Page 1 1 1. A value of ? for which ? ? ? ? 2 3isin 1 2isin is purely imaginary, is : (1) sin –1 ? ? ? ? ? ? 1 3 (2) ? 3 (3) ? 6 (4) sin –1 ? ? ? ? ? ? ? ? 3 4 Ans. (1) Sol. Z = ? ? ? ? 2 3isin 1 2isin ? Z = ? ? ? ? ? ? ? ? ? ? 2 2 3isin 1 2isin 1 4sin = ? ? ? ? ? ? ? ? 2 2 2 6sin 7isin 1 4sin for purely imaginary Z, Re(Z) = 0 ? 2 – 6sin 2 ? = 0 ? sin ? = ? 1 3 ?? ? = ±sin –1 ? ? ? ? ? ? 1 3 2. The system of linear equations x + ?y – z = 0 ?x – y – z = 0 x + y – ?z = 0 has a nontrivial solution for : (1) exactly three values of ?. (2) infinitely many values of ?. (3) exactly one value of ?. (4) exactly two values of ?. Ans. (1) Sol. ? ? ? ? ? ? ? 1 1 1 1 1 1 = 0 ?? ? ? ? ? = 0, 1, –1 PART A – MATHEMATICS JEE(MAIN) – 2016 TEST PAPER WITH SOLUTION (HELD ON SUNDAY 03 th APRIL, 2016) 3. A wire of length 2 units is cut into two parts which are bent respectively to form a square of side = x units and a circle of radius = r units. If the sum of the areas of the square and the circle so formed is minimum, then : (1) 2x = r (2) 2x = ( ? + 4)r (3) (4 – ?)x = ?r (4) x = 2r Ans. (4) Sol. Square x x r x x given that 4x + 2 ?r = 2 i.e. 2x + ?r = 1 ? r = ? ? 1 2x ..... (i) Area A = x 2 + ?r 2 = x 2 + ? 1 (2x – 1) 2 for min value of area A dA dx = 0 gives x = ? ? 2 4 ..... (ii) from (i) & (ii) r = ? ? 1 4 ..... (iii) ? x = 2r Page 2 1 1. A value of ? for which ? ? ? ? 2 3isin 1 2isin is purely imaginary, is : (1) sin –1 ? ? ? ? ? ? 1 3 (2) ? 3 (3) ? 6 (4) sin –1 ? ? ? ? ? ? ? ? 3 4 Ans. (1) Sol. Z = ? ? ? ? 2 3isin 1 2isin ? Z = ? ? ? ? ? ? ? ? ? ? 2 2 3isin 1 2isin 1 4sin = ? ? ? ? ? ? ? ? 2 2 2 6sin 7isin 1 4sin for purely imaginary Z, Re(Z) = 0 ? 2 – 6sin 2 ? = 0 ? sin ? = ? 1 3 ?? ? = ±sin –1 ? ? ? ? ? ? 1 3 2. The system of linear equations x + ?y – z = 0 ?x – y – z = 0 x + y – ?z = 0 has a nontrivial solution for : (1) exactly three values of ?. (2) infinitely many values of ?. (3) exactly one value of ?. (4) exactly two values of ?. Ans. (1) Sol. ? ? ? ? ? ? ? 1 1 1 1 1 1 = 0 ?? ? ? ? ? = 0, 1, –1 PART A – MATHEMATICS JEE(MAIN) – 2016 TEST PAPER WITH SOLUTION (HELD ON SUNDAY 03 th APRIL, 2016) 3. A wire of length 2 units is cut into two parts which are bent respectively to form a square of side = x units and a circle of radius = r units. If the sum of the areas of the square and the circle so formed is minimum, then : (1) 2x = r (2) 2x = ( ? + 4)r (3) (4 – ?)x = ?r (4) x = 2r Ans. (4) Sol. Square x x r x x given that 4x + 2 ?r = 2 i.e. 2x + ?r = 1 ? r = ? ? 1 2x ..... (i) Area A = x 2 + ?r 2 = x 2 + ? 1 (2x – 1) 2 for min value of area A dA dx = 0 gives x = ? ? 2 4 ..... (ii) from (i) & (ii) r = ? ? 1 4 ..... (iii) ? x = 2r J E E ( M A I N )  2 0 1 6 2 4. A man is walking towards a vertical pillar in a straight path, at a uniform speed. At a certain point A on the path, he observes that the angle of elevation of the top of the pillar is 30°. After walking for 10 minutes from A in the same direction, at a point B, he observes that the angle of elevation of the top of the pillar is 60°. Then the time taken (in minutes) by him, form B to reach the pillar, is : (1) 5 (2) 6 (3) 10 (4) 20 Ans. (1) Sol. 30° 60° A x B y P pillar Q h ?QPA : ? h x y = tan30° ? 3 h = x + y....