SECTION A Time allowed: 20 minutes Marks: 20

Transcription

1 Mathcity.org Merging man and maths Federal Board HSSC-II Eamination Mathematics Model Question Paper Roll No: Answer Sheet No: FBISE WE WORK FOR EXCELLENCE Signature of Candidate: Signature of Invigilator: SECTION A Time allowed: 0 minutes Marks: 0 Note: Section-A is compulsory and comprises pages -6. All parts of this section are to be answered on the question paper itself. It should be completed in the first 0 minutes and handed over to the Centre Superintendent. Deleting/overwriting is not allowed. Do not use lead pencil. Q. Insert the correct option i.e. A/B/C/D in the empty bo provided opposite each part. Each part carries one mark. i. If f() = and g() = + what is the composition of f and g? A. B. + C. ( + ) D. + ii. What is the Value of A. 0 B. 0 0 C. / D. - Limit? Page of 6 Turn Over

2 DO NOT WRITE ANYTHING HERE iii. What is the domain of +? A. R B. (, ) C. [, ) D. (,0] iv. What is the derivative of cos h? A. B. C. D. + Page of 6 Turn Over

3 v. What is the maimum value of ( Sin + Cos )? A. 0 B. 60 C. 90 D. 5 vi. What is the value of + + +?!! A. e B. Sin C. (+) n D. Cos vii. What is the value of ( + ) d? A. B. C. D. viii. What is the value of c + ) d + ( + a b ( ) d? c b A. ( + )d a a B. ( + )d b b C. ( + )d a b D. ( ± )d a Page of 6 Turn Over

4 i. What is the value of Sec ( P + q) tan( P + q) d? A. tan (P + q) + C B. tan (P + q) +C C. P Sec (P + q) C D. Sec (P + q) + C P. Which of the following points is at a distance of 5 units from (0, 0)? A. ( 76,7) B. (0, -0) C. (, 5) D. (7, 76) i. When would the pair of lines represented by a + hy + by = 0 be imaginary? A. When h = ab B. When h >ab C. When h < ab D. When ab = b ii. What would be the slope of any line perpendicular to y + K = 0 A. B. C. D. iii. Which of the following lines passes through ( 5, 6) & (, 0)? A. y + = 0 B. y 9 = 0 C. y + 7 = 0 D. + y + = 0 Page of 6 Turn Over

5 iv. What is the number of order pairs that satisfy the epression 5 + y 0? A. Finite many B. Infinite many C. Two D. Three v. What is the length of the latus rectum of parabola = y? A. 6 B. 6 C. D. vi. For what value of C would the line y = m + C be tangent to ellipse + y = 9? A. 0 B. C. 9m + D. + m vii. What is the radius of circle + y + Cosθ + y Sinθ = 8? A. B. C. D. 0 y viii. What is the condition for ellipse + = b a to be a circle? A. a > b B. a = b C. a < b D. a = 0, b = 0 Page 5 of 6 Turn Over 5

6 i. What does the epression a ( b. c) represent? A. Vector quantity B. Scalar quantity C. Area of parallelogram D. Nothing. Which of the following vectors is perpendicular to both vectors a& b? A. a B. b C. a b D. a. b For Eaminer s use only Q. No.: Total Marks: Marks Obtained: 0 Page 6 of 6 6

7 Mathcity.org Merging man and maths FBISE WE WORK FOR EXCELLENCE Federal Board HSSC-II Eamination Mathematics Model Question Paper Time allowed:.0 hours Total Marks: 80 Note: Sections B and C comprise pages - and questions therein are to be answered on the separately provided answer book. Answer any ten questions from section B and attempt any five questions from section C. Use supplementary answer sheet i.e., sheet B if required. Write your answers neatly and legibly. Note: Attempt any TEN questions. SECTION-B (Marks: 0) Q. Evaluate Limit 0 + a a () Q. Discuss the continuity of function at y = if y y < f ( y) =, y + y () Q. If y = tan (tan - dy ( + y ) ) then show that = d + () Q.5 Differentiate log a by ab initio method. () Q.6 Find the coordinates of the turning point of the curve y = 8 + and determine whether this point is the maimum or the minimum point. () K Q.7 Find the value of K if 6 ( ) d = 5 () Page of Turn Over 7

8 Q.8 Find the area bounded by curve y = and the ais. () dy y Q.9 Using the differential find when - ln = ln c. () d Q.0 Determine the value of K for which the lines y = 0, y 5 = 0 and + Ky + 8 = 0 would meet at a point. () Q. Find the equation of the perpendicular bisector of the line segment joining the points (, 5) and (9, 8). () Q. Find the mid point of chord cut off the line + y = by circle + y = 6. () Q. Find the equation of the parabola having its focus at the origin and directri parallel to y ais. () Q. Find the equation of the hyberbola with given data: () foci (±5, 0) vertices (±, 0) Q.5 Show that Cos + Cos β + Cos γ =. () SECTION C (Marks: 0) Note: Attempt any FIVE questions. Each question carries equal marks. (Marks 5 8=0) Q.6 Find the graphical solution of the functions = Sin (8) Q.7 If y = a Cos (ln ) + b Sin (ln ); prove that d y dy + + y = 0 (8) d d Q.8 The line y = a + b is parallel to line y = 6 and passes through point (-, 7). Find the value of a & b and fined the Eq of line passing through (7, -) and perpendicular to line y = 7. (8) Q.9 If a + b + c = 0 then prove that a b = b c = c a (8) Page of Turn Over 8

