ANSWER KEY 1. [A] 2. [C] 3. [B] 4. [B] 5. [C] 6. [A] 7. [B] 8. [C] 9. [A] 10. [A] 11. [D] 12. [A] 13. [D] 14. [C] 15. [B] 16. [C] 17. [D] 18.

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1 ANSWER KEY. [A]. [C]. [B] 4. [B] 5. [C] 6. [A] 7. [B] 8. [C] 9. [A]. [A]. [D]. [A]. [D] 4. [C] 5. [B] 6. [C] 7. [D] 8. [B] 9. [C]. [C]. [D]. [A]. [B] 4. [D] 5. [A] 6. [D] 7. [B] 8. [D] 9. [D]. [B]. [A]. [C]. [C] 4. [B] 5. [B] 6. [A] 7. [C] 8. [A] 9. [D] 4. [B] 4. [A] 4. [B] 4. [B] 44. [A] 45. [A] 46. [C] 47. [C] 48. [A] 49. [B] 5. [C]

2 PHCET / Free Mock CET - 8 / Mathematics / Solutions MATHEMATICS. ~ p q. p is false and q is true. A = I A A - = I A - A = A - 4. [ ] [ ] = 4 - = = /4 = ½ = = 5. We have sin A = sin B + sin C sina cosa = sin B+C cos B C sin (B + C) cos (B - C) sin (8 - A) cos (B - C) sina cosa = sina cos (B - C) cosa = cos (B - C) A = B - C 8 - (B + C) = B - C cos [8 - (B + C)] = cos (B - C) - cos (B + C) - cos (B - C) = cos (B + C) + cos (B - C) = cos B+C+B C cosb cosc = cosb cosc = cosb = OR cosc = B = 9 OR C = 9 cos B+C B+C = PHCET ( Website : ) Page

3 PHCET / Free Mock CET - 8 / Mathematics / Solutions 6. 7 cos θ + sin θ = 4 7 cos θ + ( - cos θ) = 4 7 cos θ + - cos θ = 4 4 cos θ + = 4 4 cos θ + = + 4 cos θ = cos θ = ¼ cos θ = ± ½ cos θ = ½ or cos θ = - ½ cos θ = cos π cos θ = cos π θ = nπ ± π θ = nπ ± π 7. We have ( cos - )( + cos ) = cos - = or + cos = cos = ½ or cos = -/ cos = -/ which is not possible ( - cos ) = π, 5π /5. parallel. Slope of perpendicular from AB = slope of AB = Eq n of AB is y+ =. O (, ) - = y y = 5 A (, -) B.. Power of point (, ) w.r.t. circle + y 5 = is which is. So, (, ) lies on the circle. Hence only one tangent can be drawn. PHCET ( Website : ) Page

4 PHCET / Free Mock CET - 8 / Mathematics / Solutions. Centre of intersection of + y + 4 = and - y - 5 = which is (, -). Area is 54 sq. units. so r = 7. Eq n ( - ) + (y + ) = y + 4y + 4 = 49 + y - + 4y - 44 = 4. Let P (, y) be a point on parabola, then ( - ) + (y + 4) = (+y ) ( y + 8y + 6) = + y y - 4-4y y + 6y + = + y y - 4-4y + y - y y + 46 = 5. b a = b b = a b/a = ½ e = b a = 4 = 4 = 6. a + b = 4 i + j - k a - b = - i + j - 5 k ( a + b ) ( a - b ) = (4)(-) + ()() + (-)(-5) = = ( a + b ) ( a - b ) Angle is π 7. Let given points be A, B, C, D & they are coplanar AB AC AD are coplanar AB ( AC AD ) = ( b - a ) [ (c - a) (d - a) ] = ( b - a ) [ c d - c a - a d + a a] = ( b - a ) [ c d - c a - a d + ] = b ( c d ) - b ( c a ) - b ( a d ) - a ( c d ) + a ( c a ) + a ( a d ) ] = [ b c d ] - [ b c a ] - [ b a d ] + [ a c d ] + + = [ b c d ] - [ a b c ] + [ a b d ] + [ c a d ] = [ b c d ] + [ c a d ] - [ a b d ] = [ a b c ] PHCET ( Website : ) Page

5 PHCET / Free Mock CET - 8 / Mathematics / Solutions 8. The actual d.c. s of the lines are 7, 7, 6 7 and 5, 4, If θ is the angle between them, then cos θ = ± ( 6)(5) 7 5 = ± = ± = ± The distance between the parallel planes r. ( i - j + k ) = 4 and r. ( i - j + k ) = is = 5 4 = 5 4. Since = 4 = 6 i.e. drs of the lines are proportional lines are parallel.. drs of the line joining (5,, 4) & (6, -, ) are -,, drs of the line joining (6, -, ) & (8, -7, k) are -, 6, - k both lines will be collinear if their drs are proportional - k = 4 k = - 4 = - k = = 6 = k. The equation of the plane containing (,, ) is a( - ) + b(y - ) + c(z - ) = It also contains (-,, 4) a(- - ) + b( - ) + c(4 - ) = -a + b + 4c = -a + b + c = Also the plane is perpendicular to + y - z + = By condition of perpendicularity, a + b - c = a 4 = b = c a =, b =, c = Required equation of plane ( - ) + (y - ) + (z - ) = + + z = + z = PHCET ( Website : ) Page 4