(i) ?QPB : h y = tan60° ? h = 3 y .... (ii) By (i) and (ii) : 3y = x + y ? y = x 2 ? speed is uniform Distance x in 10 mins ? Distance x 2 in 5 mins 5. Let two fair sixfaced dice A and B be thrown simultaneously. If E 1 is the event that die A shows up four, E 2 is the event that die B shows up two and E 3 is the event that the sum of numbers on both dice is odd, then which of the following statements is NOT true ? (1) E 1 , E 2 and E 3 are independent. (2) E 1 and E 2 are independent. (3) E 2 and E 3 are independent. (4) E 1 and E 3 are independent. Ans. (1) Sol. E 1 ? A shows up 4 E 2 ? B shows up 2 E 3 ? Sum is odd (i.e. even + odd or odd + even) P(E 1 ) = 6 6.6 = 1 6 P(E 2 ) = 6 6.6 = 1 6 P(E 3 ) = ? ? 3 3 2 6.6 = 1 2 P(E 1 ? E 2 ) = 1 6.6 = P(E 1 ) . P(E 2 ) ? E 1 & E 2 are independent P(E 1 ? E 3 ) = 1.3 6.6 = P(E 1 ) . P(E 3 ) ? E 1 & E 3 are independent P(E 2 ? E 3 ) = 1.3 6.6 = 1 12 = P(E 2 ) . P(E 3 ) ? E 2 & E 3 are independent P(E 1 ? E 2 ? E 3 ) = 0 ie imposible event. 6. If the standard deviation of the numbers 2, 3, a and 11 is 3.5, then which of the following is true ? (1) 3a 2 – 23a + 44 = 0 (2) 3a 2 – 26a + 55 = 0 (3) 3a 2 – 32a + 84 = 0 (4) 3a 2 – 34a + 91 = 0 Ans. (3) Sol. ? 2 2 i i x x S.D. – n n ? ? ? ? ? ? ? ? ? ? 2 2 49 4 9 a 121 16 a 4 4 4 ? ? ? ? ? ? ? ? ? ? ? ? ? 3a 2 – 32a + 84 = 0 7. For x ? R, f(x) = log2 – sinx and g(x) = f(f(x)), then : (1) g is differentiable at x = 0 and g'(0) = –sin(log2) (2) g is not differentiable at x = 0 (3) g'(0) = cos(log2) (4) g'(0) = –cos(log2) Ans. (3) Sol. In the neighbourhood of x = 0, f(x) = log2 – sinx ? g(x) = f(f(x)) = log2 – sin(f(x)) = log2 – sin(log2 – sinx) It is differentiable at x = 0, so ? g?(x) = –cos(log2 – sinx) (–cosx) ? g?(0) = cos(log2) Page 3 1 1. A value of ? for which ? ? ? ? 2 3isin 1 2isin is purely imaginary, is : (1) sin –1 ? ? ? ? ? ? 1 3 (2) ? 3 (3) ? 6 (4) sin –1 ? ? ? ? ? ? ? ? 3 4 Ans. (1) Sol. Z = ? ? ? ? 2 3isin 1 2isin ? Z = ? ? ? ? ? ? ? ? ? ? 2 2 3isin 1 2isin 1 4sin = ? ? ? ? ? ? ? ? 2 2 2 6sin 7isin 1 4sin for purely imaginary Z, Re(Z) = 0 ? 2 – 6sin 2 ? = 0 ? sin ? = ? 1 3 ?? ? = ±sin –1 ? ? ? ? ? ? 1 3 2. The system of linear equations x + ?y – z = 0 ?x – y – z = 0 x + y – ?z = 0 has a nontrivial solution for : (1) exactly three values of ?. (2) infinitely many values of ?. (3) exactly one value of ?. (4) exactly two values of ?. Ans. (1) Sol. ? ? ? ? ? ? ? 1 1 1 1 1 1 = 0 ?? ? ? ? ? = 0, 1, –1 PART A – MATHEMATICS JEE(MAIN) – 2016 TEST PAPER WITH SOLUTION (HELD ON SUNDAY 03 th APRIL, 2016) 3. A wire of length 2 units is cut into two parts which are bent respectively to form a square of side = x units and a circle of radius = r units. If the sum of the areas of the square and the circle so formed is minimum, then : (1) 2x = r (2) 2x = ( ? + 4)r (3) (4 – ?)x = ?r (4) x = 2r Ans. (4) Sol. Square x x r x x given that 4x + 2 ?r = 2 i.e. 2x + ?r = 1 ? r = ? ? 1 2x ..... (i) Area A = x 2 + ?r 2 = x 2 + ? 1 (2x – 1) 2 for min value of area A dA dx = 0 gives x = ? ? 2 4 ..... (ii) from (i) & (ii) r = ? ? 1 4 ..... (iii) ? x = 2r J E E ( M A I N )  2 0 1 6 2 4. A man is walking towards a vertical pillar in a straight path, at a uniform speed. At a certain point A on the path, he observes that the angle of elevation of the top of the pillar is 30°. After walking for 10 minutes from A in the same direction, at a point B, he observes that the angle of elevation of the top of the pillar is 60°. Then the time taken (in minutes) by him, form B to reach the pillar, is : (1) 5 (2) 6 (3) 10 (4) 20 Ans. (1) Sol. 30° 60° A x B y P pillar Q h ?QPA : ? h x y = tan30° ? 