9 d Q.0 Show that [ ln ] = ln. The diagram shows part of the curve y = ln. d Find the area of shaded region, correct to two decimal places. (8) y y = ln ' l Q. Maimize Z = + y subject to the constraints + y, + y, y, 0, y 0 (8) Q. Find the equation of a tangent to the parabola y = -6 which is parallel to the line + y + = 0. Also find the tangency. (8) NOTE Mathcity.org does not represent any official or government educational institute or board or university. And the material given on this site holds no official position in government (or in government educational institute or board or university). While using a material given on this site you agreed to the term that we (mathcity.org or person related to mathcity.org) do not take any responsibility for this material. Page of 9

10 FBISE WE WORK FOR EXCELLENCE Federal Board HSSC II Eamination Mathematics Mark Scheme SECTION A Q. i. B ii. C iii. C iv. B v. B vi. A vii. C viii. C i. D. A i. C ii. D iii. B iv. B v. A vi. C vii. B viii. B i. D. C (0 = 0) SECTION B Q. () Limit + a a ( mark) Limit + a a + a + a 0 + a + a ( mark) Limit + a/ a/ 0 + a + a Limit / 0 / ( + a + a) ( mark) = = 0 + a + a a ( mark) Q. () L.H. Limit Limit Limit f ( y) = y y y = = (I) ( mark) 0

11 R.H. Limit Limit Limit fy = y y y = 6 + = 7 (II) ( mark) Direct Limit Put y = f(y) = y + f() = () + = 7 (III) ( mark) From I, II & III L.H. Limit R.H. Limit = Direct Limit So f(y) is not continuous at y = ( mark) Q. () y = tan( tan ) tan y = tan ( mark) Diff w.r.t. dy =. /. ( marks) + y d / + dy ( + y ) = d + ( mark) Q.5 () Let y = log a y + Δy = log a ( + Δ) Δy = log a ( + Δ) log a ( mark) + Δy = log a = log + y = log+ ( mark) y = log a + ( mark) Limit y Limit = log a dy = log a e ( mark) d Q.6 () y = 8 +

12 y = 8 + Difference with reference to dy = 8 + ( / ) d / dy dy = 8, Put = 0 ( mark) d d 8 = 0 8 = = = ( mark) 8 dy = 8 d Difference again d y = d = d y = = 6 = 8 > 0 ( mark) d = So y will be minimum at = Put = = y y = 8 + = + = 6 ( mark) / Q.7 () K 6 ( ) d = 5 6( ) ( K K ) K = 5 = 5 ( mark) ( ) = 6 ( mark) ( K) ( ) = 6 ( K) + = 6 ( K) = 7 ( mark) K = K = + ( mark)

13 Q.8 () y = Put y = 0 = 0 = ± ( mark) A = yd A = ( ) d = d d A = [] ( mark) A = [ + ] [ 8 8] + 6 A = A = ( mark) A = Square Units ( mark) Q.9 () y ln = ln c y ln = ln c ( mark) Taking Differential dy d ln / d = d ln c / ( mark) dy = d ln c + d ln + d dy = d [ln c + ln + ] (I) ( mark) Dividing by d by (I) dy = ln c + ln + d ( mark) Q.0 () y = 0 y 5 = 0 + Ky + 8 = 0 Since lines are concurrent so the det. of their coeff. will be zero ( mark) 5 = 0 ( mark) K 8 Ep. by R (-8+5K)+(+5)-(K+) = 0-6+0K+7-K- = 0 7K+98 = 0

14 K = - ( marks) Q. Line is perpendicular to AB & A(, 5) B(9, 8) mid point of AB = (6, ) ( mark) y y slope of AB = m = 8 5 m = 9 m = 6 = ( mark) Line perpendicular to AB has slope, m = - Required line passes through (6, ) So its equation is: y y = m( ) ( mark) y = ( 6) y = ( 6) y = + +y 7 = 0 ( mark) () Q. () +y = 6 (I) +y = (II) y From (II) = putting in (I) y + y = y -78y+y = 0 y -78y+65 = 0 ( mark) by, y 6y + 5 = 0 y -5y-y+5 = 0 y(y-5)-(y-5) = 0 (y-)(y-5) = 0 y = 5, y = ( mark) when y = 5, then = - So A(-, 5) when y =, then = 5 So B(5, ) ( mark) AB is chord its mid point is + y + y =,

