Tiêu đề 01. Vectors Med.pdf ✅

1. n CH4 molecule, there are four C – H bounds. If two adjacent bonds are in k ˆ j ˆ i ˆ + + and k ˆ j ˆ i ˆ − − direction, then fin d the angle between these bonds. (a) sin-1       − 3 1 (b) cos-1       3 1 (c) sin-1       3 1 (d) cos-1       − 3 1 2. Two vector AandB   have magnitude 3 each i A B 5k ˆ 2 ˆ  = − +   . Find angle between A and B (a) cos-1 9 29 (b) tan-1       − 2 5 (c) sin-1       5 2 (d) sin-1         9 29 3. A particle moving eastwards with 5 ms-1 . In 10 s the velocity changes to 5 ms-1 northwards. The average acceleration in this time is (a) 1/2 ms-2 towards North east (b) 1/2 ms-2 towards North (c) 1/2 ms-2 towards North west (d) Zero 4. If a vector 8k ˆ j 3 ˆ i 2 ˆ + + is perpendicular to the vector k ˆ i 4 ˆ j 4 ˆ − +  then the value of  is (a) 1/2 (b) – 1/2 (c) 1 (d) – 1 5. A river is flowing from W to E with a speed 5m/min. A man can swim in still waters at a velocity 10 m/min. In which direction should a man swim to take the shortest path to reach the south bank? (a) 300 East of South (b) 600 East of North (c) South (d) 300 West of North 6. Electrons in a TV tube move horizontally South to North. Vertical component of earth’s magnetic field points down. The electron is deflected towards (a) West (b) No deflection (c) East (d) North to South` 7. A = j 4 ˆ i 3 ˆ + Find a vector perpendicular to A  in the plane of A  (a) k ˆ (b) j 4 ˆ i 3 ˆ − − (c) i 3 ˆ j 4 ˆ − (d) j 3 ˆ i 4 ˆ − 8. Find a vector x  which is perpendicular to both A  and B  but has magnitude equal to that of B  Rule : Inter change coeff. of i ˆ and j ˆ and change sign of one of the vectors. (a) ( k) 17ˆ j 10ˆ i ˆ 10 1 + + (b) ( k) 17ˆ j 10ˆ i ˆ 10 1 − + (c) ( k) 17ˆ j 10ˆ i ˆ 390 29 + + (d) ( k) 17ˆ j 10ˆ i ˆ 390 29 − + 9. If P + Q + R = 0    and out of these, two vectors are equal in magnitude and the third vector has magnitude 2 times that of 2any of these two vectors, then angles among the three vectors are (a) 450 , 750 , 750 (b) 450 , 900 , 1350 (c) 900 , 1350 , 1800 (d) 900 , 1350 , 1350 10. sin2 is equivalent to : (a) 1 cos 2   +      (b) 1 cos2 2   +      (c) 1 cos2 2   −      (d) cos2 1 2    −     11. –sin is not equal to : (a) cos 2    +     (b) cos 2    −     (c) sin ( – ) (d) sin ( +  ) 12. If tan  = 5 1 and  lies in the first quadrant, the value of cos  is : (a) 6 5 (b) – 6 5 (c) 6 1 (d) – 6 1 13. If f() = sin , find f ( 6  ) (a) 6  (b) 2 1 (c) 2 (d) 3  14. Calculate slope of shown line (a) 2/3 (b) – 2/3 (c) 3/2 (d) –3/2 15. Roots of the equation 2x2 + 5x – 12 = 0, are
(a) 3/2, 4 (b) 2/3, – 4 (c) 3/2, – 4 (d) 2/3, 4 16. The speed (v) of a particle moving along a straight line is given by v = t2 + 3t – 4 where v is in m/s and t in second. Find time t at which the particle will momentarily come to rest. (a) 3 (b) 4 (c) 2 (d) 1 17. y = nx + ex , then 2 2 dy dx is equal to (a) 2 1 x – e x (b) 2 1 x + ex (c) 1 x + ex (d) – 2 1 x + ex 18. y = 2u3 , u = 8x – 1 (a) dy dx = 48 (8x – 1)2 (b) dy dx = 58 (5x – 1)2 (c) dy dx = 48 (8x – 1)2 (d) dy dx = 28 (8x – 1) 19. Given s = t2 + 5t + 3 , find dt ds , at t = 1 (a) 7 (b) 9 (c) 12 (d) 15 20. If s = ut + 2 1 at2 , where u and a are constants. Obtain the value of dt ds . (a) u – at (b) u + at (c) 2u + at (d) None of these 21. The minimum value of y = 5x2 – 2x + 1 is (a) 5 1 (b) 5 2 (c) 5 4 (d) 5 3 22. y = 3x – 2 2x + 5 (a) y’ = 2 19 (3x 2) − − (b) y’ = (3x – 2) 19 (c) y’ = (3x 2) –19 + (d) y’ = 2 (3x 2) –19 + 23. A uniform metallic solid sphere is heated uniformly. Due to thermal expansion, its radius increases at the rate of 0.05 mm/second. Find its rate of change of volume with respect to time when its radius becomes 10 mm. (take