3 h = x + y....(i) ?QPB : h y = tan60° ? h = 3 y .... (ii) By (i) and (ii) : 3y = x + y ? y = x 2 ? speed is uniform Distance x in 10 mins ? Distance x 2 in 5 mins 5. Let two fair sixfaced dice A and B be thrown simultaneously. If E 1 is the event that die A shows up four, E 2 is the event that die B shows up two and E 3 is the event that the sum of numbers on both dice is odd, then which of the following statements is NOT true ? (1) E 1 , E 2 and E 3 are independent. (2) E 1 and E 2 are independent. (3) E 2 and E 3 are independent. (4) E 1 and E 3 are independent. Ans. (1) Sol. E 1 ? A shows up 4 E 2 ? B shows up 2 E 3 ? Sum is odd (i.e. even + odd or odd + even) P(E 1 ) = 6 6.6 = 1 6 P(E 2 ) = 6 6.6 = 1 6 P(E 3 ) = ? ? 3 3 2 6.6 = 1 2 P(E 1 ? E 2 ) = 1 6.6 = P(E 1 ) . P(E 2 ) ? E 1 & E 2 are independent P(E 1 ? E 3 ) = 1.3 6.6 = P(E 1 ) . P(E 3 ) ? E 1 & E 3 are independent P(E 2 ? E 3 ) = 1.3 6.6 = 1 12 = P(E 2 ) . P(E 3 ) ? E 2 & E 3 are independent P(E 1 ? E 2 ? E 3 ) = 0 ie imposible event. 6. If the standard deviation of the numbers 2, 3, a and 11 is 3.5, then which of the following is true ? (1) 3a 2 – 23a + 44 = 0 (2) 3a 2 – 26a + 55 = 0 (3) 3a 2 – 32a + 84 = 0 (4) 3a 2 – 34a + 91 = 0 Ans. (3) Sol. ? 2 2 i i x x S.D. – n n ? ? ? ? ? ? ? ? ? ? 2 2 49 4 9 a 121 16 a 4 4 4 ? ? ? ? ? ? ? ? ? ? ? ? ? 3a 2 – 32a + 84 = 0 7. For x ? R, f(x) = log2 – sinx and g(x) = f(f(x)), then : (1) g is differentiable at x = 0 and g'(0) = –sin(log2) (2) g is not differentiable at x = 0 (3) g'(0) = cos(log2) (4) g'(0) = –cos(log2) Ans. (3) Sol. In the neighbourhood of x = 0, f(x) = log2 – sinx ? g(x) = f(f(x)) = log2 – sin(f(x)) = log2 – sin(log2 – sinx) It is differentiable at x = 0, so ? g?(x) = –cos(log2 – sinx) (–cosx) ? g?(0) = cos(log2) C O D E  G 3 8. The distance of the point (1, –5, 9) from the plane x – y + z = 5 measured along the line x = y = z is : (1) 20 3 (2) 3 10 (3) 10 3 (4) 10 3 Ans. (3) Sol. Equation of line parallel to x = y = z through (1, –5, 9) is x 1 y 5 z 9 1 1 1 ? ? ? ? ? ? ? If P( ? + 1, ? ?– 5, ? ?+ 9) be point of intesection of line and plane. P (1, –5, 9) ? ? ? ?? ?? ?? ? ? ?? ?? ?? ? ? ?? ?? ?? ? ? ? ? ?? ? ?? ? Coordinates point are (–9, –15, –1) ? Required distance = 10 3 9. The eccentricity of the hyperbola whose length of the latus rectum is equal to 8 and the length of its conjugate axis is equal to half of the distance between its foci, is : (1) 3 (2) 4 3 (3) 4 3 (4) 2 3 Ans. (4) Sol. Given 2 2b 8 a ? .... (1) 2b = ae .... (2) we know b 2 = a 2 (e 2 – 1) .... (3) substitute b a = e 2 from (2) in (3) ? 2 e 4 = e 2 – 1 ??4 = 3e 2 ??e = 2 3 10. Let P be the point on the parabola, y 2 = 8x which is at a minimum distance from the cente C of the circle, x 2 + (y + 6) 2 = 1. Then the equation of the circle, passing through C and having its centre at P is : (1) x 2 + y 2 – 4x + 9y + 18 = 0 (2) x 2 + y 2 – 4x + 8y + 12 = 0 (3) x 2 + y 2 – x + 4y – 12 = 0 (4) x 2 + y 2 – x 4 + 2y – 24 = 0 Ans. (2) Sol. Circle and parabola are as shown : (0, 6)C (2t , 4t) 2 O y = 8x a =2 2 P Minimum distance occurs along common normal. Let normal to parabola be y + tx = 2.2.t + 2t 3 pass through (0, –6) : –6 = 4t + 2t 3 ??t 3 + 2t + 3 = 0 ? t = –1(only real value) ? P(2, –4) ? CP = 4 4 2 2 ? ? ? equation of circle (x – 2) 2 + (y + 4) 2 = ? ? 