15 =, = (, ) ( mark) Q. () ( mark) According to problem F(0, 0) and directri is +a = 0 Let P(, y) be any point on parabola so ( mark) PF = PM ( + a 0) + ( y 0) = ( mark) Seq on both sides () + (0) +y = (+a) / + y = / + a + a y = a(+a) ( mark) Q. () F(5, 0) F (-5, 0) c = FF c = (0) + c = 0 c = 5 We have b = c a b = 5 9 b = 6 b = (0) A(, 0) A (-, 0) a = AA a = 6 a = ( mark) ( mark) 5

16 is Hyberbola is along ais and its centre is at (0, 0) so its ( mark) y = b a y = 9 6 ( mark) equation Q.5 () Let r = i + yi + zk, r + = + y z ( mark) r y z Then =,, r r r r is unit vector in the direction of γ = OP ( mark) It is supposed that OAP is right angled length OA So Cos = Cos = OP r y z Cos β =, Cosγ = ( mark) r r Since Cos, Cosβ, Cosγ are called direction cosins of OP. So Cos +Cos β+cos y z γ = + + r r r + y + z = r r = r = ( mark) 6

17 SECTION C Q.6 (8) = Sin Let y = = Sin y =, y = Sin y = is a straight line passing through origin and bisecting Ist & IIIrd quad: ( mark) y ( marks) ( mark) These two graphs intersect at origin so = 0, y = 0 is only solution set ( mark) Q.7 (8) y = a Cos(ln ) + b Sin(ln ) Diff w.r.t. dy = a( Sin ln ) + bcos(ln ) d 7

18 dy = ( bcos ln a Sinln ) ( marks) d Diff w.r.t. again: d y = [ bcos(ln ) a Sin(ln )] + [ b Sin(ln ) acos(ln ) ] d = [ bcos(ln ) a Sin(ln ) + b Sin(ln ) + acos(ln ] ( marks) Now d y dy + + y = 0 d d bcos(ln ) + a Sin(ln ) b Sin(ln ) acos(ln ) + bcos ln a Sinln + acos(ln ) + b Sinln = 0 Proved: 0 = 0 ( marks) Q.8 (8) y = a + b (I) y = 6 (II) Comparing I&II, a = y = + b, line passing through (-, 7) 7 = - + b b = 9 Hence req. line is y = + 9 ( marks) Given that y = 7 -y = 7 Slop of line m = m = Slop of line perpendicular is -, line passing through (7, -) then eq line is y y = m( ) y + = -( 7) y + = y 0 = 0 ( marks) Q.9 (8) Given that a + b + c = 0 Taking cross mult with a a ( a + b + c) = 0 a a + a b + a c = 0 a b = ( a c) a b = c a (I) ( marks) of 8

19 Let a + b + c = 0 Taking b cross mult b ( a + b + c) = 0 b a + b b + b c = 0 b c = ( b a) b c = a b (II) From (I) & (II) a b = b c = c a ( marks) Q.0 (8) d [ ln ] = ln d L.H.S d [ ln ] d ln +. = ln = RHS ( marks) Now d ln = ( ln ) d d ln d = ( ln ) d d ln d = ln + c ( marks) [ ln ] ln d = ( mark) = (ln-) (ln-) = ln + = ln ( marks) Q. (8) Z = + y subject to: + y, + y, y, 0, y 0 Associated eq. are + y = (I) + y = (II) y = (III) = 0, y = 0 ( mark) X Intercept of I put y = 0 = A(, 0) Y Intercept of I put = 0 y = B(0, ) Put (0, 0) in given inequality 0 < so solution region lies towards origin. X Intercept of II put y = 0 = C(, 0) 9

20 Y Intercept of II put = 0 y = D(0, ) Put (0, 0) so 0 < soln region towards origin. X Intercept of III put y = 0 = E(, 0) Y Intercept of I put = 0 y = - F(0, -) ( marks) Put (0, 0) solution region towards origin as 0 < 0 & y 0 lies on st quad. So corner points are (0, ) (0, 0) (, 0) and (, ) 5 5 ( marks) + y = _ 8 + y =_6-5 = - = Put in (II) y = y = 8 5 ( mark) 5 0

21 f(, y) + y (0, 0) = 0 (0, ) = 9 (, 0 ) + 0 = 8 6, + = = So Z = + y is maimum at (0, ) ( mark) Q. (8) y = -6 & + y + = 0 Slope of required line = - = m ( mark) In parabola y = -6 a = 6 = ( mark) Eq of Tangent to parabola a y = m + ( mark) m 8 y = + + y = 8 + y = 0 Req. Eq ( mark) 8 + Put y = in y = = = = 0 6 (8 + ) = 0 = ( marks) Putting the value of in y = 8 ( ) + y = 8 y = ( mark) Hence point of Tangency is (, ) ( mark) 8