### 09 p.t (straight line + circle) solution

• 1. 1/3 GR # PT (FOR JA) # 09 (STRAIGHT LINE + CIRCLE)_SOLUTION Target : JEE (Main + Advanced) 2021 GR_PT (FOR JEE-ADV.) # 09 (Straight line + Circle) MATHEMATICS SOLUTION SECTION–I(i) 1. Ans. (B) 2 3 2 2 2 a + a + = 2 3 4 a + a + = 2 1 0 a + a - = Þ 1 5 2 + a = - or 1 5 2 - - a has only one value in first quadrant. 2. Ans. (B) Family of lines can be represented as a(x + 2y – 5) + b(3x – y – 1) = 0 Hence the family always pass through the point (1,2). Equation of other bisector must be of the form 4x + 3y = k. (Q angle bisectors are perpendicular to each other) As it pass through the point (1, 2) we get k = 10. 3. Ans. (D) Area will be maximum when the line x + y + 1 = 0 is the diameter of the given circle. Hence g 2 + ƒ 2 = 1 Hence (g, ƒ ) can take the values (0, ±1) or (±1, 0) 4. Ans. (B) We have, a + 2h + b = 0 Compairing with ax + by = h we get (x, y) as 1 1 , 2 2 æ ö - - ç ÷ è ø 5. Ans. (A) L1 : (x – y – 6) + l (2x + y + 3) Always passes through A(1, –5) and L2 : (x + 2y – 4) + m(–3x + 2y + 4) = 0 Always passes through B(2, 1) Since lines intersect at right angle so locus of point of intersection of L1 and L2 will be a circle with diametric end points A and B. Þ x2 + y2 – 3x + 4y – 3 = 0 6. Ans. (A,B,C,D) For equilateral D 2 2 2 1 2 3 1 2 2 3 3 1 z z z z z z z z z + + = + + (A) Which is satisfied by –z1– z2 & z3 also similarly |z1| = |z2| = |z3| = 1 (B) z1 + 1, z2 + 1, z3 + 1 are also satisfying the relation (C) z1 z2 z3 z +z 2 1 3 z +z 2 1 2 z +z 2 2 3 are also vertices of equilateral D. (D) 3 1 2 z z z , , 2 2 2 are also vertices of an equilateral triangle. 7. Ans. (A,D) For ax + by + c = 0 & c = –2a – 3b Þ a(x – 2) + b(y – 3) = 0 family always passes through (2, 3) Let equation of tangent is y = 1 mx m + 3 = 2m + 1 m Þ 2m2 – 3m + 1 = 0 (2m – 1)(m – 1) = 0 1 m 2 = & m = 1 Tangents are y = x 2 2 + Þ 2y = x + 4 y = x + 1 8. Ans.(A,B,C,D) ƒ '(x) = 4x3 + kx2 + 4x + k = (4x + k) (x2 + 1) (A) If x Î (0,2) then k < ƒ '(x) < 5 (8 + k) ƒ '(x) > 0 Þ k > 0 (B) ƒ '(x) < 0 Þ 5(8 + k) < 0 or k < –8 (C) If x Î (–2,2) 5(–8 + k) < ƒ '(x) < 5(8 + k) ƒ '(x) > 0 Þ 5(–8 + k) > 0 or k > 8
• 2. GR # PT (FOR JA) # 09 (STRAIGHT LINE + CIRCLE)_SOLUTION 2/3 Target : JEE (Main + Advanced) 2021 (D) ƒ '(x) < 0 Þ 5(8 + k) < 0 or k < –8 SECTION–I(iii) Paragraph for Question 9 to 10 Radius of Ci = 1 Radius of C = 2 1 + C2 C1 C3 C4 C 9. Ans. (B) Required probability ( ) 2 4. . 2 1 p = p + ( ) 4 4 3 2 2 3 2 2 = = - + 10. Ans. (C) Let the hyperbola be 2 2 2 2 x y 1 a 2a - = Passing through ( ) ( ) ( ) 2 1 , 2 1 ± + ± + Þ ( ) 2 2 2 1 2a + = lengths of L.R = 2 2.2a a 2 1 4 2 æ ö + = ç ÷ ç ÷ è ø Paragraph for Question 11 to 13 11. Ans.