151.a.We have 3 52 − = 2 13 5 0 3 52 ± = ⇒ = + − ⇒Similary, : 3 52 − = 2 13 5 ± = ⇒ 2 13 5 and 2 13 5 − = + = ∴ or vice -versa3 13) (25 41 & 19 4 26 50 2 2 = − = = + = + Thus, the equation having & as its roots is x x2 + + − = 0 10 x x 2 2 2 + + − ⇒ or 3x2 -19x + 1 = 0 152.a y = n 2 ) x 1 (x + + ; .2x ) x (1 21 1 ) x 1 n(x dx dy 1/2 2 1 n 2 + + + + = − − 2 n 2 2 2 1 n 2 x 1 x) x 1 n( x 1 x) x 1 ( ) x 1 n(x dx dy + + + = + + + + + = − ) dx dy (y ny y x 1 or ny dx dy x 1 or 1 1 2 2 = = + = + 2 2 21 2 y n )y x (1 Squaring, = + Differentiating , 1 2 21 2 1 2 .2yy n .2x y y )2y x (1 = + + (Here, y x xy )y x (1 or ) dxy d y 2 1 2 2 2 2 2 = + + = 153.c.1, log9 1) (4.3 log 2), (3 3 x 1 − + − x are in A.P. ) 1 3 . 4 ( log 1 2) (3 log 2 3 x 1 9 − + = + ⇒ − x ) 1 3 . 4 ( log 3 log 2) (3 log 3 3 x 1 3 − + = + − x 1)] [3(4.3 log 2) (3 log x 3 x 1 3 − = + − t) 3 1)(Put (4.3 3 2 3 x x x 1 = − = + − 3.3-x + 2 = 12.3x -3 : 0 3 5t 12t or 3 12t 2 t3 2 = − − − = + Hence t = 43 3 43 , 31 x = ⇒ − 4 log 1 x 4 log 3 log x or 43 log x 3 3 3 3 − = ⇒ − = = ⇒ 154.a.P (E1) = 21 , P (E2) = 31 and P (E3) = 41 ; ) E ( P ) E ( P ) E P( 1 ) UE UE P(E 3 2 1 3 2 1 − = 43 43 32 21 1 41 1 31 1 21 1 1 = × × − = − − − − = 155.b. 2 2Period ; 2 2cos 1 sin2 = = − = 156.d. R log 1) (p A log 1 log AR l 1 p − + = ⇒ = − m = ARq-1⇒ log m = log A + (q -1) log R n = ARr-1 ⇒ log n = log A +(r -1) log R Now, 0 1 r R log 1) (q A log 1 q R log 1) (q A log 1 P R log 1) (p A log 1 r n log 1 q m log 1 p l log = − + − + − + = 157.a. ; 2x x) sin 2 (1 1 lim 2x 2x cos 1 lim 2 0 x 0 x − − ⇒ − → → x | x sin | lim 2x x sin 2 lim 0 x 2 0 x → → ⇒ the function does not exist of LHS ≠ RHS 158.a.AB = 26 1) (0 1) (4 2 2 = + + + 52 1) (5 1) (3 BC 2 2 = + + + = 26 5) -(0 3) -(4 CA 2 2 = + = ; So. In isosceles triangle side AB = CA For right angled triangle. BC2 = AB2 + AC2 So, here BC = 52 or BC2 = 52 or 52 ) 26 ( ) 26 ( 2 2 = + So, given triangle is right angled and also isosceles 159.b.Total student = 100; for 70 stds. 75 × 70 = 5250 ⇒ 7200 -5250 = 1950 Average of girls = 65 30 1950 = 160.a. x ) cos ( tan ) cos ( cot 1 1 = − − − x ) cos ( tan cos 1 tan 1 1 = − − − SOLUTIONS MATHEMATICS Paper -I I -B(-1,-1)A (4, 0) C(3, 5)x cos . cos 1 1 cos cos 1 tan 1 = + − ⇒ − cos 2 cos 1 x tan x cos 2 cos 1 tan 1 − ⇒ = − ⇒ − cos 1 cos 1 x cosec or cos 1 cos 2 x cot OR −+ = − = 2tan x sin or 1 2cos 1 sin 2 (1 1 cos 1 cos 1 x sin 2 2 2 = − + − − = +− = 161.c.Order = 3, degree = 3 162.a. .....