2 2 2 ? x 2 + y 2 – 4x + 8y + 12 = 0 Page 4 1 1. A value of ? for which ? ? ? ? 2 3isin 1 2isin is purely imaginary, is : (1) sin –1 ? ? ? ? ? ? 1 3 (2) ? 3 (3) ? 6 (4) sin –1 ? ? ? ? ? ? ? ? 3 4 Ans. (1) Sol. Z = ? ? ? ? 2 3isin 1 2isin ? Z = ? ? ? ? ? ? ? ? ? ? 2 2 3isin 1 2isin 1 4sin = ? ? ? ? ? ? ? ? 2 2 2 6sin 7isin 1 4sin for purely imaginary Z, Re(Z) = 0 ? 2 – 6sin 2 ? = 0 ? sin ? = ? 1 3 ?? ? = ±sin –1 ? ? ? ? ? ? 1 3 2. The system of linear equations x + ?y – z = 0 ?x – y – z = 0 x + y – ?z = 0 has a nontrivial solution for : (1) exactly three values of ?. (2) infinitely many values of ?. (3) exactly one value of ?. (4) exactly two values of ?. Ans. (1) Sol. ? ? ? ? ? ? ? 1 1 1 1 1 1 = 0 ?? ? ? ? ? = 0, 1, –1 PART A – MATHEMATICS JEE(MAIN) – 2016 TEST PAPER WITH SOLUTION (HELD ON SUNDAY 03 th APRIL, 2016) 3. A wire of length 2 units is cut into two parts which are bent respectively to form a square of side = x units and a circle of radius = r units. If the sum of the areas of the square and the circle so formed is minimum, then : (1) 2x = r (2) 2x = ( ? + 4)r (3) (4 – ?)x = ?r (4) x = 2r Ans. (4) Sol. Square x x r x x given that 4x + 2 ?r = 2 i.e. 2x + ?r = 1 ? r = ? ? 1 2x ..... (i) Area A = x 2 + ?r 2 = x 2 + ? 1 (2x – 1) 2 for min value of area A dA dx = 0 gives x = ? ? 2 4 ..... (ii) from (i) & (ii) r = ? ? 1 4 ..... (iii) ? x = 2r J E E ( M A I N )  2 0 1 6 2 4. A man is walking towards a vertical pillar in a straight path, at a uniform speed. At a certain point A on the path, he observes that the angle of elevation of the top of the pillar is 30°. After walking for 10 minutes from A in the same direction, at a point B, he observes that the angle of elevation of the top of the pillar is 60°. Then the time taken (in minutes) by him, form B to reach the pillar, is : (1) 5 (2) 6 (3) 10 (4) 20 Ans. (1) Sol. 30° 60° A x B y P pillar Q h ?QPA : ? h x y = tan30° ? 3 h = x + y....(i) ?QPB : h y = tan60° ? h = 3 y .... (ii) By (i) and (ii) : 3y = x + y ? y = x 2 ? speed is uniform Distance x in 10 mins ? Distance x 2 in 5 mins 5. Let two fair sixfaced dice A and B be thrown simultaneously. If E 1 is the event that die A shows up four, E 2 is the event that die B shows up two and E 3 is the event that the sum of numbers on both dice is odd, then which of the following statements is NOT true ? (1) E 1 , E 2 and E 3 are independent. (2) E 1 and E 2 are independent. (3) E 2 and E 3 are independent. (4) E 1 and E 3 are independent. Ans. (1) Sol. E 1 ? A shows up 4 E 2 ? B shows up 2 E 3 ? Sum is odd (i.e. even + odd or odd + even) P(E 1 ) = 6 6.6 = 1 6 P(E 2 ) = 6 6.6 = 1 6 P(E 3 ) = ? ? 3 3 2 6.6 = 1 2 P(E 1 ? E 2 ) = 1 6.6 = P(E 1 ) . P(E 2 ) ? E 1 & E 2 are independent P(E 1 ? E 3 ) = 1.3 6.6 = P(E 1 ) . P(E 3 ) ? E 1 & E 3 are independent P(E 2 ? E 3 ) = 1.3 6.6 = 1 12 = P(E 2 ) . P(E 3 ) ? E 2 & E 3 are independent P(E 1 ? E 2 ? E 3 ) = 0 ie imposible event. 6. If the standard deviation of the numbers 2, 3, a and 11 is 3.5, then which of the following is true ? (1) 3a 2 – 23a + 44 = 0 (2) 3a 2 – 26a + 55 = 0 (3) 3a 2 – 32a + 84 = 0 (4) 3a 2 – 34a + 91 = 0 Ans. (3) Sol. ? 2 2 i i x x S.D. – n n ? ? ? ? ? ? ? ? ? ? 2 2 49 4 9 a 121 16 a 4 4 4 ? ? ? ? ? ? ? ? ? ? ? ? ? 