(D) (–2,0) A 1 0 (2,0) S1 B X S2 Y 1 (x2 + y2 + 3)2 – (4x)2 = 0 Þ S1 : x2 + y2 – 4x + 3 = 0 S2 : x2 + y2 + 4x + 3 = 0 Maximumvalueof (x1 – x2 )2 + (y1 – y2 )2 is correspondingto points A & B AB = 6 Þ (AB)2 = 36 12. Ans.(B) q=30° q=30° 1 (2,0) (0,0) 1 sin 30 2 q = Þ q = ° Þ Angle = 60° 13. Ans.. (D) Paragraph for Question 14 to 16 Foot of perpendicular from focus on any of the tangent always lies on tangent at vertex h 1 k , 2 2 + æ ö ç ÷ è ø lies on x = 0 Þ h = – 1 Þ locus is x = –1 P (h,k) L S(1,0) A (at ,2at) 2 14. Ans. (D) x = –1 15. Ans. (B) k 0, 2 æ ö ç ÷ è ø also lies on L & tangent at vertex yt = x + t 2 A º (0, t) Þ k = 2t (h, k) º (–1, 2t) y = 3x 4 3 3 + + is tangent to circle Þ Perpendicular distance = radius 2 2t 3 4 3 3 1 t 2 + - - = + Þ t 3 = Þ radius = 2 16. Ans. (C) 2 1 t 4 + = Þ t 15 = Slope of the tangent Þ 1 1 t 15 = SECTION–III(i) 1. Ans. 1 1 n 1 2 n I 0 II I x .x 1 x dx - = - ò 1 4 24 3 Applying By parts 1 1 n 1 2 3/2 n 2 2 3/2 n 0 0 x (1 x ) (n 1)x (1 x ) I dx 3 3 - - æ ö - - - = - + ç ÷ è ø ò 1 n 2 2 2 n 0 (n 1) I x (1 x ) 1 x dx 3 - - = - - ò
• 3. 3/3 GR # PT (FOR JA) # 09 (STRAIGHT LINE + CIRCLE)_SOLUTION Target : JEE (Main + Advanced) 2021 n n 2 n 3I (n 1)I (n 1)I - = - - - Þ n n 2 I n 1 I n 2 - - = + 2. Ans. 3 Let eqution of circle be (x2 + y2 + 2x – 4y – 4) + l(2x – 3y) Its centre 3 1 ,2 2 l æ ö - -l + ç ÷ è ø Centre is at minimum distance from (–5,8) Þ Centre = (–5,8) Þ l = 4 Hence equation of circle is x2 + y2 + 10x – 16y – 4 = 0 radius 93 = N 3 31 = 3. Ans. 4 2 5 Ö 90º B Ö5 D A(3,1) C ÐABC = 90º BC 2 5 = AB 9 1 5 5 = + - = Þ AC = 5 Power of point Þ AB2 = AD.AC Þ AD = 1 Þ CD = 4 SECTION–IV 1. Ans. (A)®(R); (B)®(P,Q,R,S); (C)® (P,Q,S); (D)®(T) (A) (5!6!.....9!) (10!....14!) (10!...19!) (20!....24!) (25!.....29!) In 5!,6!.........,9! exponent of 5 is one. 10!.........14! exponent of 5 is two. 15!.........19! exponent of 5 is 3. 20!.........24! exponent of 5 is 4. 25!.........29! exponent of 5 is 6. hence it is 5 5 5 10 5 15 5 20 5 30 = 5 80 (B) ( ) 2 x y 1 x = + ( ) n n 2 2 2 n 0 |x|dx xdx 2 n 1 n 1 x (1 x ) - = = + + + ò ò l A(n) = ln(1 + n 2 ) & 2 2 1 1 n A n n n æ ö + æ ö = ç ÷ ç ÷ è ø è ø l Þ A(n) – 1 A 2 nn n æ ö = + ç ÷ è ø l Þ 1 A(n) A n n 2n nn æ ö - ç ÷ è ø = l (even integer) (C) 2 2 g(x) (ƒ(x)) = Þ 3 4 g'(x) ƒ'(x) (ƒ(x)) = - Þ 2 3 4 ƒ''(x) 3(ƒ'(x)) g''(x) 4 (ƒ(x)) (ƒ(x)) æ ö = - - ç ÷ ç ÷ è ø Þ 3 2 3 4 ƒ''(x) 3(g'(x)(ƒ(x) ) g''(x) 4 (ƒ(x)) 16ƒ(x) æ ö = - - ç ÷ ç ÷ è ø = 2 2 3 ƒ''(x) 3 4 (g'(x)) (ƒ(x)) 16 (ƒ(x)) æ ö - - ç ÷ è ø l = 3 (D) Radical axis of both the circles is x – y = 0 Length of perpendicular from centre = radius 2 2 b a a b c 2 - = + - Þ (b – a) 2 = 2(a 2 + b 2 – c) Þ (a + b) 2 = 2c Þ 2 (a b) 1 2c + =
