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1 L.K.Gupta (Mathematic Classes) MOBILE: , {JEE Mai 04} Sept 0 Name: Batch (Day) Phoe No. IT IS NOT ENOUGH TO HAVE A GOOD MIND, THE MAIN THING IS TO USE IT WELL Marks: 00. If f(x) = - x, the f f x is : (a) x [Each right aswer carries 4 marks ad wrog ] (b) x f f f f( x) x ( / x) x x x. x x Time: hr. PIONEER EDUCATION (THE BEST WAY TO SUCCESS): S.C.O. 0, SECTOR 40 D, CHANDIGARH (c) x x (d) x. The relatio R defied i A= {,, } by a R b, if a b 5. Which of the followig is false? (a) R = {(, ), (, ), (, ), (, ), (, ), (, ), (, )} (b) R - = R (c) Domai of R = {,, } (d) Rage of R = {5} R = {(, ), (, ), (, ), (, ), (, ), (, ), (, )} R - = {(y, x ): (x, y)r} = {(, ), (, ), (, ), (, ), (, ), (, ), (, )} = R. Domai of R = {x : (x, y)r} = {,, }. Rage of R = {y : (x, y)r} = {,, }.. If umber of elemets i sets A ad B are m ad respectively, the the umber of

2 L.K.Gupta (Mathematic Classes) MOBILE: , 4677 relatios from A to B is: (a) m+ (b) m (c) m + (d) m o (A B) = o (a) o (b) = m o [P (AB)] = m umber of relatios form A to B is m. 4. Give f (x) = log x x ad g (x) = x x x PIONEER EDUCATION (THE BEST WAY TO SUCCESS): S.C.O. 0, SECTOR 40 D, CHANDIGARH, the (fog) (x) equals : (a) f (x) (b) f (x) (c) [f(x)] (d) oe of these x x (fog) (x) = f(g(x)) = f x = log x x x x x x ( x) log ( x) f (x) 5. Which oe of the followig is ot a fuctio? (a) {(x, y}: x, y, R, x = y} (b) {(x, y}: x, y, R, y = x} (c) {(x, y}: x, y, R, x = y} (d) {(x, y}: x, y, R, y = x} Let R = {(x, y}: x, y, R, y = x}, R is ot a fuctio; as y = x 6. If f (x) = cos (log x), the x f (x ) f(y ) - f f (x y ) has the value : y (a) - (b) - (c)/ (d) oe of these

3 L.K.Gupta (Mathematic Classes) MOBILE: , 4677 f (x ) f(y ) - x y f f (x y ) = cos (log x ) cos (log y ) x y cos log cos log(x y ) = cos (log x ) cos (log y ) - [cos (log x log y )] + cos (log x + log y ) = cos (log x ) cos (log y ) - [ cos (log x ) cos (log y )] = 0 = 7. Let A = {,,, 4} ad R be a relatio i A give by R = {(, ), (, ), (, ), (4, 4), (, ), (, ), (, ), (, )}. The R is : (a) reflexive (c) trasitive PIONEER EDUCATION (THE BEST WAY TO SUCCESS): S.C.O. 0, SECTOR 40 D, CHANDIGARH (b) symmetric (d) a equivalece relatio R = {(, ), (, ), (, ), (4, 4), (, ), (, ), (, ), (, )} As (, ), (, ), (, ), (4, 4) R, R is Reflexive. (, ) ad (, ) R, also (, ) & (, ) R, R is symmetric. Also (, ) & (, ) R (, ) R Similarly for other elemets hece R is trasitive. R is equivalet relatio. 8. Let X be the uiversal set for sets A ad B. If (a) = 00, (b) = 00 ad (A B) = 00, the (A B ) is equal to 00 provided (X) is equal to : (a) 600 (b) 700 (c) 800 (d) 900