(i) 4 4 z 5 7 y 1 4 x − = − = − a (x -4) + b ( y -7) + c ( z-4) = 0.....(ii) Line passing through point ( 3, 2, 0) a + 5x + 4c +0 .......(iii) Solving the equation we get by eqn (ii) x -y + z = 1.........(iv) 163.b. d cx 4 e y c; 2 e dx dy ; e dxy d 2x 2x 2x 2 2 + + = + − = = − − − . 164.d. x1 2 2 x 3 x x 3 5x x lim + + + + ∞ > − x1 22 x x3 x1 1 x3 x5 1 lim + + + + = ∞ > − = 1 165.a.f(x) = sin-1 3x log3 exists if -1 ≤ 1 3x log3 ≤ [1,9] x or 9 x 1 3 3x 3 1 1 ∈ ≤ ≤ ⇔ ≤ ≤ ⇔ − 166.b. 167.b. ar4 = 2 8 7 6 5 4 3 2 ar ar ar ar ar ar ar ar a × × × × × × × × 512 2 ) (ar r a 9 9 4 36 9 = = = = . 168.d. dx sinx dx sinx 10 dx | sinx | 100 0 ∫ ∫ ∫ + = 20 2 10 1] 10[1 ; x] [cos x] 10[cos 0 = × = + + = 169.b. = ∫ + 0 2 n dx x) tan x(1 tan ∫ ∫ = = 0 10 n 2 n x tan t where dt t ; dx sec x tan x ] I n[I lim ; 1 n1 I I 2 n n x 2 n n + ∞ → + + ⇒ + = + 1 n1 1 n n 1 nn 1 n1 n. lim x = + = + = + = ∞ → 170.c. 1 2 dx 0 ]dx [x dx ] [x 2 1 2 2 1 2 0 1 − = ∫ + = ∫ + ∫ 171.b. ∫ ∫ + + + ∫ = + + − − − 2 2 2 x cos 1 x sin x 2 x cos 1 2x dx x cos 1 x) sin (1 2x∫ ∫ + = + + = 0 0 2 2 x) -(cos 1 x) -(sin x) -(4 1 ; x cos 1 dx x sin x 4 0∫ ∫+ = ⇒ + = 0 0 2 2 x cos 1 sin x 41 x cos 1 sin x x) -(4 1 ∫ ∫+ = ⇒ + − 0 2 2 dx x cos 1 sin x 42I ; x cos 1 x sin x 4put cos x = t and solve it. 172.c.we have, 2 x 2f(x) xf(2) lim2 x −− →( ) ( ) ( ) ( ) 4 4 2 4 2 2f' 2 f x 2f' 2 f lim 00 2 x − = × − = − = − = → 173.b. Let |z| = | | = r ∴ i re z = = φ i re where + φ = . φ = ∴ -i re re .e re re z i i i) -i(− = − = = = ∴ φ − φ − φ 174.c.Given |z -4| < |z -2| Let z = x + iy| iy 2) (x | | iy) 4) (x | + − < + − ⇒ 2 2 2 2 y 2) (x y 4) (x + − < + − ⇒ 4 4x x 16 8x x 2 2 + − < + − ⇒ 3 Re(z) 3 x 4x 12 > ⇒ > ⇒ < ⇒ 175.b. 176.b. Let a = first term of G.P. r = common ratio of G.P. Then G.P. is a, ar, ar2 Given ∞ S = 20 20 r 1 a = − ⇒ r) 20(1 a − = ⇒ ...(i) Also 100 .....to r a r a a 4 2 2 2 2 = ∞ + + + 100 r 1 a 2 2 = − ⇒ r) r)(1 100(1 a2 + − = ⇒ ......(ii) From (i), 2 2 r) 400(1 a − = ;From (ii) and (iii), we get 100 (1-r) (1+r) = 400 2 r) (1− 3/5. r 3 5r 4r 4 r 1 = ⇒ − ⇒ − = + ⇒ 177.a. 3 3 3 3 3 3 3 3 5 3 1 9 .... 