• 4. 1/3 GR # PT (FOR JA) # 09 (STRAIGHT LINE + CIRCLE)_SOLUTION Target : JEE (Main + Advanced) 2021 GR_PT (FOR JEE-ADV.) # 09 (Straight line + Circle) MATHEMATICS SOLUTION SECTION–I(i) 1. Ans. (B) 2 3 2 2 2 a + a + = 2 3 4 a + a + = 2 1 0 a + a - = Þ 1 5 2 + a = - or 1 5 2 - - a has only one value in first quadrant. 2. Ans. (B) Family of lines can be represented as a(x + 2y – 5) + b(3x – y – 1) = 0 Hence the family always pass through the point (1,2). Equation of other bisector must be of the form 4x + 3y = k. (Q angle bisectors are perpendicular to each other) As it pass through the point (1, 2) we get k = 10. 3. Ans. (D) Area will be maximum when the line x + y + 1 = 0 is the diameter of the given circle. Hence g 2 + ƒ 2 = 1 Hence (g, ƒ ) can take the values (0, ±1) or (±1, 0) 4. Ans. (B) We have, a + 2h + b = 0 Compairing with ax + by = h we get (x, y) as 1 1 , 2 2 æ ö - - ç ÷ è ø 5. Ans. (A) L1 : (x – y – 6) + l (2x + y + 3) Always passes through A(1, –5) and L2 : (x + 2y – 4) + m(–3x + 2y + 4) = 0 Always passes through B(2, 1) Since lines intersect at right angle so locus of point of intersection of L1 and L2 will be a circle with diametric end points A and B. Þ x2 + y2 – 3x + 4y – 3 = 0 6. Ans. (A,B,C,D) For equilateral D 2 2 2 1 2 3 1 2 2 3 3 1 z z z z z z z z z + + = + + (A) Which is satisfied by –z1– z2 & z3 also similarly |z1| = |z2| = |z3| = 1 (B) z1 + 1, z2 + 1, z3 + 1 are also satisfying the relation (C) z1 z2 z3 z +z 2 1 3 z +z 2 1 2 z +z 2 2 3 are also vertices of equilateral D. (D) 3 1 2 z z z , , 2 2 2 are also vertices of an equilateral triangle. 7. Ans. (A,D) For ax + by + c = 0 & c = –2a – 3b Þ a(x – 2) + b(y – 3) = 0 family always passes through (2, 3) Let equation of tangent is y = 1 mx m + 3 = 2m + 1 m Þ 2m2 – 3m + 1 = 0 (2m – 1)(m – 1) = 0 1 m 2 = & m = 1 Tangents are y = x 2 2 + Þ 2y = x + 4 y = x + 1 8. Ans.(A,B,C,D) ƒ '(x) = 4x3 + kx2 + 4x + k = (4x + k) (x2 + 1) (A) If x Î (0,2) then k < ƒ '(x) < 5 (8 + k) ƒ '(x) > 0 Þ k > 0 (B) ƒ '(x) < 0 Þ 5(8 + k) < 0 or k < –8 (C) If x Î (–2,2) 5(–8 + k) < ƒ '(x) < 5(8 + k) ƒ '(x) > 0 Þ 5(–8 + k) > 0 or k > 8
• 5. GR # PT (FOR JA) # 09 (STRAIGHT LINE + CIRCLE)_SOLUTION 2/3 Target : JEE (Main + Advanced) 2021 (D) ƒ '(x) < 0 Þ 5(8 + k) < 0 or k < –8 SECTION–I(iii) Paragraph for Question 9 to 10 Radius of Ci = 1 Radius of C = 2 1 + C2 C1 C3 C4 C 9. Ans. (B) Required probability ( ) 2 4. . 2 1 p = p + ( ) 4 4 3 2 2 3 2 2 = = - + 10. Ans. (C) Let the hyperbola be 2 2 2 2 x y 1 a 2a - = Passing through ( ) ( ) ( ) 2 1 , 2 1 ± + ± + Þ ( ) 2 2 2 1 2a + = lengths of L.R = 2 2.2a a 2 1 4 2 æ ö + = ç ÷ ç ÷ è ø Paragraph for Question 11 to 13 11. Ans.