Board Answer Paper: October 2014

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1 Trget Pulictions Pvt. Ltd. Bord Answer Pper: Octoer 4 Mthemtics nd Sttistics SECTION I Q.. (A) Select nd write the correct nswer from the given lterntives in ech of the following su-questions: i. (D) ii..p +.q + c.r ( ) ( ) ( ). c. c c. + + c c c c c c + + c c c Mthemtics + +. c c c 3 [] (A) Let A Since, AA I A Applying R nd ( ) R 3, we get A A iii. (B) The direction rtios of the line re + 4, 3, 3 i.e., 5,, 5 5 the direction cosines of the line re ±,± ( 5) ( 5) 5 i.e., ± 5, ± 5, ± 5 5,± ( 5) [] []

2 Trget Pulictions Pvt. Ltd. Bord Answer Pper: Octoer 4 (B) Attempt ny THREE of the following: i.. Let p: 5 is n irrtionl numer. q: is comple numer. the symolic form of the given sttement is p q. Since, the truth vlue of p is T nd tht of q is F. the truth vlue of p q is F.. Consider the sttement, n N, n + 5 > Clerly, n 6, n N stisfy n + 5 >. its truth vlue is T. ii. Given, c 3.(i) Now, [ c ] ( c) ( 3 ).[From (i)] ( 3 ) ( 3 ).[ ] 3.( ) 3[ ] 3() [ c ] iii. Let n î + ĵ + ˆk ˆn n n i ˆ+ ˆj + kˆ + + i ˆ+ ˆj + kˆ i ˆ+ ˆj + kˆ 3 nd p 5 Vector eqution of plne is r ˆn p ( i ˆ+ ˆj+k ˆ ) r 3 r ( i ˆ ˆj kˆ) iv. Crtesin eqution of the given line is 3 + 6y z Dividing throughout y 6, we get y 3 3 (z ) y 3 3 z 6

3 Trget Pulictions Pvt. Ltd. Mthemtics Let e the position vector of the line pssing through the point,, 3 3. ˆ i + ˆj+ kˆ 3 3 Direction rtios of the line re,, 6. Let e the vector prllel to the line. i ˆ+ ˆj 6kˆ The vector eqution of line pssing through point with position vector nd prllel to is r +λ. vector eqution of the line is ˆ ˆ ˆ r i j k ( i ˆ ˆj 6kˆ + + +λ + ) 3 3 v. Let L nd L e the two lines with direction rtios,, nd 3, 4, respectively. Let the direction rtios of the vector perpendiculr to L nd L e,, c. + c nd c c c c Direction rtios of the required vector re 3, 5,. Q.. (A) i. Attempt ny TWO of the following: p q r p q q r (p q) r p (q r) T T T T T T T T T F T F F F T F T F T T T T F F F T T T F T T F T T T F T F F F T T F F T F T T T F F F F T T T The entries in the columns 6 nd 7 re identicl. (p q) r p (q r) [ mrk ech for column 6 nd column 7] 3

4 Trget Pulictions Pvt. Ltd. Bord Answer Pper: Octoer 4 ii. Let OA nd OB e two lines through the origin, ech mking n ngle of 3 with the line 3 + y. Let slope of OA (or OB) e m. 3 Slope of the line 3 + y is nd θ 3 Y 6 4 B y X X O A Y tn 3 3 m 3 + m 3 m + 3 3m Squring on oth sides, we get ( 3m) 3(m + 3) 4 m + 9m m + 36m + 7 3m + 48m + 3.(i) Eqution of OA (or OB) is y m, since it psses through the origin. m y Sustituting the vlue of m in (i), we get y y + 3 3y 48y y + 48y y + 3y is the required joint eqution. 4

5 Trget Pulictions Pvt. Ltd. Mthemtics iii. Let sin 3 5 sin 3 5 nd < < π cos > Now, cos sin tn sin cos 3 4 tn 3 4 sin tn 4 sin tn 4 tn tn tn tn tn sin 3 tn