4 3 2 1 + + = + + − + − 2 1 3 3 3 3 S S ) 8 ... 4 (2 ...9 − = + + + − + . For 1 6n 12n 8n 1) (2n t , S 2 3 3 n 1 − + − = − = ∑ ∑ ∑ ∑ ∑ − + − = = 1 n 2 3 n 1 6 n 12 n 8 t S n 2 1) 6n(n 6 1) 1)(2n 12n(n 4 1) (n 8n 2 2 − + + + + − + = -Here n = 5. Hence, = 1 S 2× 25× 36 -2× 5× 6× 11+3× 30-5 = 1800 -660 + 90 -5 =1890 -665 = 1225. For ∑ ∑ = = = 3 n 2 3 n 2 n 8 t S ; 8n t , S 4 1) (n 8n 2 2 + = 4) n 800.(for 25 16 2 = = × × = ∴ Required sum = 1225 -800 = 425. 178.a.Let and y, are the roots of the equations. 0 b ax x2 = + + and 0 bx x2 = + + a . a. y b, y and b a, = − = + = − = + ∴Given 2 2 (y -(-y -− = ⇒ = 4y(y 4(2 2 − + = − + ⇒ β 0 b) 4(a ) b (a 4a b 4b a 2 2 2 2 = − + − ⇒ − = − ⇒ b) a 0( 4 b a ≠ = + + ⇒ 179.c. 180.a.p + q = -p and pq = q ⇒ q (p -1) = 0 ⇒ q = 0 or p =1. If q = 0, then p = 0. i.e., p = q. ∴p = 1 and q = -2 181.a.ab + bc +ca = 1 2 1 c) b (a 2 < − + + 182.d. Required number of numbers = 5× 6× 6× 4 = 36× 20 = 720. 183.c.Required number of numbers = 3× 5× 5× 5 = 375 184.d.Required number are =5! + 5! + -4! = 216. 185.b. Required sum = (2 + 4 + 6 +.....+ 100) + (5 + 10 + 15+.....+ 100) -(10 + 20 +....+100) = 2550 + 1050 -530 = 3050. 186.a.we have q q q p 1 q p p q p 1 p x C t and x C t + + + + = = q q p p q p C C + + = . 187.c.we have n 2 = 4096 = 12 2 ⇒ n = 12; so middle term = ; t 7 924 6!6! 12! C t t 6 12 1 6 7 = = = = + . 188.c. 189.c. ; x C t 1 r 1 r 2n 2 r + + + = 1 -3r 1 -3r 2n 3r x C t = Given 1) (r -2n 2n 1 3r 2n 1 r 2n C ; C C + − + ⇒ = 4r 2n 1 3r 1 r 2n C 1 3r 2n − ⇒ − = − − ⇒ = − 190.c.we have 0 c bx b ax c bx c b b ax b a + + ++ By ); R (xR R R 2 1 3 3 + − → x) 2bx (ax 0 0 c bx c b b ax b a 2 + + − ++ = ( ) ( ) ve. ac) (b c) 2bx (ax 2 2 − = − + = − + + = 191.b. m m m 1 a 7 1 a Then 7 a Let 7. 7 a + = + < < = 14. 7 7 a 7 a m 2 1 m < + < + = ⇒ + 3 a n. 7 a So 7.; 14 a n n 1 m > ∴ ∀ < < < ⇒ + 192.a. 193.d.Equation of AB is x cos +y sin = p; 1 p/sin 1 cos p/1 1; p ysinp xcos= + ⇒ = + ⇒So co-rdinates of A and B are o , cosp and ; sinp 0, So coordinates of midpoint of AB are ( ) 1 1 1 x (let); y , x sin 2 p . cos 2 p = 2sinp y & 2cosp 1 = = cos⇒ = p/2x1 and sin = P/2y1; cos2 +sin2 =1 1 y1 x1 4 p 21 21 2 = + ⇒ Locus of (x1,y1) is 2 2 2 p4 y1 x1 = + + . 