(D) (–2,0) A 1 0 (2,0) S1 B X S2 Y 1 (x2 + y2 + 3)2 – (4x)2 = 0 Þ S1 : x2 + y2 – 4x + 3 = 0 S2 : x2 + y2 + 4x + 3 = 0 Maximumvalueof (x1 – x2 )2 + (y1 – y2 )2 is correspondingto points A & B AB = 6 Þ (AB)2 = 36 12. Ans.(B) q=30° q=30° 1 (2,0) (0,0) 1 sin 30 2 q = Þ q = ° Þ Angle = 60° 13. Ans.. (D) Paragraph for Question 14 to 16 Foot of perpendicular from focus on any of the tangent always lies on tangent at vertex h 1 k , 2 2 + æ ö ç ÷ è ø lies on x = 0 Þ h = – 1 Þ locus is x = –1 P (h,k) L S(1,0) A (at ,2at) 2 14. Ans. (D) x = –1 15. Ans. (B) k 0, 2 æ ö ç ÷ è ø also lies on L & tangent at vertex yt = x + t 2 A º (0, t) Þ k = 2t (h, k) º (–1, 2t) y = 3x 4 3 3 + + is tangent to circle Þ Perpendicular distance = radius 2 2t 3 4 3 3 1 t 2 + - - = + Þ t 3 = Þ radius = 2 16. Ans. (C) 2 1 t 4 + = Þ t 15 = Slope of the tangent Þ 1 1 t 15 = SECTION–III(i) 1. Ans. 1 1 n 1 2 n I 0 II I x .x 1 x dx - = - ò 1 4 24 3 Applying By parts 1 1 n 1 2 3/2 n 2 2 3/2 n 0 0 x (1 x ) (n 1)x (1 x ) I dx 3 3 - - æ ö - - - = - + ç ÷ è ø ò 1 n 2 2 2 n 0 (n 1) I x (1 x ) 1 x dx 3 - - = - - ò
• 6. 3/3 GR # PT (FOR JA) # 09 (STRAIGHT LINE + CIRCLE)_SOLUTION Target : JEE (Main + Advanced) 2021 n n 2 n 3I (n 1)I (n 1)I - = - - - Þ n n 2 I n 1 I n 2 - - = + 2. Ans. 3 Let eqution of circle be (x2 + y2 + 2x – 4y – 4) + l(2x – 3y) Its centre 3 1 ,2 2 l æ ö - -l + ç ÷ è ø Centre is at minimum distance from (–5,8) Þ Centre = (–5,8) Þ l = 4 Hence equation of circle is x2 + y2 + 10x – 16y – 4 = 0 radius 93 = N 3 31 = 3. Ans. 4 2 5 Ö 90º B Ö5 D A(3,1) C ÐABC = 90º BC 2 5 = AB 9 1 5 5 = + - = Þ AC = 5 Power of point Þ AB2 = AD.AC Þ AD = 1 Þ CD = 4 SECTION–IV 1. Ans. (A)®(R); (B)®(P,Q,R,S); (C)® (P,Q,S); (D)®(T) (A) (5!6!.....9!) (10!....14!) (10!...19!) (20!....24!) (25!.....29!) In 5!,6!.........,9! exponent of 5 is one. 10!.........14! exponent of 5 is two. 15!.........19! exponent of 5 is 3. 20!.........24! exponent of 5 is 4. 25!.........29! exponent of 5 is 6. hence it is 5 5 5 10 5 15 5 20 5 30 = 5 80 (B) ( ) 2 x y 1 x = + ( ) n n 2 2 2 n 0 |x|dx xdx 2 n 1 n 1 x (1 x ) - = = + + + ò ò l A(n) = ln(1 + n 2 ) & 2 2 1 1 n A n n n æ ö + æ ö = ç ÷ ç ÷ è ø è ø l Þ A(n) – 1 A 2 nn n æ ö = + ç ÷ è ø l Þ 1 A(n) A n n 2n nn æ ö - ç ÷ è ø = l (even integer) (C) 2 2 g(x) (ƒ(x)) = Þ 3 4 g'(x) ƒ'(x) (ƒ(x)) = - Þ 2 3 4 ƒ''(x) 3(ƒ'(x)) g''(x) 4 (ƒ(x)) (ƒ(x)) æ ö = - - ç ÷ ç ÷ è ø Þ 3 2 3 4 ƒ''(x) 3(g'(x)(ƒ(x) ) g''(x) 4 (ƒ(x)) 16ƒ(x) æ ö = - - ç ÷ ç ÷ è ø = 2 2 3 ƒ''(x) 3 4 (g'(x)) (ƒ(x)) 16 (ƒ(x)) æ ö - - ç ÷ è ø l = 3 (D) Radical axis of both the circles is x – y = 0 Length of perpendicular from centre = radius 2 2 b a a b c 2 - = + - Þ (b – a) 2 = 2(a 2 + b 2 – c) Þ (a + b) 2 = 2c Þ 2 (a b) 1 2c + =
