Looking for a Tutor Near You?

Post Learning Requirement »
x

Choose Country Code

x

Direction

x

Ask a Question

x

x
x
For better view click here
x
Hire a Tutor
DISCOVER

Notes On JEE Mains Mathematics Solution

Loading...

Published in: Mathematics
870 Views

In this Note I have made solution of previous paper of JEE mains, mainly 7 April 2019 Shift 1. I have provided detailed solution for each question of Mathematics part. Please see the note very carefully, it will benefit every student who are aspiring for JEE.

Samir C / Kolkata

7 years of teaching experience

Qualification: M.Sc (IISc, Bangalore - 2015)

Teaches: Algebra, KVPY Exam, Mathematics, B.Sc Tuition, M.Sc Tuition, IIT JAM, IIT JEE Advanced, WBJEE

Contact this Tutor
  1. 7 April, 2019 Shift 1 JEE Solution Samir Ch Mandal QI. Let R be a relation on Z X Z by (a, b)R(c, d) a —d = b — c, where Z is the set of all integers, then R is A. Reflexive but neither symmetric nor transitive. B. Symmetric but neither reflexive nor transitive C. Transitive but neither reflexive nor symmetri D. Symmetric, transitive but not reflexive. Solution: Checking of Reflexive: 1 — 2 * 2 — 1 we see that (1,2) ( , 2). So, R is not reflexive. Checking of Symmetry: Let (a, b R(c, d). So we ave a For symmetry we need (c, d)R a, b or c.b = d — a which is certainly true. So R is Symmetric. Checking of ransitivit',c et (a, b) (c, d) and (c, f). Therefore we have : a —d = W— c ; & c —f = d — e Our target: a — — e it is not true. For example, (1,2)R(3,2) and (3,2)R( 6) but (1,2) is not related to (5,6). So R is not Transitive. Hence the correct option is B. Q2. Let —1) be any complex number such that Izl = 1 and arg(z) Then the imaginary part of the is z(l + i) sin 9
  2. Solution: z = eL9. So z(1+Z) 2. 4. cos 29 — isin 29 cos 9 cos 9 sin 9 Z(l—z) e-19-1 2 + 2 COS 9 19+1) i9+1) -1 sin 29 Therefore the required imaginary part is e-219-1 2 sin 9.cos9 9 4 cos2— 2 4 sing cos— cos 9 2 2 9 4 cos2— 2 So the correct option is 3. 2(I+COS 9) cos 9. Q3. The set of all real values of a for hic the equation, Ix—2 • x Has real solutions for x. 1. (-00, o] u [2, 00) 3. [-1 -us, 1 Glo[ u [2,1+VS] Solution• Clearl we have so ution for x, if — 2) 2 0. Because then 2 2 = 2 + — 2) has real solution for x. x or x Hence correct optionas 1. Q4. Let A bea 2 2 matrix such that A2 + A +1 = 0, where I = 12. Then ladj( (I — A) 6) | is equal to 34 2. 1. 33 Solution: unity, (D. 3. 36 39 4. As we do in algebra we have A3 = I, compare A with cube root of Now I and A commutes with each other so we can apply binomial theorem. Therefore
  3. = 1 - 6CIA + 6C2A2 - 6C3A3 + 6C4A4 - 6C5A5 + A6 = 1 - 6A + 15A2 - 201 + 15A - 6A2 +1 = —181 + 914 -k 9A2 = -181 - 91 = -271 Now as A isa 2 x 2 matrix, it is easy to see adj(—271) = -271 Therefore the required determinant —27 • —27 = 36. So the correct option is 3. Q5. If the system of linear equations 2x + 7 y — 4z = a —x — 5y + 5z — ß Has infinitely many solutions, then the ordered paw a, 73 2. Solution: This is a rank problem Let A = 1 4 7 ) cannot take value: (-3,3) 4. -23 —4 and B = 5 1 4 2 7 =rank(B)=2. —4 a then for in inite solution we must have rank(A) Then we must find a, b such that a 1,4, —3,2) + b(2,7, —4, a) = (—1, —5,5,ß) a + 2b Therefore we need 'o olve for and we get 4 a + 7b Therefore equating the last component we should have —6 + a = ß or a — ß = 6. Therefore theuast option, 4 is the correct one. Q6. Let A be the,set of all 3-digit natural numbers and B = {x e A: H.C.F. (x, 15) Then the number of elements of B is 1. 