4 L.K.Gupta (Mathematic Classes) MOBILE: , 4677 We have (A B) = (a) + (b) (A B). (A B) = = 400 Also (A B ) = ((A B) = (X) (A B) 00 = (X) or (X) = Suppose A, A,.., A0 are thirty sets each with five elemets ad B, B,., B are sets each with three elemets. Let 0 Ai Bj S. Assume that each elemets of S belogs i j to exactly 0 of the Ai s ad exactly 9 of Bj s. The value of must be: (a) 0 (b) 40 (c) 45 (d) 50 Sice S = = 5. Also S = we have, (S) = 0 Ai ad each elemet of S is i 0 Ai s, we have, (s) = i Bj ad each elemet of S is i 9 Bj s, j (B j) 9 j 5 = ( ) = (A i ) i (0 5) 0 0 PIONEER EDUCATION (THE BEST WAY TO SUCCESS): S.C.O. 0, SECTOR 40 D, CHANDIGARH 4

5 L.K.Gupta (Mathematic Classes) MOBILE: , The void relatio i a set A is: (a) reflexive (c) reflexive ad trasitive Let R deotes the void relatio. Let x A. (x, x) R because R = ϕ. R is ot reflexive. (b) reflexive ad symmetric The relatio R is trivially symmetric ad trasitive (d) trivially symmetric ad trasitive. If ta θ = ta ϕ +, the cos θ+ si ϕ is equal to (a) (b) (c) - (d) oe of these We have, cos θ = ta θ ta θ (ta ) ta ta ta ta sec = - si ϕ Hece (d) is correct aswer.. The value of si x + cos x is cos θ si 0. ta θ ta (a) (b) (c) (d) PIONEER EDUCATION (THE BEST WAY TO SUCCESS): S.C.O. 0, SECTOR 40 D, CHANDIGARH 5

6 L.K.Gupta (Mathematic Classes) MOBILE: , 4677 Sice AM of two positive quatities their GM. six cos x si x cosx. π cos x sixcosx 4 si x + cos x =. Hece (b) is the correct aswer. If a cos θ+ b si θ = c has α ad β as its solutios, the ta α + ta β ad ta α ta β is equal to (a) b c a, c a c a (b) c a, c a b c a We have a cos θ + b si θ = c a (cos θ - si θ) + b si θ cos θ = c PIONEER EDUCATION (THE BEST WAY TO SUCCESS): S.C.O. 0, SECTOR 40 D, CHANDIGARH 6 (c) a ( ta θ) + b ta θ = c sec θ = c ( + ta θ) ta θ (c+a) b taθ + c a = 0. This equatio has ta α ad ta β as its roots. ta α + ta β = b, ta α ta β = (c a) c a c a b c a, c a c a Hece (a) is the correct aswer. π 4π 4. If x = y cos z cos, the xy + yz + zx is equal to : (a) (b) 0 (c) (d) (d) b c a, c a c a

7 L.K.Gupta (Mathematic Classes) MOBILE: , 4677 x = y cos π = z cos 4 π x y z y = z = -x xy + yz + zx = -x + 4x x = 0 5. The miimum value of 7 cos x + 8 si x is equal to : (a) (b) 7 cos x + 8 si x = cos x + 4 si x. =. / ( cosx + 4 si x). /(-5) =. -5/ =. -. -/ = 9 cos x 4si x. Thus miimum value of give expressio is 9 PIONEER EDUCATION (THE BEST WAY TO SUCCESS): S.C.O. 0, SECTOR 40 D, CHANDIGARH 7 (c) 9 (d) oe of these

8 L.K.Gupta (Mathematic Classes) MOBILE: , cos x. si x = (a) = 5, a = ar. si (rx), x R, the: x0 (b) = 5, a = 4 cos x. six = cos x. cos x. si x = cosx si x cos x = 4 ( cosx) (six + six) six si x (six.cosx) (cosx.si x) 4 six si x (si5x si x) (six si x) 4 si x six si5x 4 (c) = 5, a = 8 a =,a, If x + y = π + z, the si x + si y - si z is equal to : (a) si x. si y. si z.siy. cos z si x + si y - si z = si x + si (y + z). si (y - z) (d) = 5, a = 4 (b) cos x. cos y. cos z (c) si x. cos y. cos z (d) si x = si x + si (y + z). si (π - x) PIONEER EDUCATION (THE BEST WAY TO SUCCESS): S.C.O. 0, SECTOR 40 D, CHANDIGARH 8