• 7. 1/3 GR # PT (FOR JA) # 09 (STRAIGHT LINE + CIRCLE)_SOLUTION Target : JEE (Main + Advanced) 2021 GR_PT (FOR JEE-ADV.) # 09 (Straight line + Circle) MATHEMATICS SOLUTION SECTION–I(i) 1. Ans. (B) 2 3 2 2 2 a + a + = 2 3 4 a + a + = 2 1 0 a + a - = Þ 1 5 2 + a = - or 1 5 2 - - a has only one value in first quadrant. 2. Ans. (B) Family of lines can be represented as a(x + 2y – 5) + b(3x – y – 1) = 0 Hence the family always pass through the point (1,2). Equation of other bisector must be of the form 4x + 3y = k. (Q angle bisectors are perpendicular to each other) As it pass through the point (1, 2) we get k = 10. 3. Ans. (D) Area will be maximum when the line x + y + 1 = 0 is the diameter of the given circle. Hence g 2 + ƒ 2 = 1 Hence (g, ƒ ) can take the values (0, ±1) or (±1, 0) 4. Ans. (B) We have, a + 2h + b = 0 Compairing with ax + by = h we get (x, y) as 1 1 , 2 2 æ ö - - ç ÷ è ø 5. Ans. (A) L1 : (x – y – 6) + l (2x + y + 3) Always passes through A(1, –5) and L2 : (x + 2y – 4) + m(–3x + 2y + 4) = 0 Always passes through B(2, 1) Since lines intersect at right angle so locus of point of intersection of L1 and L2 will be a circle with diametric end points A and B. Þ x2 + y2 – 3x + 4y – 3 = 0 6. Ans. (A,B,C,D) For equilateral D 2 2 2 1 2 3 1 2 2 3 3 1 z z z z z z z z z + + = + + (A) Which is satisfied by –z1– z2 & z3 also similarly |z1| = |z2| = |z3| = 1 (B) z1 + 1, z2 + 1, z3 + 1 are also satisfying the relation (C) z1 z2 z3 z +z 2 1 3 z +z 2 1 2 z +z 2 2 3 are also vertices of equilateral D. (D) 3 1 2 z z z , , 2 2 2 are also vertices of an equilateral triangle. 7. Ans. (A,D) For ax + by + c = 0 & c = –2a – 3b Þ a(x – 2) + b(y – 3) = 0 family always passes through (2, 3) Let equation of tangent is y = 1 mx m + 3 = 2m + 1 m Þ 2m2 – 3m + 1 = 0 (2m – 1)(m – 1) = 0 1 m 2 = & m = 1 Tangents are y = x 2 2 + Þ 2y = x + 4 y = x + 1 8. Ans.(A,B,C,D) ƒ '(x) = 4x3 + kx2 + 4x + k = (4x + k) (x2 + 1) (A) If x Î (0,2) then k < ƒ '(x) < 5 (8 + k) ƒ '(x) > 0 Þ k > 0 (B) ƒ '(x) < 0 Þ 5(8 + k) < 0 or k < –8 (C) If x Î (–2,2) 5(–8 + k) < ƒ '(x) < 5(8 + k) ƒ '(x) > 0 Þ 5(–8 + k) > 0 or k > 8
• 8. GR # PT (FOR JA) # 09 (STRAIGHT LINE + CIRCLE)_SOLUTION 2/3 Target : JEE (Main + Advanced) 2021 (D) ƒ '(x) < 0 Þ 5(8 + k) < 0 or k < –8 SECTION–I(iii) Paragraph for Question 9 to 10 Radius of Ci = 1 Radius of C = 2 1 + C2 C1 C3 C4 C 9. Ans. (B) Required probability ( ) 2 4. . 2 1 p = p + ( ) 4 4 3 2 2 3 2 2 = = - + 10. Ans. (C) Let the hyperbola be 2 2 2 2 x y 1 a 2a - = Passing through ( ) ( ) ( ) 2 1 , 2 1 ± + ± + Þ ( ) 2 2 2 1 2a + = lengths of L.R = 2 2.2a a 2 1 4 2 æ ö + = ç ÷ ç ÷ è ø Paragraph for Question 11 to 13 11. Ans.