194.a. 195.a. 0 2 3a a 0 2 a 3a 2 2 = − + ⇒ = − + ; 2 17 3 2 8 9 3 a ± − = + ± − = ⇒ 196.c.Equation of circles 2 2 2 (1) 1 y x = = + 2mxy; x m x mx) (y y x 2 2 2 2 2 2 − = ⇒ − = + ⇒ 0 2mxy ) m (1 x 2 2 = + − ⇒ ; m 1 2m m 1 0 m 2 45 tan 2 2 2 − ± = − − ± = ⇒ 1-m2 = ± 2m 0 1 2m m2 = − ± ⇒ . 2 1 2 2 2 2 2 4 4 2 m ± − = ± − = + ± − = ⇒ 197.a.Let (h, k)be the centre of any such circle. Equation of such circle is (x -h)2 + (y -k)2 = 32 Since (h, k) lies on x2 +y2 =25, ∴ h2 + k2 = 25. 9; 25 2yk) (2xh y x 2 2 = + + − + Locus of (h, k) is 16 2 2 = + y x , which clearly satisfies (a). 198.b. 199.c.Let ABC be an equilateral triangle, whose median is AD. ) y , M (x 1 1 -Given AD =3a. In ∆ ABD, AB2 = AD2 +BD2; /4) (x 9a x 2 2 2 + = ⇒ where AB = BC = AC = x. 12a x 9a x 43 2 2 2 = ⇒ = 2. In ∆ OBD, OB2 = OD2 + BD2 ; 3a r 6ar 9a r 4 x r) (3a r 2 2 2 2 4 2 2 + + − = ⇒ + − = ⇒ 2a r 12a 6ar 2 = ⇒ = ⇒So equation of circle is 2 2 2 4a y x = + 200.b. Any tangent to the parabola y2 = 8ax is i) .........( m2a mx y + = If (i) is a tangent to the circle, 2 2 2 2a y x = + then, 1 m m 2a 2 2 + ± = a 2 ) m (1 m 2 2 = + ⇒0; 1) 2)(m ( 2 2 = − + ⇒ m 1. m ± = ⇒ So, from (i), y = 2a). (+ ± 201.a. ; c sb s a s r r r 3 2 1 − > − > − ⇒ > > c b a c; b a c s b s a s > > ⇒ − < − < − ⇒ − < − < − ⇒ 202.b. The given equation is tanx + secx = 2cos; ⇒ sinx + 1 = 2cos2x ⇒ sinx + 1 = 2(1 -sin2x); ⇒ 2sin2x + sinx -1 = 0; ⇒ (2sinx-1)(sinx +1) = 0 ⇒ sinx = 1.; , 21 − . 270 , 150 , 30 x 0 0 0 = ⇒ 203.b. 204.a.we have ; n n ... 2 1 lim 1 p p p p n + ∞ → + + +∫ + = + = = ∑ + = ∞ → 10 10 1 p p pp n1 r n 1 p1 1 px dx x .n nr lim 205.d.Since | x | limO x→ does not exist, hence the required limit does not exist. 206.a. 1 -x 1 -f(x) lim1 n→ 00 form Using L’ Hospital’s rule = 2. 12 f(1) (1) f' x 1/2 (x) f' f(x) 2 1 lim x = = = ∞ → 207.b. 208.d.∴f’ (x) -g”(x) = 0 Integrating, f’(x) -g (x) = c; 2 c c 2 4 c (1) g' (1) f' = ⇒ = − ⇒ = − ⇒∴f’(x) -g’(x) = 2; Integrating, f (x) -g (x) = 2x + c1 1 1 c 4 3 9 c 4 g(2) f(2) + = − ⇒ + = − ⇒ ; 2 2x (x) g (x) f 2 c1 + = − ∴ = ⇒ At x = 3/2, f (x) -g(x) = 3 + 2 = 5. 