Date: 8..5 Q. (A) SECTION - I (i) (d) A () (ii) (c) A A I 6 6 6 A I 64 I I A A 6 (iii) (a) fg cos A cos HSC - BOARD - 5 MATHEMATICS (4) - SOLUTIONS cos cos ch () hy g fy c...(i) Comparing with A Hy By G Fy C, we get h g f A, B, C c, H, G, F A H G If equation (i) represents a pairs of straight line then H B F G F C h g h f g f c ()
Q. (B) h ch fg g f 4 4 ch fgh fgh 4 8 8 ch fgh 4 4 ch ch fg fgh (i) Converse :If the areas of two triangles are equal then they are congruent. () Contrapositive : If the area of two triangles are not equal then they are not congruent. () (ii) Comparing the equation ky y with a hy by, we get, a, h k, b. () Let m, and m be the slopes of the lines represented by ky y h k k m m b and m m a b Now, m m m m k k () (iii) The angle between the planes r n p and r n p is given by cos n n n n...(i) Here, n ˆ ˆ ˆ ˆ ˆ ˆ i j k and n i j k () ˆ ˆ ˆ ˆ ˆ ˆ n n i j k i j k Also, n 4 6 n 4 6
from we get, cos 6 6 Q. (A) () (iv) 6 y z 6 y z y z () 6 Line is passing through the point,, with dr s,, i. e.,, 6 6 Vector equation of line is r i j k i j 6k (v) c a yb 4iˆ ˆj iˆ ˆj y iˆ ˆj () 4iˆ j y iˆ y ˆj Comparing on both side 4 y & y y 4 (i) y (ii) Solving equation (i) and (ii) we get and y () (i) A,,, B,,, C,,, D,, 4 Let a, b, c, d at the position vectors of the point A, B, C, D respectively w.r.t. O, then a iˆ ˆj kˆ b iˆ ˆj kˆ c iˆ ˆj kˆ d iˆ ˆj 4kˆ AB iˆ kˆ AC ˆ i ˆj kˆ AD iˆ ˆj kˆ () ()
Volume of parallelopiped = AB AC AD AB AC AD () Q. (B) (ii) p q q 4 4 () () () (4) (5) 5 cubic units () (6) (7) p q p q p q p q p q q T T F F F T T T F F T F T T F T T F F T T F F T T T F F Column No. 5 ( Mark) Column No. 7 ( Mark) The above statement is contigency. () (iii) Let a, b & c be the position vector of the points A, B, C respectively w.r.t. O. AR m As R is a point on the line segment AB A R Band, RB n () mrb n ARand, AE and RB are in same direction. mrb n AR mob OR nor OA () mb r nr a mb na r m n (i) Let a and b be the vectors along the lines whose direction ratios are,, and, 4, respectively. a iˆ ˆj kˆ and b iˆ 4 ˆj kˆ () A vector perpendicular to bth a and b is given by () a iˆ ˆj kˆ b 4 () 4 4
4 i ˆ ˆ j 8 k ˆ iˆ 5 ˆj kˆ the direction ratios of the required line are, 5, () Now, 9 5 55 Direction cosine of the line are (ii) By the sine rule a b c k sin A sin B sin C 5,, 55 55 55 a k sin A, b k sin B, c k sin C () Now, a, b, c are in A.P. b a c b b a c b b a c b () ac ac () k sin B k A k C sin sin cos B sin B sin Asin C cos B sin B cos B sin A sin C sin B sin A C cot B... A B C () sin Asin C sin AC sin Asin C cot B sin A cosc cos A sin C sin A sin C cot B sin A cosc cos A sin C sin A sin C sin A sin C cot B cosc cos A cot B sin C sin A cot A cot C cot B () Hence, cot A, cot B, cot C are in A. P 5 5
(iii) Let the three numbers be, y, z y z 6 z y 5 y yz OR y z 6 y z 5 5y 4z () The equations can be written in matri forms 6 y 5 5 4 z () R R R, R R 5R 6 4 y 8 9 z 7 y z 6 4y 8 y 9z 7 z In equation y z 6 6 Three numbers are,, () [Note: If the student has started writing answer, then full marks will be given.] Q. (A) (i) Let m and m be the slopes of the lines represented by the equation a hy by....() Then their separate equations are y m and y m their combined equation is ( m y)( m y) i.e., mm ( m m ) y y...() Since () and () represent the same two lines, comparing the coefficients, we get, mm ( m m ) a h b h a m m and mm () b b ( m m ) ( m m ) 4m m 4h 4a 4( h ab) b b b 6 () 6
h ab b m m () If is the acute angle between the lines, then tan m m m m, if m m tan ( ) / h ab b ( a / b) a, if b tan h ab a b, if a b. () (ii) The lines y y z z y y z z and intersect,if a b c a b c y y z z a b c a b c The equations of the given lines are y z y k z and 4, y, z,, y k, z () a, b, c 4, a, b, c () Since these lines intersect, we get k 4 k 8 4 4 k k k 9 k () 7 7
(iii) Let p : switch s is closed p : switch s is open q : switch s is closed q : switch s is open r : switch s is closed r : switch s is open () Now, p v( p q) S S S ( q r) v p S S () S p p q q r p S S S S S () S L Q. (B) (i) cos sin Dividing by cos sin () cos cos sin sin 4 4 cos cos 4 4...(i) 8 8