(D) (–2,0) A 1 0 (2,0) S1 B X S2 Y 1 (x2 + y2 + 3)2 – (4x)2 = 0 Þ S1 : x2 + y2 – 4x + 3 = 0 S2 : x2 + y2 + 4x + 3 = 0 Maximumvalueof (x1 – x2 )2 + (y1 – y2 )2 is correspondingto points A & B AB = 6 Þ (AB)2 = 36 12. Ans.(B) q=30° q=30° 1 (2,0) (0,0) 1 sin 30 2 q = Þ q = ° Þ Angle = 60° 13. Ans.. (D) Paragraph for Question 14 to 16 Foot of perpendicular from focus on any of the tangent always lies on tangent at vertex h 1 k , 2 2 + æ ö ç ÷ è ø lies on x = 0 Þ h = – 1 Þ locus is x = –1 P (h,k) L S(1,0) A (at ,2at) 2 14. Ans. (D) x = –1 15. Ans. (B) k 0, 2 æ ö ç ÷ è ø also lies on L & tangent at vertex yt = x + t 2 A º (0, t) Þ k = 2t (h, k) º (–1, 2t) y = 3x 4 3 3 + + is tangent to circle Þ Perpendicular distance = radius 2 2t 3 4 3 3 1 t 2 + - - = + Þ t 3 = Þ radius = 2 16. Ans. (C) 2 1 t 4 + = Þ t 15 = Slope of the tangent Þ 1 1 t 15 = SECTION–III(i) 1. Ans. 1 1 n 1 2 n I 0 II I x .x 1 x dx - = - ò 1 4 24 3 Applying By parts 1 1 n 1 2 3/2 n 2 2 3/2 n 0 0 x (1 x ) (n 1)x (1 x ) I dx 3 3 - - æ ö - - - = - + ç ÷ è ø ò 1 n 2 2 2 n 0 (n 1) I x (1 x ) 1 x dx 3 - - = - - ò
• 9. 3/3 GR # PT (FOR JA) # 09 (STRAIGHT LINE + CIRCLE)_SOLUTION Target : JEE (Main + Advanced) 2021 n n 2 n 3I (n 1)I (n 1)I - = - - - Þ n n 2 I n 1 I n 2 - - = + 2. Ans. 3 Let eqution of circle be (x2 + y2 + 2x – 4y – 4) + l(2x – 3y) Its centre 3 1 ,2 2 l æ ö - -l + ç ÷ è ø Centre is at minimum distance from (–5,8) Þ Centre = (–5,8) Þ l = 4 Hence equation of circle is x2 + y2 + 10x – 16y – 4 = 0 radius 93 = N 3 31 = 3. Ans. 4 2 5 Ö 90º B Ö5 D A(3,1) C ÐABC = 90º BC 2 5 = AB 9 1 5 5 = + - = Þ AC = 5 Power of point Þ AB2 = AD.AC Þ AD = 1 Þ CD = 4 SECTION–IV 1. Ans. (A)®(R); (B)®(P,Q,R,S); (C)® (P,Q,S); (D)®(T) (A) (5!6!.....9!) (10!....14!) (10!...19!) (20!....24!) (25!.....29!) In 5!,6!.........,9! exponent of 5 is one. 10!.........14! exponent of 5 is two. 15!.........19! exponent of 5 is 3. 20!.........24! exponent of 5 is 4. 25!.........29! exponent of 5 is 6. hence it is 5 5 5 10 5 15 5 20 5 30 = 5 80 (B) ( ) 2 x y 1 x = + ( ) n n 2 2 2 n 0 |x|dx xdx 2 n 1 n 1 x (1 x ) - = = + + + ò ò l A(n) = ln(1 + n 2 ) & 2 2 1 1 n A n n n æ ö + æ ö = ç ÷ ç ÷ è ø è ø l Þ A(n) – 1 A 2 nn n æ ö = + ç ÷ è ø l Þ 1 A(n) A n n 2n nn æ ö - ç ÷ è ø = l (even integer) (C) 2 2 g(x) (ƒ(x)) = Þ 3 4 g'(x) ƒ'(x) (ƒ(x)) = - Þ 2 3 4 ƒ''(x) 3(ƒ'(x)) g''(x) 4 (ƒ(x)) (ƒ(x)) æ ö = - - ç ÷ ç ÷ è ø Þ 3 2 3 4 ƒ''(x) 3(g'(x)(ƒ(x) ) g''(x) 4 (ƒ(x)) 16ƒ(x) æ ö = - - ç ÷ ç ÷ è ø = 2 2 3 ƒ''(x) 3 4 (g'(x)) (ƒ(x)) 16 (ƒ(x)) æ ö - - ç ÷ è ø l = 3 (D) Radical axis of both the circles is x – y = 0 Length of perpendicular from centre = radius 2 2 b a a b c 2 - = + - Þ (b – a) 2 = 2(a 2 + b 2 – c) Þ (a + b) 2 = 2c Þ 2 (a b) 1 2c + =