6 Trget Pulictions Pvt. Ltd. Bord Answer Pper: Octoer 4 (B) Attempt ny TWO of the following: i. Mtri form of the given system of equtions is 3 y z This is of the form AX B, where, A 3, X y, B z Applying R R, we get 3 8 y 3 4 z Applying R R R, R 3 R 3 3R, we get y z 3 Applying R 3 R 3 R, we get y 5 8 z 8 Hence, the originl mtri A is reduced to n upper tringulr mtri. By equlity of mtrices, we get + y + 3z 8...(i) 5y 5z 5 i.e., y + z 3...(ii) 8z 8 i.e., z Sustituting z in eqution (ii), we get y + 3 y Sustituting y nd z in eqution (i), we get + () + 3() 8, y, z is the required solution. ii. c A 6 O θ P B c C D Let, nd c e the position vectors of points A, B nd C respectively with respect to origin O. Complete the prllelopiped s shown in the figure with OA, OB nd OC s its coterminous edges.

7 Trget Pulictions Pvt. Ltd. Mthemtics AP is perpendiculr drwn to the plne of nd c. Let θ e the ngle mde y AP with OA. Volume of prllelopiped (Are of prllelogrm OCDB) (height) Now, re of prllelogrm OCDB c.(i) Height of prllelopiped l(ap) l(oa) cos θ OA cos θ cos θ.(ii) From (i) nd (ii), we get volume of prllelopiped c cos θ ( c) Volume of prllelopiped c Let ˆi+ ˆj, ˆ j+ k ˆ, c kˆ + ˆi c ( ) ( ) + + Volume of the prllelopiped is cuic units. iii. To drw the fesile region, construct tle s follows: Inequlity + y 3 + y Corresponding eqution (of line) + y 3 + y Intersection of line with X-is (, ) (3, ), Intersection of line with Y-is (, 3) (, ) Region Origin side Origin side Origin side 5 4 Y X (, )D O C 7, 3 3 B(, ) A(, ) 3 4 X + y 3 + y 3 4 Y 7

8 Trget Pulictions Pvt. Ltd. Bord Answer Pper: Octoer 4 Shded portion OABCD is the fesile region, whose vertices re O(, ), A(, ), B, C nd D(, ). B is the point of intersection of the lines nd + y 3. Putting in + y 3, we get y B (, ) C is the point of intersection of the lines + y 3 nd + y. Solving the ove equtions, we get Q.3. (A) 8 3, y C, 3 3 Here, the ojective function is Z 6 + 4y. Z t O(, ) 6() + 4() Z t A(, ) 6() + 4() Z t B(, ) 6() + 4() Z t C, Z t D(, ) 6() + 4() 4 Z hs mimum vlue 6 t B(, ). Z is mimum, when nd y Attempt ny TWO of the following: i. L.H.S. sin C Α + csin cosc cosa + c cos C + c c cos A ( + c) ( cos C + c cos A) + c...[by projection rule] R.H.S. ii.. It is time ut it is not wrm.. If it is not time then it is wrm. c. It is wrm if nd only if it is time. y y z z y y z z iii. The lines nd c c re intersecting, if shortest distnce is zero. y y z z i.e., if c c

9 Trget Pulictions Pvt. Ltd. Mthemtics Equtions of the given lines re y+ z 3 k z nd y 3 4 Here,, y, z, 3, y k, z,, 3, c 4,,, c Since, the given lines intersect. k+ 3 4 (3 8) (k + )( 4) (4 3) + k + k 9 k 9 (B) Attempt ny TWO of the following: i. The eqution r + λ + µ c represents plne pssing through point hving position vector nd prllel to vectors nd c. Here, i ˆ + kˆ, î nd c ˆ i + j ˆ 3k ˆ The given plne is perpendiculr to the vector n ˆi ˆj kˆ c 3 î() ĵ( 3 ) + ˆk( ) 3 ĵ + ˆk Norml to the plne ( ) n 3j ˆ + kˆ Vector eqution of the required plne is r n n r ( 3j ˆ + k ˆ ) ( i ˆ + k ˆ ) ( 3j ˆ + k ˆ ) r ( 3j ˆ kˆ) r ( 3j ˆ kˆ) + () + (3) + () +.(i) Eqution of the plne in norml form is r.(3j ˆ+ k) ˆ For crtesin form: Sustituting r ˆi+ y ˆj+ zkˆ in (i), we get ˆi + y ˆj + zk ˆ ˆ + kˆ ( ) ( ) 3y + z 9

