos e sin d + c e (sin cos) I + c I e (sin cos) + c e I (sin cos ) + c. y y A 5

6 9 [ 7 ] sq. units.. Equation of circle is + ( y k) - () ( y k) 9 Differentiate () + ( y k) d d d uuur uuur uuur 5. Now AB OB OA i + ( ) j + k uuur uuur uuur AC OC OA i + j k uuur uuur uuur AD OD OA i + j k Now uuur uuur uuur AB. ( AC AD) ( + 9) ( ) ( - + 9) + ( ) a + b + c a + b + c. a + b + c. Now ( ) ( ) r a + a. b + a. c + b. a + b + b. c + c. a + c. b + c r a + b + c + ( a. b + b. c + c. a) (+ 6 + ) a. b + b. c + c. a 7. a i + j + k, b i j + k a i j k b i + j + k a a i j k ( a a ). ( b b ) d b b d 6

7 8. ( ) + ( ) + ( ) ( ) ( ) ( + ) d (,),(,), (,),(,5),(,6) (,),(,)...(,6) (,)... (,6) E (,)... (,6) (5,)...(5,6) (6,)......(6,6) F { (, ), (, ), (, )} E F {, ), (, )} P(E), P( F) P( E F) P( E F) 6 P( F E) P( E) 5 6 PART -D 9. Let y be an arbitrary element of range f. Then y for some in N, which implies that y ( + ) + 6 ( y 6) This gives, as y 6 ( y ) Let us define g : s N by g(y) 6 Now gof() g(f()) g( + + 5) g( +) + 6) ( ) ( + ) and fog(y) (( y ) ) ( y 6) ( y 6) f ( ) y 6 y y 7

8 . Hence gof I N and fog I S This implies that f is invertible with f - g. A A A A A A A A A I A y B z A - B A - Adj A T 9 6 A 9 6 adj A A , y 5, z.. y (tan - ) Differentiating w.r.t ; y tan - + ( + )y tan - Differentiating again w.r.t ; ( + )y + y () + 8

9 ( + ) y + (+ )y. a) P perimeter ( + y). Let dp d + dt dt dt d Given 5 cm / min ( is decreasing) dt cm / min dt di ( 5 ) cm / min dt + b) Area y da dt 8() + 6( 5) cm /min. I a d. da d + y when 8 cm, y 6 cm dt dt dt a I a du I a d a a + a a d a a a d a a I a a a d a d a d a a I a d a log + a + C () I d ( ) d ( -) - Using (); ( ) log 8 7 I c 9

10 5. Given equation of sides of the triangle are y +, y + and on solving these equations, we obtain the vertices of triangle a A(, ), B(, ) and C(, 9) Required area (shown in shaded region) Area ( OLBAO) Area (OLCAO) ( + ) d ( + ) d sq unit. 6. Diveide the D.E by os. we have tan d + cos cos d + This is of the form Py Q d +. y i. e sec. y tan sec Here P sec and Q tan. sec. Both are functions of. Linear in y and solution is y ( I. F) Q( I. F) d + c pol I. F e e tan sec d e t tan tan tan y. e tan.sec. e d + c t t. e dt + c t d( e t ) + c + + t t t t t e e dt c te e c i.e. ye tan tan. e tan - e tan + c y tan + C e -tan 7. Let a plane pass through a point A with position vector a r and perpendicular to the vector N r. Let be position vector of any point P (, y, z) (,) A dt sec in the plane. Then the point P lies in plane if and only if AP N. 8. Let represents the number of bulbs that will fuse after 5 days of use in an eperiment X of s trials. The trials are Bernoulli trials. P P(success).5 and q p has a binomial distribution with n 5, p.5 and q.95. ' y O y' d y + y + Z a r O B (, ) A L C (, 9) N r P(, y, z) Y

11 (i) Required probability P( ) 5 C P q 5 q 5 (.5) 5 (ii) Required probability P( ) P() + P() 5 C P q C p q q (q + 5P) (.95) ( (.5) (.95. (vi) Required probability P( > ) {P () + P()} (.95). 9. (a) a a f ( ) du f ( a ) du PART E Proof : Put u a on R.H.S. Then du -du when, u a and when, u a f ( u)( du) a b a f ( u) du ( Q f ( ) d ( f ( ) d) a a b a b b f ( u) du Q f ( u) du f ( ) d a a (Let π / π / I (log sin logsin ) d ( log sin log sin ) I d π / sin log d sin π / π / sin tan log d log d...() cos π / tan( π / log d. π / cot log d...() + π / tan cot I log + log d

12 π / tan cot I log. d π / I log d I log [ ] π / π I log π I log. (b) yz y y z y y z z y yz z z z y ( ) ( ) ( ) ( + + ) yz LHS y y yz yz z z yz Using, R R R yr R zr yz yz y z y z y y z z Epanding corresponding to c y y z z (y ) (z ) (z ) (y )] ( y) ( y z) (z ) (y + yz + z) RHS. (using R R R and R R R ) 5. + y.() + y () + 5y..(), y 6, y y y, y, 6 y, 5

13 8 + y + 5y 6 C X + y Cornor points Z 6 + y A(, ) B(, 6) C(, ) D(6, ) E(5, ) 6 maimum 6 Maimum value of z and it occurs at and y Minimum value of z 6 and it occurs at and y. E 6 D 8

Rao IIT Academy/ 2015/ XII - CBSE - Board Mathematics Code(65 /2 /MT) Set-2 / Solutions XII - CBSE BOARD CODE (65/2/MT) SET - 2

Rao IIT Academy/ 2015/ XII - CBSE - Board Mathematics Code(65 /2 /MT) Set-2 / Solutions XII - CBSE BOARD CODE (65/2/MT) SET - 2 Rao IIT Academ/ 5/ XII - CBSE - Board Mathematics Code(65 / /MT) Set- / Solutions XII - CBSE BOARD CODE (65//MT) SET - Date: 8.3.5 MATHEMATICS - SOLUTIONS. Let a iˆ 3iˆ kˆ b iˆ ˆj and a b 3 5, b a b Projection