• 10. 1/3 GR # PT (FOR JA) # 09 (STRAIGHT LINE + CIRCLE)_SOLUTION Target : JEE (Main + Advanced) 2021 GR_PT (FOR JEE-ADV.) # 09 (Straight line + Circle) MATHEMATICS SOLUTION SECTION–I(i) 1. Ans. (B) 2 3 2 2 2 a + a + = 2 3 4 a + a + = 2 1 0 a + a - = Þ 1 5 2 + a = - or 1 5 2 - - a has only one value in first quadrant. 2. Ans. (B) Family of lines can be represented as a(x + 2y – 5) + b(3x – y – 1) = 0 Hence the family always pass through the point (1,2). Equation of other bisector must be of the form 4x + 3y = k. (Q angle bisectors are perpendicular to each other) As it pass through the point (1, 2) we get k = 10. 3. Ans. (D) Area will be maximum when the line x + y + 1 = 0 is the diameter of the given circle. Hence g 2 + ƒ 2 = 1 Hence (g, ƒ ) can take the values (0, ±1) or (±1, 0) 4. Ans. (B) We have, a + 2h + b = 0 Compairing with ax + by = h we get (x, y) as 1 1 , 2 2 æ ö - - ç ÷ è ø 5. Ans. (A) L1 : (x – y – 6) + l (2x + y + 3) Always passes through A(1, –5) and L2 : (x + 2y – 4) + m(–3x + 2y + 4) = 0 Always passes through B(2, 1) Since lines intersect at right angle so locus of point of intersection of L1 and L2 will be a circle with diametric end points A and B. Þ x2 + y2 – 3x + 4y – 3 = 0 6. Ans. (A,B,C,D) For equilateral D 2 2 2 1 2 3 1 2 2 3 3 1 z z z z z z z z z + + = + + (A) Which is satisfied by –z1– z2 & z3 also similarly |z1| = |z2| = |z3| = 1 (B) z1 + 1, z2 + 1, z3 + 1 are also satisfying the relation (C) z1 z2 z3 z +z 2 1 3 z +z 2 1 2 z +z 2 2 3 are also vertices of equilateral D. (D) 3 1 2 z z z , , 2 2 2 are also vertices of an equilateral triangle. 7. Ans. (A,D) For ax + by + c = 0 & c = –2a – 3b Þ a(x – 2) + b(y – 3) = 0 family always passes through (2, 3) Let equation of tangent is y = 1 mx m + 3 = 2m + 1 m Þ 2m2 – 3m + 1 = 0 (2m – 1)(m – 1) = 0 1 m 2 = & m = 1 Tangents are y = x 2 2 + Þ 2y = x + 4 y = x + 1 8. Ans.(A,B,C,D) ƒ '(x) = 4x3 + kx2 + 4x + k = (4x + k) (x2 + 1) (A) If x Î (0,2) then k < ƒ '(x) < 5 (8 + k) ƒ '(x) > 0 Þ k > 0 (B) ƒ '(x) < 0 Þ 5(8 + k) < 0 or k < –8 (C) If x Î (–2,2) 5(–8 + k) < ƒ '(x) < 5(8 + k) ƒ '(x) > 0 Þ 5(–8 + k) > 0 or k > 8
• 11. GR # PT (FOR JA) # 09 (STRAIGHT LINE + CIRCLE)_SOLUTION 2/3 Target : JEE (Main + Advanced) 2021 (D) ƒ '(x) < 0 Þ 5(8 + k) < 0 or k < –8 SECTION–I(iii) Paragraph for Question 9 to 10 Radius of Ci = 1 Radius of C = 2 1 + C2 C1 C3 C4 C 9. Ans. (B) Required probability ( ) 2 4. . 2 1 p = p + ( ) 4 4 3 2 2 3 2 2 = = - + 10. Ans. (C) Let the hyperbola be 2 2 2 2 x y 1 a 2a - = Passing through ( ) ( ) ( ) 2 1 , 2 1 ± + ± + Þ ( ) 2 2 2 1 2a + = lengths of L.R = 2 2.2a a 2 1 4 2 æ ö + = ç ÷ ç ÷ è ø Paragraph for Question 11 to 13 11. Ans.