3a 2 – 32a + 84 = 0 7. For x ? R, f(x) = log2 – sinx and g(x) = f(f(x)), then : (1) g is differentiable at x = 0 and g'(0) = –sin(log2) (2) g is not differentiable at x = 0 (3) g'(0) = cos(log2) (4) g'(0) = –cos(log2) Ans. (3) Sol. In the neighbourhood of x = 0, f(x) = log2 – sinx ? g(x) = f(f(x)) = log2 – sin(f(x)) = log2 – sin(log2 – sinx) It is differentiable at x = 0, so ? g?(x) = –cos(log2 – sinx) (–cosx) ? g?(0) = cos(log2) C O D E  G 3 8. The distance of the point (1, –5, 9) from the plane x – y + z = 5 measured along the line x = y = z is : (1) 20 3 (2) 3 10 (3) 10 3 (4) 10 3 Ans. (3) Sol. Equation of line parallel to x = y = z through (1, –5, 9) is x 1 y 5 z 9 1 1 1 ? ? ? ? ? ? ? If P( ? + 1, ? ?– 5, ? ?+ 9) be point of intesection of line and plane. P (1, –5, 9) ? ? ? ?? ?? ?? ? ? ?? ?? ?? ? ? ?? ?? ?? ? ? ? ? ?? ? ?? ? Coordinates point are (–9, –15, –1) ? Required distance = 10 3 9. The eccentricity of the hyperbola whose length of the latus rectum is equal to 8 and the length of its conjugate axis is equal to half of the distance between its foci, is : (1) 3 (2) 4 3 (3) 4 3 (4) 2 3 Ans. (4) Sol. Given 2 2b 8 a ? .... (1) 2b = ae .... (2) we know b 2 = a 2 (e 2 – 1) .... (3) substitute b a = e 2 from (2) in (3) ? 2 e 4 = e 2 – 1 ??4 = 3e 2 ??e = 2 3 10. Let P be the point on the parabola, y 2 = 8x which is at a minimum distance from the cente C of the circle, x 2 + (y + 6) 2 = 1. Then the equation of the circle, passing through C and having its centre at P is : (1) x 2 + y 2 – 4x + 9y + 18 = 0 (2) x 2 + y 2 – 4x + 8y + 12 = 0 (3) x 2 + y 2 – x + 4y – 12 = 0 (4) x 2 + y 2 – x 4 + 2y – 24 = 0 Ans. (2) Sol. Circle and parabola are as shown : (0, 6)C (2t , 4t) 2 O y = 8x a =2 2 P Minimum distance occurs along common normal. Let normal to parabola be y + tx = 2.2.t + 2t 3 pass through (0, –6) : –6 = 4t + 2t 3 ??t 3 + 2t + 3 = 0 ? t = –1(only real value) ? P(2, –4) ? CP = 4 4 2 2 ? ? ? equation of circle (x – 2) 2 + (y + 4) 2 = ? ? 2 2 2 ? x 2 + y 2 – 4x + 8y + 12 = 0 J E E ( M A I N )  2 0 1 6 4 11. If A = ? ? ? ? ? ? ? 5a b 3 2 and A adj A = A A T , then 5a + b is equal to : (1) 13 (2) –1 (3) 5 (4) 4 Ans. (3) Sol. A = 5a –b 3 2 ? ? ? ? ? ? and A T = 5a 3 –b 2 ? ? ? ? ? ? AA T = 2 2 25a b 15a – 2b 15a 2b 13 ? ? ? ? ? ? ? ? Now, A adj A = AI 2 = 10a 3b 0 0 10a 3b ? ? ? ? ? ? ? ? Given AA T = A. adj A 15a –2b = 0 ...(1) 10a + 3b = 13 ...(2) Solving we get 5a = 2 and b = 3 ? 5a + b = 5 12. Consider f(x) = tan –1 ? ? ? ? ? ? ? ? ? ? 1 sin x 1 sin x , x ? ? ? ? ? ? ? ? 0, 2 . A normal to y = f(x) at x = ? 6 also passes through the point : (1) ? ? ? ? ? ? ? ,0 4 (2) (0, 0) (3) ? ? ? ? ? ? ? 2 0, 3 (4) ? ? ? ? ? ? ? ,0 6 Ans. (3) Sol. 1 1 sin x f (x) tan 1–sin x ? ? ? ? ? ? ? ? ? ? ? where x 0, 2 ? ? ? ? ? ? ? ? = 2 1 2 (1 sin x) tan 1– sin x ? ? ? ? ? ? ? ? ? ? 1 1 sin x tan  cos x  ? ? ? ? ? ? ? ? ? 1 1 sin x tan as x 0, cos x 2 ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? 2 1 2 2 x x cos sin 2 2 tan x x cos sin 2 2 ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? 1 x tan tan 4 2 ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? f(x) x 4 2 ? ? ? as x 0, 2 ? ? ? ? ? ? ? ? ?? 1 f 6 2 ? ? ? ? ? ? ? ? ? ? Equation of normal y – 2 x 3 6 ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? which passes through 2 0, 3 ? ? ? ? ? ? ? 13. Two sides of a rhombus are along the lines, x – y + 1 = 0 and 7x – y – 5 = 0. If its diagonals intersect at (–1, –2), then which one of the following is a vertex of this rhombus ? (1) ? ? ? ? ? ? ? ? 