10 Trget Pulictions Pvt. Ltd. Bord Answer Pper: Octoer 4 ii. Let θ e the cute ngle etween the pir of lines + hy + y. tn θ h +.(i) Compring the eqution 5y + 3y with + hy + y, we get, h 5, 3 Let α e the cute ngle etween the lines 5y + 3y. tn α 5 ()(3) tn α 5.(ii) But θ α tn θ tn α.(given) h + 5.[From (i) nd (ii)] Squring on oth sides, we get 4(h ) ( + ) 5 (h ) ( + ) iii. sin tn tn sin + sin tn tn + sin tn (sin ) + (sin ) (sin ) (tn + ) sin or tn π π sin sin or tn tn 4 tn π π 4 tn 3π 4 n Since, sin θ sin α implies θ nπ + ( ) α nd tn θ tn α implies θ nπ + α, n Z. the required generl solution is nπ + ( ) n π or mπ + 3 π, where n, m Z. 4

11 Trget Pulictions Pvt. Ltd. Mthemtics Q.4. (A) SECTION II Select nd write the correct nswer from the given lterntives in ech of the following su-questions: i. (B) y c e + c e.(i) d e c e d y d e + c e d y y d.[from (i)] [] ii. (C) 3 P() k k 3k Since, the function is p.m.f. P( i ) k + k + 3k k 6 [] iii. (D) sec + y y + y y sec ( ) (sy) + y y ( + ) y ( ) y + y c, where c + y d c d y () d y d d y. (i).[from (i)] y y []

12 Trget Pulictions Pvt. Ltd. Bord Answer Pper: Octoer 4 (B) Attempt ny THREE of the following: i. y sin (3) + sec 3 sin (3) + cos (3). cos ( ) sec π y. sin cos + d ii. Let I.log d log d d ( log ) d d d log d log d I log 4 + c iii. h + 8 h π d 6 π d h tn π d 6 h tn ( ) 8 h 6 π π 6 tn h tn 4 4 π 6 tn π h 4 6 π 4, h π tn 4 ( ) ( ) ( ) tn ( h) h

13 Trget Pulictions Pvt. Ltd. Mthemtics iv. Let X e the numer of tested component survive. P(component survive the check test) p.5, q.5.5 Given, n 4 X ~ B (4,.5) The p.m.f. of X is given y P(X ) P() 4 C (.5) (.5) 4 P(X ) 4 C (.5) (.5) 6(.5) 4 6(.65).375 v. Required re π/ y d π/ π/ π/ π/ sin d sin d + π/ sin d [ ] π cos π + [ cos ] / X π π π Y cos cos cos cos ( ) ( ) + sq. units π Y y sin O X Q.5. (A) Attempt ny TWO of the following: i. f(8) 8.(given) log log8 lim f() lim Put 8 h, then 8 + h, s 8, h log (h + 8) log8 lim f() lim 8 h h h+8 log 8 lim h h h log + 8 lim h h 8 8 h log + 8 lim 8 h h 8 8 (). h log( ) + h, nd lim 8 Since, 8 f(8) lim f() f(8), f is discontinuous t

14 Trget Pulictions Pvt. Ltd. Bord Answer Pper: Octoer 4 ii. φ(t) is differentile function of t. d φ (t) dt Put f ( )d g() d [g()] f() d Consider, d dt [g()] d d [g()] d dt.(by chin rule) f () d dt f[φ(t)] φ (t) By definition of integrtion, g() f[ φ() t ] φ' () t dt f( )d f[ () t ] φ' () t dt iii. 3e tn y d + ( + e ) sec y 3e + e sec y tn y Integrting on oth sides, we get 3e + e sec y tn y c 3 log + e + log tn y log c, where c log c log + e 3 + log tn y log c log ( + e ) 3 tn y log c ( + e ) 3 tn y c When, y π ( + e ) 3 tn π c c the prticulr solution is ( + e ) 3 tn y. (B) Attempt ny TWO of the following: i. Let S e the position of source of light. Let BD e the position of the mn t time t. Let AB nd BC length of the shdow y ft/min S Now, d dt From the figure, ASC BDC AS BD AC BC y y 5y + y y 4 3 D A B y 6 C d dt 4dt 4 () 5 4