420 2. 360 3. 480 240 Solution: Clearly n(A) = 999 — 99 = 900. Let C and D are the sets of multiples of 3 and multiples of 5 respectively, then we want n(A — C U D).
  4. Solution: 98/12 gives remainder 2. So 985/12 gives remainder as in 25/12 or in 32/12 which is 8. So we see r = 8. C) 16.15.14 the coefficient of x 3 is 16C3 • = 70. 6.8 So the correct option is 4. Q8. In an increasing geometri eries, the sum of the first and the sixth term is 66 and the product of the second and the fifth terms is 128. Then the sum of the first 6 terms of the series js • 300, where [ ] is the box function. n(D) 180 and — 60. So we have n(C U D) = n(C) + n(D) — n(C n D) 300 + 180 — 60 = 420. Hence the required number is 900 — 420 = 480. So the correct option is 3. Q7. If r is the remainder obtained on dividing (98)5 by 12, then the coefficient of x 3 in the binomial expansion of 1 + is : is : 55 1. 2 91 2. 2 102 3. 70 4. 1. 127 2. 129 3. 126 4. 128 Solution: We have a ar 4 an ar. ar a2r5 = 128. Therefore 66. ora2 - + 128 = O eliminating r we geta Solving we get, a = 2,64. Certainly 64 would be too large to get the sum of 128 5 first nd sixth term as 66. So we must take a = 2. Therefore r = 32 = r 6—1 25 . r = 2. We needto find a. r—l Hence the correct option is 3. Q9. The sum of the infinite series I + 2 4 26-1 = 126. 2-1 10 14 9 4. 2 Solution: This is a Arithmetic-Geometric series. The general term is given by tn , we have ignored the first two terms. —1 Now we know (1 — x)
  5. Diff w.r.t. x, we get (I-x)2 Multiplying by x, we get (I—x)2 = x + + 3 x 3 + 4x4 + . n=1 Putting x - we get So we have n=1 4.3 = 3-1=2 Hence the total sum is 5. So the correct option is 2. Q10. lim is equal to: x-.»3 (x—3) cos(x—3) •vTE6-3 _ Solution: We see x-3 Now separating the limits we see that required limit is -. Therefore the correct option IS 4 QII. Let f be a continuous unction defined by a, in 2X. b cosx (x) 2b cosx Then the value of a + b is : 3 Solution: We know for continuity f (x) = f (x) = f Now, lim 2b cos y+ lim 2b cosx f (x) = lirp . Writing x — y, we have lim f (x) 2b sin y — lim = 2b = 4 This implies that b = 2.
  6. b sin y 2a sin 2y = lim — lim 2y = 2a—2 Therefore, 2a — 2 = 4 a = 3. Hence the correct option is 3. Q12. Let y be an implicit function of x defined by On the other hand, a sin 2x — b cosx lim f (x) = lim x If y(0) = —1, then = lim = lim asin(2y + IT) —b cos y + 2 —a sin 2y + b sin y 1 1. 2 1 1 at x dx 1 2. 2 2 2 1 + 12 y 3. 4 4 4. 5 Solution: Calculating the determinant,v e get 2 — 4 ) 2 (x -k y — 2) -k (2 — x — y) -k 12 y xy y —.94) — 3(x + y— 2) + 12y = 0 Differentiating .r. . , we o tain, (1 + y l) + 12 y ' = O Now putting 0, y = —1, we have (—1)(—2 3 (1 -k y 1) + 12 y ' = O Simplifying we get (1 + y')(2 — 3 — 3) + 12y' So the correct option is 1. 1 2