10 7 , 3 3 (2) (–3, –9) (3) (–3, –8) (4) ? ? ? ? ? ? ? 1 8 , 3 3 Ans. (4) Sol. Equation of angle bisector of the lines x – y + 1 = 0 and 7x – y – 5 = 0 is given by x y 1 7x y 5 2 5 2 ? ? ? ? ? ? ? ?5(x – y + 1) = 7x – y – 5 and 5(x – y + 1) = –7x + y + 5 ? 2x + 4y – 10 = 0 ? x + 2y – 5 = 0 and 12x – 6y = 0 ??2x – y = 0 Now equation of diagonals are (x + 1) + 2(y + 2) = 0 ? ?x + 2y + 5 = 0 ...(1) and 2(x + 1) – (y + 2) = 0 ? ?2x – y = 0 ...(2) Clearly ? ? ? ? ? ? ? 1 8 , 3 3 lies on (1) Page 5 1 1. A value of ? for which ? ? ? ? 2 3isin 1 2isin is purely imaginary, is : (1) sin –1 ? ? ? ? ? ? 1 3 (2) ? 3 (3) ? 6 (4) sin –1 ? ? ? ? ? ? ? ? 3 4 Ans. (1) Sol. Z = ? ? ? ? 2 3isin 1 2isin ? Z = ? ? ? ? ? ? ? ? ? ? 2 2 3isin 1 2isin 1 4sin = ? ? ? ? ? ? ? ? 2 2 2 6sin 7isin 1 4sin for purely imaginary Z, Re(Z) = 0 ? 2 – 6sin 2 ? = 0 ? sin ? = ? 1 3 ?? ? = ±sin –1 ? ? ? ? ? ? 1 3 2. The system of linear equations x + ?y – z = 0 ?x – y – z = 0 x + y – ?z = 0 has a nontrivial solution for : (1) exactly three values of ?. (2) infinitely many values of ?. (3) exactly one value of ?. (4) exactly two values of ?. Ans. (1) Sol. ? ? ? ? ? ? ? 1 1 1 1 1 1 = 0 ?? ? ? ? ? = 0, 1, –1 PART A – MATHEMATICS JEE(MAIN) – 2016 TEST PAPER WITH SOLUTION (HELD ON SUNDAY 03 th APRIL, 2016) 3. A wire of length 2 units is cut into two parts which are bent respectively to form a square of side = x units and a circle of radius = r units. If the sum of the areas of the square and the circle so formed is minimum, then : (1) 2x = r (2) 2x = ( ? + 4)r (3) (4 – ?)x = ?r (4) x = 2r Ans. (4) Sol. Square x x r x x given that 4x + 2 ?r = 2 i.e. 2x + ?r = 1 ? r = ? ? 1 2x ..... (i) Area A = x 2 + ?r 2 = x 2 + ? 1 (2x – 1) 2 for min value of area A dA dx = 0 gives x = ? ? 2 4 ..... (ii) from (i) & (ii) r = ? ? 1 4 ..... (iii) ? x = 2r J E E ( M A I N )  2 0 1 6 2 4. A man is walking towards a vertical pillar in a straight path, at a uniform speed. At a certain point A on the path, he observes that the angle of elevation of the top of the pillar is 30°. After walking for 10 minutes from A in the same direction, at a point B, he observes that the angle of elevation of the top of the pillar is 60°. Then the time taken (in minutes) by him, form B to reach the pillar, is : (1) 5 (2) 6 (3) 10 (4) 20 Ans. (1) Sol. 30° 60° A x B y P pillar Q h ?QPA : ? h x y = tan30° ? 3 h = x + y....(i) ?QPB : h y = tan60° ? h = 3 y .... (ii) By (i) and (ii) : 3y = x + y ? y = x 2 ? speed is uniform Distance x in 10 mins ? Distance x 2 in 5 mins 5. Let two fair sixfaced dice A and B be thrown simultaneously. If E 1 is the event that die A shows up four, E 2 is the event that die B shows up two and E 3 is the event that the sum of numbers on both dice is odd, then which of the following statements is NOT true ? (1) E 1 , E 2 and E 3 are independent. (2) E 1 and E 2 are independent. (3) E 2 and E 3 are independent. (4) E 1 and E 3 are independent. Ans. (1) Sol. E 1 ? A shows up 4 E 2 ? B shows up 2 E 3 ? Sum is odd (i.e. even + odd or odd + even) P(E 1 ) = 6 6.6 = 1 6 P(E 2 ) = 6 6.6 = 1 6 P(E 3 ) = ? ? 3 3 2 6.6 = 1 2 P(E 1 ? E 2 ) = 1 6.6 = P(E 1 ) . P(E 2 ) ? E 1 & E 2 are independent P(E 1 ? E 3 ) = 1.3 6.6 = P(E 1 ) . P(E 3 ) ? E 1 & E 3 are independent P(E 2 ? E 3 ) = 1.3 6.6 = 1 12 = P(E 2 ) . P(E 3 ) ? E 2 & E 3 are independent P(E 1 ? E 2 ? E 3 ) = 0 ie imposible event. 