9 L.K.Gupta (Mathematic Classes) MOBILE: , 4677 = si x {si x + si (y+z)} = si x {si (π +z-y) + si (y+z)} = si x{si (y+z) si (z-y)} = si x {si (y + z)-si(z-y)} = si x.cos z.si y 8. A quadratic equatio whose roots are cosec θ ad sec θ, ca be : (a) x x + = 0 (b) x x + = 0 (c) x 5x + 5 = 0 (d) oe of these 4 sec θ + cosec θ = 4 cos θ si θ si θ 4 Also, sec θ. cosec θ = 4 si.θ Thus, required quadratic equatio will be x tx + t = 0, where t 4. Hece x 5x + 5 = 0 ca be the solutio si x cos x sixsi x 9. The equatio = 8 is satisfied for the values of x give by (a) cos x = 0, ta x = - (b) ta x = 0 (c) ta x = (d) oe of these The give equatio is or si x cos x sixsi x si x cos x (sixcos x) 7 y + y = 8, where y = 8 = 8 sixcos x y 8y + 7 = 0 PIONEER EDUCATION (THE BEST WAY TO SUCCESS): S.C.O. 0, SECTOR 40 D, CHANDIGARH 9

10 L.K.Gupta (Mathematic Classes) MOBILE: , 4677 y = 7 or If y = 7, the sixcos x si x + cos x = = si x + cosx = si x = cos x = which is ot possible for ay value of x ad so y 7. also we have y =. sixcos x = = o si x + cos x = 0 cos x (si x + cos x) = 0 Either cos x = 0 or ta x = Hece (a) is the correct aswer. 0. A triagle ABC is such that si (A + B) =. If A, B, C are i A.P., the the values of A, B, C are (a) π, π, 5π 4 si (A + B) = = si (b) π 6 π π π,, 6 π A + B = π + (-) 6 Also A+B+C = π ad B = A + C B = π B = π () (c) π, π, π 4 4 5π π 5π From (), for =, A + B = A C. 6 4 Hece (a) is the correct aswer. (d) oe of these PIONEER EDUCATION (THE BEST WAY TO SUCCESS): S.C.O. 0, SECTOR 40 D, CHANDIGARH 0

11 L.K.Gupta (Mathematic Classes) MOBILE: , The umber of values of x for which si x + cos 4x = is : (a) 0 (b) (c) (d) ifiite x + cos 4x = si x =, cos 4x = si x = or = (absurd). Thus o solutio.. The umber of solutio of cos x = si x, 0 x 4 π is : (a) 8 (b) 4 (c) (d) oe of these Clearly form graph four solutio.. If x, y [0, π], the total umber of ordered pairs (x, y) satisfyig si x. cos y = is equal to : (a) (b) (c) 5 (d) 7 si x. cos y = si x =, cos y = or si x = -, cos y = - If si x =, cos y = x = π, y = 0, π PIONEER EDUCATION (THE BEST WAY TO SUCCESS): S.C.O. 0, SECTOR 40 D, CHANDIGARH

12 L.K.Gupta (Mathematic Classes) MOBILE: , 4677 If si x = -, cos y = - Thus possible ordered pairs are π π π,0,, π,, π. x = π, y = π 4. If cos x = cos y, where x, y (0, π), the ta x. cot y is equal to cos y (a) (b) (c) cos x = cos y cosy ta x ( ta y /) ta y / x ta y / ta ta y / y x 6ta ta x y ta.cot Hece (b) is correct aswer. PIONEER EDUCATION (THE BEST WAY TO SUCCESS): S.C.O. 0, SECTOR 40 D, CHANDIGARH (d)