6 PHCET / Free Mock CET - 8 / Mathematics / Solutions. The equation of the plane passing through A(a) and parallel to B(b) and C(c) is ( r - a ) ( b - a ) ( c - a ) = Now b - a = - j ( r - a ) ( b - a ) ( c - a ) = and c - a = - i - 4 j - 6 k y z 4 6 ( - )( - ) - (y - )( - ) + (z - )( - 4) = z + 4 = - 4z - 8 = - z - = = 4. The minimum value of z at is. 5. OABC is the feasible region Z() = 5() + () = (, 6) Z(A) = 5() + (5) = 5 Z(B) = 5(6/5) + (/5) = = 6 A(, 5) B 6 5, 5 Z(C) = 5(4) + () = 6 Maimum value of Z is 6. O C (4, ) (5/, ) 6. u tan, v sin 4 sin, sin u sin cos tan, v sin sin u sin v sin du dv, du dv du / dv / PHCET ( Website : ) Page 5

7 PHCET / Free Mock CET - 8 / Mathematics / Solutions 7. As required circle touches y-ais at the origin. Let Centre of the circle is d (a, ) and radius is a Equation of circle will be, ( a) ( y ) a a a y a y a (i) By differentiating above equation w.r.t., we get y a a y (ii) From (i) and (ii), y y y y y y 8. Given, f ( ) e (sin cos ) f '( ) e [cos sin ] [sin cos ] e f '( ) e sin To verify Rolle s Theorem. f '( c) c e sin c sin c c 9. I Let I I 8 9 ( ) I sin c PHCET ( Website : ) Page 6

8 PHCET / Free Mock CET - 8 / Mathematics / Solutions. As given,, 5 5 f( ), otherwise Now, probability of waiting time not more than 4 is sin for f( ) k for lim sin k k. log( )sin for As given, f( ) k for Is continuous at = log( ) sin lim k sin log( ) 8 lim lim 8 8 k 8 = k k 9. By using anti differentiation method, We will get to know that, Option (C) is correct. i.e. a + tan log a + c PHCET ( Website : ) Page 7

9 PHCET / Free Mock CET - 8 / Mathematics / Solutions 4. Degree Order 5. ( ) A A 8 A 4 A 4 sq units. 6. Given f( ) log [log sin ] C log (sin ) By using anti differentiation method,we will get d log [log sin ] c cos log (sin ) sin cot f ( ) cot log (sin ) 7. Let I / n sec (i) n sec n cos ec I / n n sec sec n cos ec I / n cos ec (ii) n cos ec n sec Adding equation (i) and (ii) / I I / I 4 PHCET ( Website : ) Page 8

10 PHCET / Free Mock CET - 8 / Mathematics / Solutions 8. y( log ) log log log y Integrating on both side log log y log( log ) log y log C log ( log ) log ( y c) log y c (i) As e, y e e e c c e Putting c in eq (i) we get e log y e y e log 9. Given. I.F of py Q is sin P e sin P ln(sin ) By anti-differentiation method, we will get d P [ln(sin )] P cot cos sin 4. Given n = 5, q p p( 4) p( ) 5 C PHCET ( Website : ) Page 9

11 PHCET / Free Mock CET - 8 / Mathematics / Solutions 4. Σ P() = P() + P() + P() + P(4) = k = K = 4. P 4 / = f () /4 / = ( - ) /4 = [ - ] = / /4 =.7 4. P (more heads than tails) = P ( = 5 or 6 or 7 or 8) = P ( = 5) + P(6) + P(7) + P(8) = 8 8 C + 8 C + 8 C + 8 C = = log ( + ) i.e. = log ( + ) Integrating y = log ( + ) C y = ( + ) log ( + ) - + C solution is y = ( + ) log ( + ) y = a ( - y) Diff w.r.t = + y +. a + a + y +y PHCET ( Website : ) Page

12 PHCET / Free Mock CET - 8 / Mathematics / Solutions But from the given relation y = y = a ( - y) y = y = a ( - y) y = + y = a y = y + a + y = a + - = a + a + y +y = = + y + Clearly it is a differential equation of first order and first degree. 46. The area bounded by the circle + y = a and the line + y = a is given by a 4 (π - ) sq. units = 6 4 (π - ) sq. units = 9(π - ) sq. units 47. By area of the region rounded by y = 4a and = 4by is 6 ab π 48. I = I = π cos θ dθ sin θ + g g sin θ cos θ sin θ +cos θ + sin θ cos θ +sin θ dθ π π I = dθ I = 4 PHCET ( Website : ) Page

13 PHCET / Free Mock CET - 8 / Mathematics / Solutions 49. ( ) is equal to z dz z + z Putting + / + = z = = z dz = dz +z = tan - z + C = tan C 5. + y = g + y = = 5 Y 5 P(, 5 ) The equation of tangent is ( y - 5 ) = 5 ( - ) A The equation of normal is ( y - 5 ) = 5 ( - ) X The tangent and normal intersect at -ais at the points A(9/, ) and (, ) respectively. A (Δ OAP) = 9 5 = PHCET ( Website : ) Page

y mx 25m 25 4 circle. Then the perpendicular distance of tangent from the centre (0, 0) is the radius. Since tangent

y mx 25m 25 4 circle. Then the perpendicular distance of tangent from the centre (0, 0) is the radius. Since tangent Mathematics. The sides AB, BC and CA of ABC have, 4 and 5 interior points respectively on them as shown in the figure. The number of triangles that can be formed using these interior points is () 80 ()