  7. Q13. Let f: R — {0) * R be defined by f (x) = alogelxl + bx3 + x2. If x —1 and x = 1 are the critical points of f (x), then : 3. both x = 1 and x = —1 are local minima for f (x) = —1 is a local maxima of f (x). 4. x = 1 isa local minima and x Solution: Let us calculates the derivatives f (x) = alogelxl + bx3 + x2 a — + 3bX2 + 2X x a 2 x f = a + 3b 2 and f = —a + 3b BAs 1 and are ritical points we have f '(1) = 0 = f '(—1) = 0 and a = Now f "(1) = 2+2 = 4 and f "(—1) 2 = 4. Therefore both x 1 and x = —lare local minima*lence the correct option is 3. Q14. Let f: [0,5] + R be continuous function such that 3 for all x e [0,5] and = 3. Then hen value of can be: 10 4. 12 Solution: we know I f(t)dtl < 103 If (t) Idt J 3 3dt = 9. Then we certainly discard the options and 4. By the same logic f 3 f(t)dt 6. So if we take the first ption then e must have = — = 3 — (—4) —9 and it crosses the bound 6. So the correct option is 2. Q15. 1. 2. 3. 4. sec x dx sin x • cos5x 5 2 (tan + - (tan + C 5 (tan + - (tan + C 5 2 (tan + - (tan + C 5 — - (tan + C 2 2 (tan x) 2 5
  8. Solution: Let I = dx —3 | -He—X f dx — Now further we can write —3 1-I-ex —x and denominator by e Therefore 21 = J dx — -3 So the correct option is Solution: We can write sec x dx sin x • cos5x sec4 x dx — tan x = 35 1 dx — sinx • cos7 x 1 dx tan x • cos8 x (1 + tan2 x) sec2 x dx tan x Now we put z = tan x, so we get dz = sec2 x dx. So the integral becomes (1 + z2)dz dz —+ z3/2dz = 2zä + —zi + C Putting the value of z we see that our answer is option 1. 4 Q16. The value of J dx is : 1. 2(35) 2.35 4 35 3. dx, [replacing x 4. 34 — x], as we know 3 e-X5x4 x, multiplying numerator 5 = 2.35 Hence I Q17. The area (Irvsq. units) above the x-axis bounded by the parabola, 0 and the line x -y -3=0 is: 8 1. 3 10 13 4. Solution: Wr•te the equations in standard form y 3 —x— 1 and y Let us draw the figure 2 0 The required area is given by 2 3 4 5 6
  9. 2 2 — Y2 + 2)dy 1 -.22 2 1 3 8 3 10 3 Hence the correct option is 2. Q18. Let the tangent be drawn at any point P(x, y) on a curve intersect th e x and y axes at two distinct points A and B respectively. If AP: PB = 5: 1, and the curve passes through the point (2,2) then the equation of the curve is: 1. xy5 = 26 2. xy4 = 25 3. x 4 y = 25 Solution: Let us draw a rough figure. Let the tangent drawn at on the curve meet at x-axis at point A(a, 0) and y-axis at point B(O, ß) respectively. We know AP: PB = 5: 1 so from section — (x, y . Hence formula we have P 6y —. The slope of the we get a = 6x and ß 5 tangent at P is given by the formula dy ß — 0 dx 0 —a 5dy- dx Simplifying we get — 0. Integrating we 5 get 5 In y + n x = In C IS the integrating constant. Therefore we have xy 5 = 26 C. As the curve passes through the point (2,2) we must have C = 2.2 So the correct option is 10 Q19. Let A(1,3) and C(5,1) be two opposite vertices of a rectangle. The other two vertices B(a, b) and D(c, d) lie on the line y = 2x + k for some k. Then the value of (a + b)(c + d) is: 2. 16 3. 24 4. 32 Solution: The midpoint of A and C is O is given byo (3,2). It must pass through the line y = 2x + k. Therefore the value of k is given by k = 2—2 x 3 So the equation of the line BD becomes y = 2x — 4. So let us take, D(c, d) — 2.7} k: o B(a, b)
