HSC - BOARD MATHEMATICS (40) - SOLUTIONS

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1 Date: 8..5 Q. (A) SECTION - I (i) (d) A () (ii) (c) A A I 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 ()

2 Q. (B) h ch fg g f 4 4 ch fgh fgh 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

3 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 () ()

4 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

5 4 i ˆ ˆ j 8 k ˆ iˆ 5 ˆj kˆ the direction ratios of the required line are, 5, () Now, Direction cosine of the line are (ii) By the sine rule a b c k sin A sin B sin C 5,, 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

6 (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 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

7 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 k k k 9 k () 7 7

8 (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 (i) 8 8

9 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 (,, ) () ( ) () () ( ) () () 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 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

10 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 D 5 B 4 P X O A C X () Y 7 + = 64 y = 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 z(c) 6 () 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. ()

11 Q.4 (A) SECTION - II (i) (a).8 () E p (ii) (c) () I d 8 (iii) (a) 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

12 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 ()

13 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 / / () 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 ()

14 (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

15 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

16 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

17 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

18 (iii) p ,,,,...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) V npq () [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 ' () for maima or minima f ' or 6 () Now f '' 4 f '' 4 Hence f has a maimum at =, Maimum value of f = f 6 () Again f '' f has minimum at =6 Minimum value at = 6 = f () (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

19 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 ()

20 Thus, F, R 9 9 Consider P X F () Range of X is, P X P P X F () 9 P X F F () 9