6. If the standard deviation of the numbers 2, 3, a and 11 is 3.5, then which of the following is true ? (1) 3a 2 – 23a + 44 = 0 (2) 3a 2 – 26a + 55 = 0 (3) 3a 2 – 32a + 84 = 0 (4) 3a 2 – 34a + 91 = 0 Ans. (3) Sol. ? 2 2 i i x x S.D. – n n ? ? ? ? ? ? ? ? ? ? 2 2 49 4 9 a 121 16 a 4 4 4 ? ? ? ? ? ? ? ? ? ? ? ? ? 3a 2 – 32a + 84 = 0 7. For x ? R, f(x) = log2 – sinx and g(x) = f(f(x)), then : (1) g is differentiable at x = 0 and g'(0) = –sin(log2) (2) g is not differentiable at x = 0 (3) g'(0) = cos(log2) (4) g'(0) = –cos(log2) Ans. (3) Sol. In the neighbourhood of x = 0, f(x) = log2 – sinx ? g(x) = f(f(x)) = log2 – sin(f(x)) = log2 – sin(log2 – sinx) It is differentiable at x = 0, so ? g?(x) = –cos(log2 – sinx) (–cosx) ? g?(0) = cos(log2) C O D E  G 3 8. The distance of the point (1, –5, 9) from the plane x – y + z = 5 measured along the line x = y = z is : (1) 20 3 (2) 3 10 (3) 10 3 (4) 10 3 Ans. (3) Sol. Equation of line parallel to x = y = z through (1, –5, 9) is x 1 y 5 z 9 1 1 1 ? ? ? ? ? ? ? If P( ? + 1, ? ?– 5, ? ?+ 9) be point of intesection of line and plane. P (1, –5, 9) ? ? ? ?? ?? ?? ? ? ?? ?? ?? ? ? ?? ?? ?? ? ? ? ? ?? ? ?? ? Coordinates point are (–9, –15, –1) ? Required distance = 10 3 9. The eccentricity of the hyperbola whose length of the latus rectum is equal to 8 and the length of its conjugate axis is equal to half of the distance between its foci, is : (1) 3 (2) 4 3 (3) 4 3 (4) 2 3 Ans. (4) Sol. Given 2 2b 8 a ? .... (1) 2b = ae .... (2) we know b 2 = a 2 (e 2 – 1) .... (3) substitute b a = e 2 from (2) in (3) ? 2 e 4 = e 2 – 1 ??4 = 3e 2 ??e = 2 3 10. Let P be the point on the parabola, y 2 = 8x which is at a minimum distance from the cente C of the circle, x 2 + (y + 6) 2 = 1. Then the equation of the circle, passing through C and having its centre at P is : (1) x 2 + y 2 – 4x + 9y + 18 = 0 (2) x 2 + y 2 – 4x + 8y + 12 = 0 (3) x 2 + y 2 – x + 4y – 12 = 0 (4) x 2 + y 2 – x 4 + 2y – 24 = 0 Ans. (2) Sol. Circle and parabola are as shown : (0, 6)C (2t , 4t) 2 O y = 8x a =2 2 P Minimum distance occurs along common normal. Let normal to parabola be y + tx = 2.2.t + 2t 3 pass through (0, –6) : –6 = 4t + 2t 3 ??t 3 + 2t + 3 = 0 ? t = –1(only real value) ? P(2, –4) ? CP = 4 4 2 2 ? ? ? equation of circle (x – 2) 2 + (y + 4) 2 = ? ? 2 2 2 ? x 2 + y 2 – 4x + 8y + 12 = 0 J E E ( M A I N )  2 0 1 6 4 11. If A = ? ? ? ? ? ? ? 5a b 3 2 and A adj A = A A T , then 5a + b is equal to : (1) 13 (2) –1 (3) 5 (4) 4 Ans. (3) Sol. A = 5a –b 3 2 ? ? ? ? ? ? and A T = 5a 3 –b 2 ? ? ? ? ? ? AA T = 2 2 25a b 15a – 2b 15a 2b 13 ? ? ? ? ? ? ? ? Now, A adj A = AI 2 = 10a 3b 0 0 10a 3b ? ? ? ? ? ? ? ? Given AA T = A. adj A 15a –2b = 0 ...(1) 10a + 3b = 13 ...(2) Solving we get 5a = 2 and b = 3 ? 5a + b = 5 12. Consider f(x) = tan –1 ? ? ? ? ? ? ? ? ? ? 1 sin x 1 sin x , x ? ? ? ? ? ? ? ? 0, 2 . A normal to y = f(x) at x = ? 6 also passes through the point : (1) ? ? ? ? ? ? ? ,0 4 (2) (0, 0) (3) ? ? ? ? ? ? ? 2 0, 3 (4) ? ? ? ? ? ? ? ,0 6 Ans. (3) Sol. 1 1 sin x f (x) tan 1–sin x ? ? ? ? ? ? ? ? ? ? ? where x 0, 2 ? ? ? ? ? ? ? ? = 2 1 2 (1 sin x) tan 1– sin x ? ? ? ? ? ? ? ? ? ? 1 1 sin x tan  cos x  ? ? ? ? ? ? ? ? ? 1 1 sin x tan as x 0, cos x 2 ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? 