The Equation Of The Tangent To The Curve X 2 2xy Y 2 2x Y 6 0 At 2 2 Is 1) 2x + y - 6 = 0 2) 2y + x - 6 = 0 3) x + 3y - 8 = 0 4) 3x + y - 8 = 0 5) x + y - 4 = 0 Solution: (1) 2x + y - 6 = 0 y = x2 -... View Article
The Equation Of The Tangent To The Curve Y 4e X 4 At The Point Where The Curves Crosses Y Axis Is Equal To 1) 3x + 4y = 16 2) 4x + y = 4 3) x + y = 4 4) 4x - 3y = -12 5) x - y = -4 Solution: (3) x + y = 4 Since the curve crosses the... View Article
If The Curves X 2 A 2 Y 2 16 1 And Y 3 8x Interest At Right Angle Then The Value Of A 2 Is 1) 16 2) 12 3) 8 4) 4 5) 2 Solution: (4) 4 (x2 / a2) + (y2 / b2) = 1 (2x / a2) + (2yy’ / 12) = 0 y’ = -12x / a2y = m1 y3 =... View Article
The Equation Of The Tangent To The Curve Y 4 X E X At 1 4 E Is 1) y = - 1 2) - (4/e) 3) x = - 1 4) x = -4/e Solution: (2) - (4 / e) m = dy / dx m = d (4xex) / dx = 4 [xex + ex] m(-1, -4 / e)=... View Article
The Shortest Distance Between The Line Y X 1 And The X Y 2 2 Is 1) 3√2 / 8 2) 2√3 / 8 3) 3√2 / 5 4) √3 / 4 Solution: (1) 3v2 / 8 y2 = x y2 = 4ax a = 1 / 4 Points will be (t2 / 4, t / 2) yy1... View Article
Let P X X 4 A X 3 B X 2 C X D Such That X 0 Is The Only Real Root Of P Dash X Equal To 0 1) P(-1) is the minimum and P(1) is the maximum of P 2) P(-1) is not minimum but P(1) is the maximum of P 3) P(-1) is minimum and P(1) is... View Article
The Minimum Value Of F X E X 4 X 3 X 2 Is 1) e 2) -e 3) 1 4) -1 Solution: (3) 1 f (x) = e(x^4 - x^3 + x^2) f’ (x) = e(x^4 - x^3 + x^2) (4x3 - 3x2 + 2x) f’ (x) = 0... View Article
If The Line Ax By C 0 Is A Tangent To The Curve Xy 4 Then 1) a < 0, b > 0 2) a = 0, b > 0 3) a < 0, b < 0 4) a = 0, b < 0 Solution: (3) a < 0, b < 0 xy = 4 y = 4 /... View Article
Let F R R Be Defined By F X If F Has A Local Minimum At X Equals 1 Then A Possible Value Of K Is 1) 1 2) 0 3) -(1/2) 4) -1 Solution: (4) -1 f (x) = lim x=-1- f (x) = lim x=-1+ f (x) LHL = RHL lim x=-1- (k - 2x) = lim... View Article
For Any Integer N Greater Than 1 The Sum Sigma N Eq 1 To N K K Plus 2 Is Equal To For any integer n = 1, the sum Sk=1n k(k+2) is equal to (1) n(n+1)(n+2)/2 (2) n(n+1)(2n+1)/6 (3) n(n+1)(2n+7)/6 (4) n(n+1)(2n+9)/6... View Article
If A A1 A2 A2n B Are In Arithmetic Progression And A G1 G2 G2n B Are In Geometric Progression And Is The Harmonic Mean Of A And B Then (1) 2nh (2) n/h (3) nh (4) 2n/h Solution: Given a, a1, a2, ,...,a2n, b are in arithmetic progression. a1, a2, ,...,a2n are AMs... View Article
If Am And Hm Between Two Numbers Are 27 And 12 Respectively Then Their Gm Is (1) 9 (2) 18 (3) 24 (4) 36 Solution: Given AM = 27 HM = 12 We know GM2 = AM × HM = 27×12 = 324 GM = v324 = 18 Hence option... View Article
If A1 A2 A3 An Are In Hp Then The Expression A1a2 Plus A2a3 Plus An 1an Is Equal To (1) (n - 1) (a1 - an) (2) na1an (3) (n - 1)a1an (4) n(a1 - an) Solution: Given a1, a2, a3,...,an are in HP. So 1/a1, 1/a2, 1/a3... View Article
If The 7th Term Of Hp Is 1 By 10 And The 12th Term Is 1 By 25 Then The 20th Term Is (1) 1/41 (2) 1/45 (3) 1/49 (4) 1/37 Solution: Given 7th term of HP = 1/10 12th term = 1/25 7th term of AP = 10 12th term of AP... View Article
If A1 A2 A3 Are In A Harmonic Progression With A1 Eq 5 And A20 Eq 25 The Least Positive Integer N For Which An Less Than 0 Is (1) 22 (2) 23 (3) 24 (4) 25 Solution: Given a1, a2, a3,...are in HP. So 1/a1, 1/a2, 1/a3 are in AP. Let d be the common... View Article
In A Geometric Progression Consisting Of Positive Terms Each Term Equals The Sum Of The Next Two Terms Then The Common Ratio Of This Progression Equals (1) 1/2(1 - v5) (2) 1/2(v5) (3) v5 (4) 1/2(v5 - 1) Solution: Let a, ar, ar2 be the terms of the GP. Given a = ar+ar2 1 = r+r2... View Article
If Sum Of The Series Sigma N Eq 0 To Infinity Rn Eq S For R Less Than 1 Then The Sum Of The Series (1) S2 (2) S2/(2S + 1) (3) 2S/(S2 - 1) (4) S2/(2S - 1) Solution: S = Sn=08 rn = 1+r+r2+...8 = 1/(1-r) r = (S-1)/S Sn=08 r2n... View Article
A Man Saves Rs 200 In Each Of The First Three Months Of His Service In Each Of The Subsequent Months (1) 19 months (2) 20 months (3) 21 months (4) 18 months Solution: Saving for first 3 months = 3×200 = 600 Let the time taken to... View Article
Let A1 A2 A3 A100 Be An Arithmetic Progression With A1 Eq 3 And Sp Eq Sigma 1 To P For Any Integer N With 1 Let M Eq 5n If Sm By Sn Does Not Depend On N Then A2 Is (1) 9 (2) 2 or 4 (3) 4 or 16 (4) None of these Solution: Sum of n terms of AP = (n/2)(2a+(n-1)d Given m = 5n a = 3 Sm/Sn =... View Article
In A Triangle The Lengths Of Two Larger Sides Are 10 Cm And 9 Cm If The Angles Of The Triangle Are In Ap Then The Length Of The Third Side Is (1) v5 - v6 (2) v5 + v6 (3) v5 ± v6 (4) 5 ± v6 Solution: Let x-d, x, x+d be the 3 angles of triangle ABC in AP. Since sum of... View Article