  10. = (a, 2a — 4) and D (c, d) = (c, 2c — 4). The midpoint = (3,2). Therefore a + c = 6. becomes So our requirement is (a + b)(c + d) This is same as (3a — 4)(3c — 4) = 9ac — 12(a + c) + 16. Now line DA is perpendicular to line AB. Therefore we get, c —1 a—I (d — 3)(b — 3) + (a — I(c — 1) = 0 (2c - 15(a + c) + 50 5 ac —3 •6 + 10=0 (1) So the answer is 9 x 8 — 12 6+16 = 16. So the correct option is 2. Q20. Let the abscissae of two points A an B on a circle be the roots of x2 + 2x —4 = 0 and the prdinates of Wand B be he roots of Y2 + 4y = 16. If AB is a diameter of this circle, then the radius of this circle is: 2.6 Solution: Let A (a, b 2 a, c are roots o t e equation x SimilarlVforb, dxwe have 'bå = —16 AB = c, d). According to the question we have, x —4 = 0. Therefore we have O 4 + 16 16 -k 64 = 10 Therefore the radius of the circle is 5. So correct option is 1. Q21. Let P be the point on the parabola Y2 = 3x such that OP makes an angle of — with the x-axis, where O is the origin. A normal is drawn to the parabola at
  11. P intersecting the axis of the parabola at Q. If S is the focus of the parabola, then SQ is equal to : 39 1. 2 39 2. 4 3.9 41 4. 4 Solution: Let us write the equation of the parabola in standard form, Y2 = 4 • -x. Therefore the focus is 3t2 3t S (2, O). Let P . Also let us 4 assume that the normal drawn at P meet the axis at Q(h, 0). The slope of the line OP is given by tan 300 3 t2 2 Hence t = SOP = (9,3vfi). 3t2 p 4 3t 2) o = 300 s(3/4, o) 12 dy dx We get, 2y— = 3 or — dx dy - -2Vä. 3 2y 3 Also we see that the slope of the normal PQ is gi\/F9 by —. So equating them, = 2Nfi. Solving we get = 10. . Hence SQ = 10.5 - 0.75 = we get — — 39 9.75 = —. So the correct answer IS option 2. 4 Q22. Let the ellipse x + 1 16 be inscribed in a rectangle whose sides are parallel tot e coor inate axes„lf the rectangle is inscribed in another ellipse that passes through the point (16,0), then the equation of the outer ellipse is . 2 —+256! 1. 2. x2-+ 248y2 3. X2 + 240)/ 4. x2 + 232 y — 162 2 = 162 = 162 = 1. Here a = 4, b = 1. Then surely, the Solution: The given ellipse is + 1 co-ordinate of D is (4, 1). The required ellipse must pass through the point D. Let the required ellipse is given by x2 + ky 2 162 As it passes through (4,1), we have k = 162 16 = 256- 16 = 240. Hence the correct option is 3. s c
  12. Q23. Let the planes x — 2y + kz = 0 and x + 5y —z = 0 be perpendicular. Then the plane through the point (2, —2, —2) and perpendicular to the given planes also passes through the point: 2. (0,5, -8) 3. (-1,0, -7) Solution: As the given planes are perpendicular so their normals. Hence we have (1, —2, k) • (1,5, —1) = 0 it is dot product. Therefore 1 — 10 — k = 0. Or k = —9. Now the direction ratio of the plane which is perpendicular to both the given planes is given by the cross produt of the vectors (1, —2, —9) and (1,5, —1) which is found to be (47, —8,7). So the equation of the required plane is given by 47x — 8y + 7 z = 47 >< 2 — 8 x €2) + -1 7 x = 94 + 16 - 14 = 96. Or 47x - 8y+7 through (1,0,7). So the correct option is 4. Q24. If the two lines x —2m y z —4 and = 96. It certainly passes z — 2m 2m +5 - 8m 2 1 — 3m -1 Are parallel for some m e R, then t e distance between t em is: 3. 