2 1 2 2 x x cos sin 2 2 tan x x cos sin 2 2 ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? 1 x tan tan 4 2 ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? f(x) x 4 2 ? ? ? as x 0, 2 ? ? ? ? ? ? ? ? ?? 1 f 6 2 ? ? ? ? ? ? ? ? ? ? Equation of normal y – 2 x 3 6 ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? which passes through 2 0, 3 ? ? ? ? ? ? ? 13. Two sides of a rhombus are along the lines, x – y + 1 = 0 and 7x – y – 5 = 0. If its diagonals intersect at (–1, –2), then which one of the following is a vertex of this rhombus ? (1) ? ? ? ? ? ? ? ? 10 7 , 3 3 (2) (–3, –9) (3) (–3, –8) (4) ? ? ? ? ? ? ? 1 8 , 3 3 Ans. (4) Sol. Equation of angle bisector of the lines x – y + 1 = 0 and 7x – y – 5 = 0 is given by x y 1 7x y 5 2 5 2 ? ? ? ? ? ? ? ?5(x – y + 1) = 7x – y – 5 and 5(x – y + 1) = –7x + y + 5 ? 2x + 4y – 10 = 0 ? x + 2y – 5 = 0 and 12x – 6y = 0 ??2x – y = 0 Now equation of diagonals are (x + 1) + 2(y + 2) = 0 ? ?x + 2y + 5 = 0 ...(1) and 2(x + 1) – (y + 2) = 0 ? ?2x – y = 0 ...(2) Clearly ? ? ? ? ? ? ? 1 8 , 3 3 lies on (1) C O D E  G 5 14. If a curve y = f(x) passes through the point (1, –1) and satisfies the differential equation, y(1 + xy) dx = x dy, then f ? ? ? ? ? ? ? 1 2 is equal to : (1) 4 5 (2) ? 2 5 (3) ? 4 5 (4) 2 5 Ans. (1) Sol. Given differential equation ydx + xy 2 dx = xdy ? 2 xdy – ydx y = xdx 2 x x d d y 2 ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? Integrating we get 2 x x – C y 2 ? ? ??It passes through (1, –1) ? 1 1 1 C C 2 2 ? ? ? ? ? 2 2 2x 2x x 1 0 y y x 1 ? ? ? ? ? ? ? ? 1 4 f – 2 5 ? ? ? ? ? ? ? 15. If all the words (with or without meaning) having five letters, formed using the letters of the word SMALL and arranged as in a dictionary; then the position of the word SMALL is : (1) 58 th (2) 46 th (3) 59 th (4) 52 nd Ans. (1) Sol. Total number of words which can be formed using all the letters of the word 'SMALL' = 5! 60 2! ? Now, 60 th word is ??SMLLA 59 th word is ??SMLAL 58 th word is ??SMALL 16. If the 2 nd , 5 th and 9 th terms of a nonconstant A.P. are in G.P., then the common ratio of this G.P. is : (1) 7 4 (2) 8 5 (3) 4 3 (4) 1 Ans. (3) Sol. Let 'a' be the first term and d be the common difference 2 nd term = a + d, 5 th term = a + 4d, 9 th term = 4 + 8d ? ? ? ? ? ? ? ? ??Common ratio = a 4d a 8d a d a 4d ? ? ? ? ? = 4d 4 3d 3 ? 17. If the number of terms in the expansion of n 2 2 4 1 ,x 0 x x ? ? ? ? ? ? ? ? ? , is 28, then the sum of the coefficients of all the terms in this expansion, is : (1) 729 (2) 64 (3) 2187 (4) 243 Ans. (1 or Bonus) Sol. Number of terms in the expansion of n 2 2 4 1– x x ? ? ? ? ? ? ? is n + 2 C 2 (considering 1 x and 2 1 x distinct) ? n + 2 C 2 = 28 ??n = 6 ? Sum of coefficients = (1 – 2 + 4) 6 = 729 But number of dissimilar terms actually will be 2n + 1 (as 1 x and 2 1 x are functions as same variable) Hence it contains error, so a bonus can be expected. 18. If the sum of the first ten terms of the series 2 2 2 2 2 3 2 1 4 1 2 3 4 4 ..., 5 5 5 5 ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? is 16 m, 5 then m is equal to : (1) 99 (2) 102 (3) 101 (4) 100 Ans. (3) Sol. Given series is 2 2 2 2 2 8 12 16 S ...10 terms 5 5 5 ? ? ? ? ? = 2 2 2 2 2 4 (2 3 4 ....10 terms) 5 ? ? ? = 16 11.12.23 1 25 6 ? ? ? ? ? ? ? = 16 505 25 ? ? m = 101Read More
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