(D) (–2,0) A 1 0 (2,0) S1 B X S2 Y 1 (x2 + y2 + 3)2 – (4x)2 = 0 Þ S1 : x2 + y2 – 4x + 3 = 0 S2 : x2 + y2 + 4x + 3 = 0 Maximumvalueof (x1 – x2 )2 + (y1 – y2 )2 is correspondingto points A & B AB = 6 Þ (AB)2 = 36 12. Ans.(B) q=30° q=30° 1 (2,0) (0,0) 1 sin 30 2 q = Þ q = ° Þ Angle = 60° 13. Ans.. (D) Paragraph for Question 14 to 16 Foot of perpendicular from focus on any of the tangent always lies on tangent at vertex h 1 k , 2 2 + æ ö ç ÷ è ø lies on x = 0 Þ h = – 1 Þ locus is x = –1 P (h,k) L S(1,0) A (at ,2at) 2 14. Ans. (D) x = –1 15. Ans. (B) k 0, 2 æ ö ç ÷ è ø also lies on L & tangent at vertex yt = x + t 2 A º (0, t) Þ k = 2t (h, k) º (–1, 2t) y = 3x 4 3 3 + + is tangent to circle Þ Perpendicular distance = radius 2 2t 3 4 3 3 1 t 2 + - - = + Þ t 3 = Þ radius = 2 16. Ans. (C) 2 1 t 4 + = Þ t 15 = Slope of the tangent Þ 1 1 t 15 = SECTION–III(i) 1. Ans. 1 1 n 1 2 n I 0 II I x .x 1 x dx - = - ò 1 4 24 3 Applying By parts 1 1 n 1 2 3/2 n 2 2 3/2 n 0 0 x (1 x ) (n 1)x (1 x ) I dx 3 3 - - æ ö - - - = - + ç ÷ è ø ò 1 n 2 2 2 n 0 (n 1) I x (1 x ) 1 x dx 3 - - = - - ò
• 12. 3/3 GR # PT (FOR JA) # 09 (STRAIGHT LINE + CIRCLE)_SOLUTION Target : JEE (Main + Advanced) 2021 n n 2 n 3I (n 1)I (n 1)I - = - - - Þ n n 2 I n 1 I n 2 - - = + 2. Ans. 3 Let eqution of circle be (x2 + y2 + 2x – 4y – 4) + l(2x – 3y) Its centre 3 1 ,2 2 l æ ö - -l + ç ÷ è ø Centre is at minimum distance from (–5,8) Þ Centre = (–5,8) Þ l = 4 Hence equation of circle is x2 + y2 + 10x – 16y – 4 = 0 radius 93 = N 3 31 = 3. Ans. 4 2 5 Ö 90º B Ö5 D A(3,1) C ÐABC = 90º BC 2 5 = AB 9 1 5 5 = + - = Þ AC = 5 Power of point Þ AB2 = AD.AC Þ AD = 1 Þ CD = 4 SECTION–IV 1. Ans. (A)®(R); (B)®(P,Q,R,S); (C)® (P,Q,S); (D)®(T) (A) (5!6!.....9!) (10!....14!) (10!...19!) (20!....24!) (25!.....29!) In 5!,6!.........,9! exponent of 5 is one. 10!.........14! exponent of 5 is two. 15!.........19! exponent of 5 is 3. 20!.........24! exponent of 5 is 4. 25!.........29! exponent of 5 is 6. hence it is 5 5 5 10 5 15 5 20 5 30 = 5 80 (B) ( ) 2 x y 1 x = + ( ) n n 2 2 2 n 0 |x|dx xdx 2 n 1 n 1 x (1 x ) - = = + + + ò ò l A(n) = ln(1 + n 2 ) & 2 2 1 1 n A n n n æ ö + æ ö = ç ÷ ç ÷ è ø è ø l Þ A(n) – 1 A 2 nn n æ ö = + ç ÷ è ø l Þ 1 A(n) A n n 2n nn æ ö - ç ÷ è ø = l (even integer) (C) 2 2 g(x) (ƒ(x)) = Þ 3 4 g'(x) ƒ'(x) (ƒ(x)) = - Þ 2 3 4 ƒ''(x) 3(ƒ'(x)) g''(x) 4 (ƒ(x)) (ƒ(x)) æ ö = - - ç ÷ ç ÷ è ø Þ 3 2 3 4 ƒ''(x) 3(g'(x)(ƒ(x) ) g''(x) 4 (ƒ(x)) 16ƒ(x) æ ö = - - ç ÷ ç ÷ è ø = 2 2 3 ƒ''(x) 3 4 (g'(x)) (ƒ(x)) 16 (ƒ(x)) æ ö - - ç ÷ è ø l = 3 (D) Radical axis of both the circles is x – y = 0 Length of perpendicular from centre = radius 2 2 b a a b c 2 - = + - Þ (b – a) 2 = 2(a 2 + b 2 – c) Þ (a + b) 2 = 2c Þ 2 (a b) 1 2c + =