29 Solution: As the lines are parallel we ave 1 2 . Solving 1st pair 1—3m of equations, we get m = - an Solving last pair of equations we have m = 2 1 1 1 - and — So the common value we g€! is m = 2 Therefore the dirction ratio of any line is given by (6,4,2) or (3,2,1). Two given points on the two lines are respectively A(1,0,4) and B(2,0,1). The vector AB is given by (1,0,— ) Now if we take the dot product of AB with the direction ratio of any line is O. So AB is perpendicular to given lines. So the shortest distance IS the length of AB (i 3k)
  13. Q25. If three vectors VI = at + j + k, = t+ßj— 2k and 1/3 = t + j are coplanar, and VI and 1/3 are perpendicular, then the vector VI x 14 is: 2. 4. Solution: As VI L we have VI •1/3 = 0. Hence we have a + 1 = 0 or a — —1. As the three vectors are coplanar we have 1 = 0 . Solving we getß = —3. Now it (s straig t forward 0 calculation to find the required cross product. The correct ption is 2. Q26. If Xl,X2, ... , n x be the observed data such tha Xi — 2n = 180 and i=l i=l Then the mean of the data (Xl — 2.8 3 - 711 —go, — 3) is equal to: 16 4. 3 Solution: Subtracting we get, 5n = 150. So n = 30. Putting this value we get 240 30 P(AIB) Y—1Xi = 180 +60 = 0 . So tm, required mean is— — option is 1. Q27 If two events A and B, in a random experiment, 1 P(BIA)— -, then (AIA U B) is equal to: 3 = 5. Correct 4 = -and 5 4 5 1. 16 5 2. 17 16 3. 17 11 4. 16 Solution: We have PCB IA). P(A) = n B) Putting the given values we get
  14. 4 5 1 4 = 2p(A) 4 So comparing we get, B) = Ip(A). 4 Now P(AIA U B) . This gives P (B) = AP(A) and n 16 Now if we put all the values in terms of P (A), P(A) will get cancelle and we get our answer. The correct option is 3. Q28. If 6 cos2 9 — 2 cos 29 —3 = 0, then tan 1 1. 3 Solution: 39 is equal to• 9 4 2 6 cos29— 2 cos 29 —3 —6 os29 — (2 cos?9 — 1) 1 COS 29 = 4 1 2 herefore we can take = —. Now tan2 39 — 1. So the correct option is 4 2. Q29. If a, b and c(all distinct) are the sides of a triangle ABC opposite to the angles A, B and C respectively, then c sin(A — B) b sin(C — A) a 2 _ b2 is equal to: 2 2 c
  15. Solution: First we calculate, c sin(A — B) a2 _ b2 c sin(A — B) 4R2(sinA + sin — sin B) c sin (A — B) 4R2. 2 sin 2 c sin (A 4R2 sin(A + B) c 4R2 sin C 1 4R2 - 2R • cos 2 -B) sin(A — B) • 2 cos 2 By the similar calculation, the next term is also —Seth Q30. The expression e q) is equivalent to: 3. p Aæq 4. (æ p A q) V (æ q Ap Solution: Let us try to find truth table P • sin 2 orregoption is 1. s Vt p T F q F F s At F F F F Hence the correct option is 4.
  16. YIS

Need a Tutor or Coaching Class?

Post an enquiry and get instant responses from qualified and experienced tutors.

Post Requirement

Related Study Notes

Upload Study Notes

If you have your own Study Notes which you think can benefit others, please upload on LearnPick. For each approved Study Note you will get 100 Credit Points and 100 Activity Score which will increase your profile visibility.

Upload Now
Home > Study Notes > Notes On JEE Mains Mathematics Solution