13 L.K.Gupta (Mathematic Classes) MOBILE: , If si x + si x =. The the value of cos x + cos 0 x + cos 8 x + cos 6 x is equal to (a) (b) - (c) (d) - I = cos x + cos 0 x + cos 8 x + cos 6 x = cos x +. cos 8 x (cos x + ) + cos 6 x = (cos 4 x) + cos 6 x(cos 4 x + cos x) + (cos x) = (cos 4 x + cos x) = (cos x (cos x + )) We have si x = si x = cos x 5 si x cos x 5 5 I Hece (a) is correct aswer. 6. If x - 0 ad x x 0, the x lies i the iterval set: (a) (-, ) (b) (-, ) (c) (, ) (d) {-} x - 0 ad x x 0 PIONEER EDUCATION (THE BEST WAY TO SUCCESS): S.C.O. 0, SECTOR 40 D, CHANDIGARH

14 L.K.Gupta (Mathematic Classes) MOBILE: , 4677 x ad (x - ) (x + ) 0 x ad x (, ] [, ) {- }. 7. Solutio of x + < is : (a) [-, -/] (b) {-/, -} (c) (-, -/) (d) oe of these x + < - < x + < - < x < - - < x < - 8. Solutio of x - < x + is : (a) ( -, /) (b) (/, 5) (c) (5, ) (d) (-,/) (5, ) x - < x + squarig both sides, (x - ) (x + ) < 0 (x - 5) (x - ) < 0 x, Solutio of (x - ) (x + 4) < 0 is : (a) (, ) (b) (-, 4) (c) (, 4) (d) (, 4) Usig umber lie Rule x (, 4) 0. Solutio of x is : (a) (, 4) (b) (-4, -) (c) ( 4, ) (, 4) (d) [-4, -] [, 4] PIONEER EDUCATION (THE BEST WAY TO SUCCESS): S.C.O. 0, SECTOR 40 D, CHANDIGARH 4

15 L.K.Gupta (Mathematic Classes) MOBILE: , 4677 x x x 6 x 4. either x 4 or 4 x x [ 4, ] [, 4].. If x+ = 6 log, the x is : (a) (b) (c) log (d) log x+ = log log = log x log.. Solutio of x < 4 is : x (a) (-, + )(--, -+ ) (b) R ( -, + ) (c) R (, -+ ) (d) oe of these x (clearly x 0) x x x ( x + > 0) x x x + > x x x + > 0 ( x - ) > 0 PIONEER EDUCATION (THE BEST WAY TO SUCCESS): S.C.O. 0, SECTOR 40 D, CHANDIGARH 5

16 L.K.Gupta (Mathematic Classes) MOBILE: , 4677 x x -, x R {, 0, }.. If ( ) x + ( ) x = ( ) x/, the the umber of values of x is : (a) (b) 4 (c) (d) oe of these x/ + x/ = x/ x/ x/. x which of the form cos x/ α + si x/ α =.. 4. The product of three positive real s is ad their sum is greater tha sum of their reciprocals. Exactly oe of them is greater tha (a) 0 (b) (c) - (d) - Let three positive reals be a, b ad b. We are give a + b + a ab. () ab a b Now (a - ) (b - ) ab PIONEER EDUCATION (THE BEST WAY TO SUCCESS): S.C.O. 0, SECTOR 40 D, CHANDIGARH 6