p = 3.14) (a) 31.4 mm3 /second (b) 62.8 mm3 /second (c) 3.14 mm3 /second (d) 6.28 mm3 /second 24. If y = 3t2 – 4t ; then minima of y will be at : (a) 3/2 (b) 3/4 (c) 2/3 (d) 4/3 25. If y = sin(t2) , then 2 2 d y dt will be - (a) 2t cos(t2 ) (b) 2 cos (t2 ) – 4t2 sin (t2 ) (c) 4t2 sin (t2 ) (d) 2 cos (t2 ) 26. p = 5 + cot q 1 , find dq dp (a) sec2 q (b) sec3 q (c) sec q (d) tan2 q 27. p = (1 + cosec q) cos q, find dq dp (a) sin q – cosec2 q (b) – sin q – cosec2 q (c) – sin q – cos2 q (d) sec q – cosec2 q 28. The displacement of a body at any time t after starting is given by s = 15t – 0.4t2 . The velocity of the body will be 7 ms–1 after time : (a) 20 s (b) 15 s (c) 10 s (d) 5 s 29. If velocity of particle is given by v = 2t 4 then its acceleration (dv/dt) at any time t will be given by : (a) 8t3 (b) 8t (c) –8t3 (d) t2 30. The maximum value of xy subject to x + y = 8, is : (a) 8 (b) 16 (c) 20 (d) 24 31. The slope of graph as shown in figure at points 1, 2 and is m1, m2 and m3 respectively then
(a) m1 > m2 > m3 (b) m1 < m2 < m3 (c) m1 = m2 = m3 (d) m1 = m2 > m3 32. Slope of the shown graph. (a) First increases then decreases (b) First decrease then increases (c) increase (d) decrease 33. y = – x 2 + 3 (a) dx dy = – 2x, 2 2 dx d y = –2 (b) dx dy = 2x, 2 2 dx d y = –2 (c) dx dy = – 2x, 2 2 dx d y = 2 (d) none of these 34. y = 3 x 3 + 2 x 2 + 4 x (a) dx dy = x2 –x + 4 1 , 2 2 dx d y = 2x + 3 (b) dx dy = x2 + x – 4 1 , 2 2 dx d y = 2x + 1 (c) dx dy = x2 + x + 4 1 , 2 2 dx d y = 2x + 1 (d) dx dy = x2 + x + 4 1 , 2 2 dx d y = 2x – 1 35. y = 4 – 2x – x –3 (a) dx dy = 2 + 3x–4 , 2 2 dx d y = – 12x–5 (b) dx dy = – 2 + 3x–4 , 2 2 dx d y = – 12x–5 (c) dx dy = – 2 + 3x–4 , 2 2 dx d y = 12x–5 (d) dx dy = – 2 – 3x–4 , 2 2 dx d y = – 12x–5 36. y = sin3 x, find the dx dy (a) 3 sin2 x (cosx)(b) 3 sin3 x (cosx) (c) 3 sin x (cos x)2 (d) sin x (cos x) 37. y = 5 cos–4 x, find dx dy (a) 20 sin x cos–5 x (b) 10 sin x cos–5 x (c) 20 sin x cos–3 x (d) 20 sin x sin–5 x 38. ( ) 2 x – 2x 1 dx c + +  (a) 3 x 3 + x2 – x – c (b) 3 x 3 + x + x + c (c) 3 x + x 2 + x – c (d) 3 x 3 – x 2 + x + c 39. dx x 1  x         + (a) 3 2 x + 2 x – c (b) 3 2 x 2 –2 x + c (c) 3 2 x 3 + 2 x +c (d) 2 2 x +2 x – c 40.  3x 1 dx (a) 3 1 nx + c (b) 3 1 nx (c) 2 1 nx + c (d) 3 1 nx + c
41. 2 x sin(2x )  dx, (use , u = 2x2 ) (a) – 1 4 cos (2x2 ) + C (b) 1 4 cos (2x2 ) + C (c) – 2 1 cos (2x) + C (d) – 3 1 cos (3x2 ) + C 42. 2 3 (2 x) −  dx (a) 3 2 x − + C (b) C 2 – x 2 + (c) C 2 – x 3 + (d) C 2 x 3 + + 43. 2sin(x)dx  is equal to : (a) –2cos x + C (b) 2 cosx + C (c) –2 cos x (d) 2 cosx 44. (sinx cosx)dx +  is equal to : (a) –cosx + sinx (b) – cosx + sinx + C (c) cosx – sinx + C (d) – cosx – sinx + C 45. 2 3 4 (x x x x )dx +++  is equal to : (a) 1+2x+3x2+4x3 + C (b) 1+2x+3x2+4x3 (c) 2 3 4 5 x x x x 2 3 4 5 + + + + C (d) 2 3 4 5 x x x x 2 3 4 5 + + + 46. If y = sin(ax+b) , then y dx  will be : (a) cos(ax b) a + + C (b) – cos(ax b) a + + C (c) a cos(ax+b)+C (d) – a cos(ax+b)+C 47. If y = x2 sin(x3 ) , then y dx  will be : (a) –cos(x3 ) + C (b) 3 cosx C 3     − +     (c) cos(x3 ) + C (d) 3 cosx C 3 + 48.  (x +1)dx (a) 2x – C 2 x 2 + (b) x C 2 x 2 + + (c) – x C 2 x 2 + (d) – x – C 2 x 2 49.  (5 – 6x)dx (a) 5x – x 2 + C (b) x – 3x2 – C (c) 5x + 3x2 + C (d) 5x – 3x2 + C 50.        + 2 t 3t 2 dt (a) t2 + 4 t 2 – C (b) t2 + 4 t 2 + C (c) t3 – 4 t 2 – C (d) 6 t 3 + t4 + C 51. dx x 2 2 x          + (a) 3 x 3 / 2 + 4x1/2 + C (b) 3 x 3 / 2 + x1/2 + C (c) 3 x 3 / 2 + 4x2/5 + C (d) 3 x 3 / 2 + 4x2 + C 52.          dy y 2 8y – 1/ 4 (a) 4y2 – 3 8 y 3/4 + C (b) 4y2 + 3 8 y 3/4 + C (c) y2 – 3 8 y 3/4 + C (d) 4y2 – 3 8 y 1/3 + C 53. 2x(1– x )dx –3 (a) x + x 2 – C (b) x2 + x 2 + C