The general solution of cos cos is n ; n z 4 () The general solution of equation (i) given by n ; n z () 4 4 n ; n ; n z () (ii) The required equation of the plane parallel to the plane y z 4 is y z Now, distance of this plane from the point (,, ) () ( ) () () ( ) () 4 6 4 4 () or or or 6 () Hence, the equations of the required planes are y z and y z 6. () (iii) Let units of food F and y units of food F be included in the diet of the sick person. Then their total cost is z Rs. (6 y) () This is the objective function which is to be minimized. The constraints are as per the following table : Food F Food F Minimum ( ) ( y) requirement Vitamin A 6 8 48 Vitamin B 7 64 From this table, the constraints are 6 8y 48, 7 y 64 Also, the number of units of foods F and F cannot be negative., y. the mathematical formulation of given LPP is Minimize z 6 y, subjected to 6 8y 48, 7 y 64,, y. () 9 9
First we draw the lines AB and CD whose equations are 6 8y 48 and 7 y 64 respectively. Line Equation Point on the X-ais Point on the Y-ais AB 6 8y 48 A (8, ) B (, 6) 64 6 CD 7 y 64 C, D, 7 Y 8 7 6 D 5 B 4 P X O A C 4 5 6 7 8 9 X () Y 7 + = 64 y 6 + 8 = 48 y The feasible region is shaded in the figure. 64 The vertices of the feasible region are C,, P and B(, 6). 7 P is the point of intersection of the lines 6 8y 48 and 7 y 64 Solving these eqautions we get 4, y P (4, ) The values of the objective function z 6 y at these vertices are 64 84 z(c) 6 () 54.85 7 7 z(p) 6(4) () 4 54 z(b) 6() (6) 6 the minimum value of z is 54 at the point (4, ). Hence, 4 units of food F and units of food F should be included in the diet of the sick person to meet the minimal nutritional requirements, in order to have the minimum cost of Rs. 54. ()
Q.4 (A) SECTION - II (i) (a).8 () E p.......8 (ii) (c) () I d 8 (iii) (a) 8 8 4 dy dy y d d y c c diff.w.r.t. dy c d c dy d Putting in equation () y = dy d dy d y dy 4 dy d d 4 dy dy y d d
Q.4 (B) (i) sin I e d sin e d e sin d () d d e sin sin d I e sin c () (ii) y sin sin sin... y sin y () y y sin Diff. w.r.t. dy dy y cos d d dy y cos d dy d cos y y () (iii) I I / / d cos d () cos / / sec / / tan d tan tan 4 ()
a (iv) y e Taking log both side log y = a log e = a log y a () Diff. w.r.t. dy. log y y d dy y log y d () (v) Let = Number of heads n = 5 p = / q = / () Q.5 (A) P( = heads) (i) I sec d sec.sec C p q n n 5 C / / 5 4 5.5 () d 5 6 d sec. sec sec. sec d () d d d = sec.tan sec sec sec.tan sec tan tan d d sec.tan sec sec d I sec.tan sec d sec d sec.tan I log sec tan c () I sec. tan log sec tan c I (sec.tan log sec tan ) c ()
(ii) y tan dy d d d tan tan tan dy d Again differenitating w.r.t., we get tan () d y dy + () d d d y dy d d dy d d () (iii) f is continous at = lim f f () R. H. S f k... given / tan tan L. H. S lim tan lim 4 4 tan tan 4 / Q.5 (B) / / tan tan tan lim lim / tan tan tan e e e, tan / from k e () () (i) Equation of the curve : y 4... dy 4 d () 4 Slope of the tangent at (, y) Equation of the line is y = Its slope is () 4 4 4 () 4 4
From (), if =, then 4 y 4 and if =, then y The required points are (,) and (,) () (ii) Let I a d a.d d d a. d ( a ). d d () a.. d () a a.. d a a a a d a d () a a d a a a I a log a c log I a a a c I a a log a c a c a d a log a c, where c. () 5 5
Q.6 (A) (iii) Let I sin d sin sin a a d By using f d f a d sin () sin sin d sin sin sin d d sin sin 6 sin I d sin () sin sin d sin sin sin sin cos d sin sin d cos cos sec tan tan d sec tan sec d sec tan () I I (i) f is continous on, f is continous at every point of, a, f is continous at = () lim f lim f f () 6
lim sin lim a+sin+ sin asin b a b a+b b, f is continous at = lim f lim f f () lim a sin b lim cos a sin b cos a+b solving and, a=, b= () (ii) Given: y log y y y y y 99 y Differentiating w.r.t., we get dy 99 y d dy 99 y d dy y 99 d () () dy 99 () d y 7 7
(iii) p 45 4 9 9 4,,,,...4 n p p q Comparing with n 5 4 n 4, p, q () 9 9 E 5 np 4 () 9 9 Q.6 (B) 5 4 8 V npq 4 9 9 8 () [Note: If the student attempts this question or even writes the question number they will get full credit] f 6 (i) Diff. w.r.t. f ' 6 4 6 () for maima or minima f ' 6 4 6 6 6 or 6 () Now f '' 4 f '' 4 Hence f has a maimum at =, Maimum value of f = f 6 () Again f '' 6 6 4 f has minimum at =6 Minimum value at = 6 = f 6 6 6 66 8 () (ii) The given equation can be written as y d ydy i. e dy y. d y Above equation is homogeneous differential equation. To solve it, we substitute y v () 8 8
dy v d dv d Equation becomes dv v v v d v v v v dv v v v d v v v v dv d Which is in variables separatable form Integrating both sides, we get v v dv d c () log v log log c () log v log c Resubstitution v c v y y y, we get c c i.e. y c, which is the requried general solution. () (iii) c.d.f of X is given by F f y dy y y dy 9 () 9 9 9 9
Thus, F, R 9 9 Consider P X F () 9 9 9 Range of X is, P X P P X F 9 9 8 () 9 P X F F 8 9 9 9 9 7 () 9