17 L.K.Gupta (Mathematic Classes) MOBILE: , 4677 = + = a b ab ab a b a b ab 0, usig () ab a b either all three a, b ad are positive or exactly oe of them is positive. ab But a >, b > ad ab Thus, exactly oe of a, b, exceed. ab Hece (b) is correct aswer. 5. If log x + log y = + log ad log (x + y) = the (a) x =, y = 8 (b) x = 8, y = (c) x =, y = 6 (d) x = 9 y = (c) The first equatio ca be writte as log xy = log + log = log 8 so xy = 8 ad secod equatio is x + y = = 9. Solvig we get x =, y = 6 or x = 6, y =. 6. If a = log 8, b = log4 54 the the value of ab + 5(a - b) is (a) 0 (b) 4 (c) (d) oe of these (c) We have log 8 log a = log 8 = ad log log log54 log b = log4 54 = log 4 log Puttig x = log, we have 7. The value of x satisfyig log (x - ) = log/ x is (a) / (b) (c) / (d) PIONEER EDUCATION (THE BEST WAY TO SUCCESS): S.C.O. 0, SECTOR 40 D, CHANDIGARH 7

18 L.K.Gupta (Mathematic Classes) MOBILE: , 4677 (d) log x log (x - ) = log/ x = log = log x- x = x - x x = x = or x = - /. But log (x - ) ad log/ x are meaigful if x > /. Hece =. 8. If log x log x = log log 6 the x equals (a) 9 (b) (c) 4 (d) 5 (a) Clearly x > 0. The give equatio ca be writte as log x log log x = log log log log x = log x = The umber of solutios of log4 (x - ) = log (x - ) is (a) (b) (c) (d) 0 (b) The give equality is meaigful if x > 0, x > 0, x, ad x x > ad x 4. The give equality ca be writte as log(x ) log (x ) log 4 log log (x - ) = log (x - ) (x - ) = (x ) x 7x + 0 = 0 (x - 5) (x - ) = 0 x = 5 or. But x > so x = 5. PIONEER EDUCATION (THE BEST WAY TO SUCCESS): S.C.O. 0, SECTOR 40 D, CHANDIGARH 8

19 L.K.Gupta (Mathematic Classes) MOBILE: , If log0.5 (x - ) < log0.5 (x - ), the x lies i the iterval. (a) (, ) (b) (, ) (c) (-, 0) (d) (0, ) (a) log0.5 (x - ) < log0.5 (x ) log (x ) log (x ) 0.5 (0.5) log (x) log (x ) log0.5 (x - ) log (0.5) log0.5 (x - ) < 0 x > x (, ) 4. If α + β = π/ ad β+ γ = α, the ta α equals (a) (ta β + ta γ ) (b) ta β + ta γ (c) ta β + ta γ (d) ta β + ta γ (c) α + β = π/ α = π/ - β ta α ta β = Next, β + γ = α ta α = cot β PIONEER EDUCATION (THE BEST WAY TO SUCCESS): S.C.O. 0, SECTOR 40 D, CHANDIGARH 9

20 L.K.Gupta (Mathematic Classes) MOBILE: , 4677 taα taβ α β γ ta γ taα taβ taα taβ ta γ taα taβ ta γ 4. If A ad B are acute agles such that si A = si B, cos A = cos B; the (a) A = π/6 (b) A = π/ (c) B = π/4 (d) B = π/ (a) ad (c) From the give coditios ( si A) = ( si B) = ( - sia) si A si A + = 0 (sia ) (si A ) = 0 si A = or si A = ½ A = π/ or π/6 But sice a is acute, we have A = π/6 si B = si (π/6) = ½ si B = / B = π/4 4. If a a cos x + = 674 abd ta (x/) = 7 the the itegral value of a is (a) 5 (b) 49 (c) 67 (d) 74 (a) 674 = a - a = a a ta (x /) ta (x /) = a +a 50 5a + 48a 67 5 = 0 PIONEER EDUCATION (THE BEST WAY TO SUCCESS): S.C.O. 0, SECTOR 40 D, CHANDIGARH 0