209.c.f (x + y) = f (x) × f(y) Differentiate with respect to x, treating y as constant f’(x + y) = f’ (x) f (y) Putting x = 0 any = x, we get f’(x) = f’(0) f(x); ⇒ f’(5) = 3 f’(5) = 3 × 2 = 6 210.a.Distance of origin from (x, y) = 2 2 y x + − − + = b at t cos 2ab b a 2 2 ab 2 b a 2 2 − + =b a 1 b at t cos max. − = = − 211.a.Let f(x) = ⇒ + + cx 2 bx 3 ax 2 3 f(0) = 0 and f(1) = 0. 6 6c 3b 2a c 2b 3a = + + = + + Also f (x) is continuous and differentiable in [0,1] and [0, 1]. So by Rolle’s theorem, f’(x) = 0 i.e ax2 + bx + c = 0 has at least one root in [0,1]. 212.d. we have ∫ = 20 ; 43 (x)dx f Now, ∫ ∫ ∫− = 20 20 20 f(x)dx (x)dx f' x (x)dx xf' ; 43 (2) 2f 43 f(x)] [x 20 − = − = . 43 0) f(2) ( 43 -0 − = = = 213.a. 214.b.we have, 6cos | b | | a | b . a = = . 3 4 23 2 4 = × × Now, ; b a ) b . a ( ) b a ( 2 2 2 2 = + × ( ) 4 16 48 b a 2 × = + × ⇒ ( ) 16 b a 2 = × ⇒ 215.a.we have, ( )( ) ( ) { }a c c b b a ] a c c b b a [ × × × × = × × × ( ) ( ) { } ) c b m (Where ) a c . m c a . m . ) b a ( × = − × ={ }{ } 16. 4 ] c b a [ ) c b ( . a ( . c ). b a ( 2 2 = = = × × = 216.a. a c b 0 c b a − = + ⇒ = + + 2 ) c b ( + ⇒ = 2 ) a (2 2 2 7 c . b 2 3 5 = + + = 15; 34 49 cos | c | | b | 2 = − = ⇒ 15; 3cos 5 2 = × × ⇒ -0 60 31/2; cos = = ⇒ = ⇒ 217.a.we have, 0 ) c b a ( 0 c b a 2 = + + ⇒ = + + 0 ) a . c c . b b . a 2( | c | | b | | a | 2 2 2 = + + + + + ⇒ 0 ) a . c c . b b . a 2( 9 16 25 = + + + + + ⇒ -25 ) a . c c . b b . a ( = + + ⇒ 25 | a . c c . b b . a | = + + ∴ 218.b. 219.b. we have c k 39 b a = = × Also 39; | c | , 45 | b | , 34 | a | = = = 39 : 45 : 34 | c | : | b | : | a | = ∴ 220.c. 221.a. B); P(A P(B) P(A) B) P(A ∩ − + = ∪ 41 P(B) ) A P( 1 43 − + − = ⇒ ; 32 P(B) P(B) 32 1 1 = ⇒ + − = ⇒Now, 12 5 41 -32 B) P(A P(B) B) A P( = = ∩ − = ∩ . 222.d. The event follows binomial distribution with n = 5, p= 3/6 = 1/2. q = 1 -p= 1/2,; variance = npq = 5/4. 223.b. Equation of plane through (1, 0, 0) is a (x -1) +by +cz = 0 ...........(i) (i) Passes through (0, 1, 0). -a + b = 0 ⇒ b = a ; Also, cos ) c 2(2a a a 45 2 2 0 + + = . 2a c c 2a c 2a 2a 2 2 2 2 = ⇒ = ⇒ + = ⇒So d.r of normal are a, a 2a i.e. 1, 1, 2 . 224.a.Let two forces be p and Q. Given P + Q = 18 and b ˆQ cˆ 12 aˆ P ; cˆ 12 b ˆQ aˆ P − = − ⇒ = + ; P) (18 Q 144 P 2 2 2 − = = + ⇒ 2 2 p 36P 324 144 P + − = + ⇒ 13. Q and 5 P 180 36P = = ⇒ = ⇒(where aand b are unit vectors along P and Q). 225.a. →b →a →c12 P -