15 Trget Pulictions Pvt. Ltd. Mthemtics the shdow of the mn is lengthening t the rte 5 ft/min. The tip of shdow is t C. Let AC z BC z AS BD AC BC 3 6 z z 5z 5 z z 5 4 dz dt 5 d 5 () 5 4 dt 4 the tip of the shdow is moving t the rte 5 ft/min. ii. log Let I ( + log ) d Put log t e t d e t dt t I + t e t dt ( ) ( ) ( + t) t t+ e dt t t+ e ( + t) ( + t) dt t e t + ( + t) dt Put f(t) + t + t t e f t + f t dt e t f(t) + c e t + t + c f (t) ( ) I () () I + log + c iii. nd y re differentile functions of t. Let there e smll increment δt in the vlue of t. Correspondingly, there should e smll increments δ, δy in the vlues of nd y respectively. As δt, δ, δy 5

16 Trget Pulictions Pvt. Ltd. Bord Answer Pper: Octoer 4 δy δy Consider, δt δ δ δt Tking lim on oth sides, we get δ t δy lim δy δ t lim δt δ t δ δ lim δ t δt Since, nd y re differentile functions of t, δy lim d y eists nd is finite. δ t δt dt δ lim d eists nd is finite. δ t δt dt δy lim dt δ t δ d dt δy lim δ δ dt d dt (s δt, δ ) Limits on right hnd side eists nd re finite. Limits on the left hnd side should lso eists nd e finite. δy lim δ δ y eists nd is finite. d d dt, d dt dt cos t nd y sin t d sin t nd cos t dt dt d dt d cot t sint dt 6 Q.6. (A) Attempt ny TWO of the following: i. First, we need to verify tht cosine function is continuous. We know tht, f() cos is defined for every rel numer. Let c e rel numer. Put c + h, then s c, h lim f() lim cos lim cos (c + h) lim (cos c cos h sin c sin h) c Thus, lim c c h h lim (cos c cos h) lim (sin c sin h) cos c() sin c() cos c f(c) h f() f(c) h

17 Trget Pulictions Pvt. Ltd. Mthemtics Hence, cosine function is continuous function. Let g() cos Then, g() cos, for cos < cos, for cos Since, cos is continuous, cos (i.e., k cos, where k ) is lso continuous. g() cos is continuous function. ii. d y+ + y d y + + y Put y u y + + y...(i)...(ii) y u Differentiting w.r.t., we get d u + du...(iii) d Sustituting (ii) nd (iii) in (i), we get u + du d u + + u du d + u du d + u Integrting on oth sides, we get du d + u + c log u+ + u log + log c, where c log c log u+ + u log c u + y + + u c y + + y y+ + y c c + y c iii. Given X B(n, p), n nd E(X) Since, E(X) np p p q p 7

18 Trget Pulictions Pvt. Ltd. Bord Answer Pper: Octoer 4 Vr.(X) npq 5 S.D.(X) Vr(X) 5 (B) Attempt ny TWO of the following: i. Let X demnd for ech type of cke (ccording to the profit) P(X 3) %. 5 P(X.5) 5%.5 P(X ) %. P(X.5) 5% 5.5 P(X ) 5% 5.5 The proility distriution tle is s follows: X P(X) E(X) P( i i) 3(.) +.5(.5) + (.) +.5(.5) + (.5) Epected profit per cke `.73 ii. f() +, [, 3] As f() is polynomil function,. f() is continuous on [, 3]. f() is differentile on (, 3) Thus, ll the conditions of LMVT re stisfied. To verify LMVT, we need to find c [, 3] such tht f (c) f(3) f() 3.(i) Now, f(3) nd f() + f () f (c) c From (i), we get ( ) ( ) f 3 f 3 c 8

19 Trget Pulictions Pvt. Ltd. Mthemtics 3 3 c 4 3 c 3 c 3 c c 3 c 3 c 3 c 3 lies etween nd 3. Thus, LMVT is verified. iii. Let I f( )d Put + t d dt When, t nd when, t I f( + t)( dt) f(+ t)dt f(+ t)dt. f( )d f( )d f( )d f(+ )d. f( )d f(t)dt Let I I f( ) d f ( ) + f( + ) f(+ ) d f(+ ) + f(+ (+ )) f(+ ) I d f( + ) + f( ) Adding (i) nd (ii), we get f ( ) + f( + ) I d f( ) + f( + ) I d I [ ] I.(i) 9.(ii)

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