21 L.K.Gupta (Mathematic Classes) MOBILE: , 4677 (a - 5) (5a -67) = 0 a = 5 (Takig the itegral value of a). 44. si x cos x = si x if (a) x = π + π/6 (b) x = π - π/6 (c) x = π (d) x = π + π/ (I) (a), (b) ad (c) The give equatio ca be writte as si x (cos x - ) = 0. That is, either si x = 0 or cos x =/ = cos (π/). Hece x = π or x =π ±(π/), i.e. x = π ± π/6 45. Let X= {,,, 4, 5} ad Y = {,, 5, 7, 9}, which of the followig is ot relatio from X to Y? (a) R = {(x, y) : y = x +, x X, y Y} (b) R = {(, ), (, ), (, ), (4, ), (5, 5)} (c) R = {(, ), (, ), (, 5), (, 7), (5, 7)}(d) = R4 = {(, ), (, 5), (, 4), (7, 9)} R is a relatio from X to Y because R X Y. R is a relatio from X to Y because R X Y. R is a relatio from X to Y because R X Y. R4 is ot a relatio from X to Y because (, 4), (7, 9) X Y. 46. If (a) = 4 ad (b) = 7, the the miimum ad maximum value of (A B) are respectively : (a) 4, (b) 4, 7 (c) 7, (d) oe of these (A B) is miimum whe A B. I this case A B = B ad we have (A B) = (a) 4. (A B) is maximum whe A B = ϕ. I this case (A B) = =. = (a) + (B) PIONEER EDUCATION (THE BEST WAY TO SUCCESS): S.C.O. 0, SECTOR 40 D, CHANDIGARH

22 L.K.Gupta (Mathematic Classes) MOBILE: , x If ta α ad ta β x x is : (a) 0 (b) π/4 (c) π/ (d) π/ x taα taβ we have, ta α β x x taα taβ x. x x = x(x ) (x ) x x (x )(x ) x x x = = ta π 4, π α β 4 π 4π 5π 48. The value of cos cos cos is : (a) / (b) /4 (c) /8 (d) /8 We have π 4 π 5π π π 4π cos cos cos cos cos cos π π π as, cos cos π cos π π π 4π si cos cos cos π si π 4π 8π si cos si π 7 7 π si 8si π π si π si. π 7 π 8si 8si PIONEER EDUCATION (THE BEST WAY TO SUCCESS): S.C.O. 0, SECTOR 40 D, CHANDIGARH

23 L.K.Gupta (Mathematic Classes) MOBILE: , 4677 π 49. The value of cos8θ, where 0 < θ, is equal to 8 (a) cos θ (b) cos θ (c) si θ (d) - cos θ We have, cos 8θ (4cos 4θ) cos4θ ( cos4θ) (cos θ) cosθ ( cosθ) = (cos θ) cosθ. 8θ cos8θcos cos4θ cos θ 50. The value of cos A cos A cos A cos A. cos - A is equal to PIONEER EDUCATION (THE BEST WAY TO SUCCESS): S.C.O. 0, SECTOR 40 D, CHANDIGARH

24 L.K.Gupta (Mathematic Classes) MOBILE: , 4677 (a) cos A cos A We have, si A (b) si A cosa cosa cos A cos A.. cos - A = si A (c) cos A si A [( si A cos A) cos A cos A cos A. cos - A] [( si A cos A) cos A cos A.. cos - A] si A [( si Acos A) cos A cos A cos - A] si A [si(a). cos A cos A cos - A] si A [( si A cos A) cos A cos - A] si A [si (. A)cos A cos - A] si A [(si A cos A cos 4 A cos - A)] si A =. [si - A cos - A] si A [ si - A cos - A] si A si (. - A) si A PIONEER EDUCATION (THE BEST WAY TO SUCCESS): S.C.O. 0, SECTOR 40 D, CHANDIGARH 4 (d) cos A si A

25 L.K.Gupta (Mathematic Classes) MOBILE: , 4677 Aswers GOOD LUCK PIONEER EDUCATION (THE BEST WAY TO SUCCESS): S.C.O. 0, SECTOR 40 D, CHANDIGARH 5

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