MATHEMATICS-04-06- 11th (PQRS) Space

REVIEW TEST-1 Class : XI Time : 90 min Max. Marks : 75 General Remarks: INSTRUCTIONS 1. The question paper contain two parts. Part-A contains short question and Part-B contains subjective questions. All questions are compulsory. Paper contains 10 questions in Part-A and 9 questions in Part-B. 2. You are advised NOT to spend more than 30 minutes in any case for part-A. 3. Each question should be done only in the space provided for it, otherwise the solution will not be checked. 4. Use of Calculator, Log table and Mobile is not permitted. 5. Legibility and clarity in answering the question will be appreciated. 6. Put a cross ( × ) on the rough work done by you. Name : Roll No. Batch Class : XI Invigilator's Full Name For Office Use ……………………………. Total Marks Obtained………………… Part-A Part-B Q.No 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 Marks PART-A Q.1 Find the number of real solution(s) of the equation logx9 – log3x2 = 3. [3] Q.2 Simplify: cos x · sin(y – z) + cos y · sin(z – x) + cos z · sin (x – y) where x, y, z  R. [3] Q.3 If logx–3(2x – 3) is a meaningful quantity then find the interval in which x must lie. [3] Q.4 If x = 1 and x = 2 are solutions of the equation x3 + ax2 + bx + c = 0 and a + b = 1, then find the value of b. [3] Q.5 tan  = 1 2 + 1 2 + 1 where   (0, 2), find the possible value of . [3] 2 +O  Q.6 Find the exact value of cos12  cos 72 sin 72 + sin12 . [3] Q.7 If the tangent of DAB is expressed as a ratio of positive integers a b in lowest term, then find the value of (a + b). [3] 12 A Q.8 If sin A = 13 . Find the value of tan 2 . [3] Q.9 Let x = (0.15)20. Find the characteristic and mantissa in the logarithm of x, to the base 10. Assume log102 = 0.301 and log103 = 0.477. [3] Q.10 The figure (not drawn to scale) shows a regular octagon ABCDEFGH with diagonal AF = 1. Find the numerical value of the side of the octagon. [3] PART-B cos3  + cot(3 + ) sec(  3) cosec 3   Q.11 Simplify     tan2 (  ) sin(  2)     . [4] Q.12 Find the sum of the solutions of the equation 2e2x – 5ex + 4 = 0. [4] Q.13 Prove the identity cot A + cosec A 1 = cot A . [4] cot A  cosec A +1 2 Q.14 If log (log ( log x)) = log (log ( log y)) = 0 then find the value of (x + y). [5]  1 2   1  Q.15 If log25 = a and log 225 = b, then find the value of log   + log  in terms of a and  9    2250  b (base of the log is 10 everywhere). [5] Q.16 Prove that the expression sin2 + sin2(120° + ) + sin2(120° – ) remains constant    R. Find also the value of the constant. [5] Q.17 Suppose that x and y are positive numbers for which log9x = log12y = log16(x + y). If the value of y x = 2 cos , where   (0,  2) find . [6] tan  1 cot  Q.18 If tan   tan 3 = 3 , find the value of cot   cot 3 . [6]  2 3 4 Q.19 Let S = sec 0 + sec + sec + sec + sec and P = tan  · tan 2 · tan 4 then prove 5 5 5 5 9 9 9 S + P = 2 2 sin 7 [6] 12 EXTRA PAGE

Comments

Post a Comment

Popular posts from this blog

PHYSICS-15-10- 11th (PQRS) SOLUTION

XI (PQRS) PHYSICS REVIEW TEST-5 Select the correct alternative. (Only one is correct) [15 × 3 = 45] There is NEGATIVE marking. 1 mark will be deducted for each wrong answer. Q.31 Two trains, which are moving along different tracks in opposite directions towards each other, are put on the same track due to a mistake. Their drivers, on noticing the mistake, start slowing down the trains when the trains are 300 m apart. Graphs given below show their velocities as function of time as the trains slow down. The separation between the trains when both have stopped, is: (A) 120 m (B) 280 m (C) 60 m (D*) 20 m 1 [Sol: x1 = 2 1 × 10 × 40 = 200 m; x2 = – 2 × 8 × 20 = –80 m  xreletive = x1 + x2 = 280  separation = 300 – 280 = 20 m (D) ] Q.32 One end of a thin uniform rod of length L and mass M is riveted to the centre of a uniform circular disc of radius r and mass 2 M so that the rod is normal to the disc. The centre of mass of the combination from the centre of the disc is at dist
Read more

8-Circular Motion

8.1 (D) The maximum angular speed of the hoop corresponds to the situation when the bead is just about to slide upwards. The free body diagram of the bead is For the bead not to slide upwards. m2 (r sin 45°) cos 45° – mg sin 45° < N (1) where N = mg cos 45° + m2 (r sin 45ยบ) sin 45° (2) From 1 and 2 we get.  = rad / s. 8.2 (C) Let v be the speed of particle at B, just when it is about to loose contact. From application of Newton's second law to the particle normal to the spherical surface. mv2 = mg sin  (1) r Applying conservation of energy as the block moves from A to B.. 1 mv2 = mg (r cos  – r sin ) (2) 2 Solving 1 and 2 we get 3 sin  = 2 cos  8.3 (A) As the mass is at the verge of slipping ∴ mg sin37 –  mg cos37 = m2r 6 – 8 = 4.5   = 3 16 8.4 (B) As when they collide vt  1  72v 2  2 2  25R   R = vt  t =   5R 6v Now angle covered by A =   vt R 11 Put t  angle covered by A = 6 8.5 (C) The acceleration vector s
Read more

4. THEORY-Current Electrictricity

CURRENT ELECTRICITY  1 . ELECTRIC CURRENT (a) Time rate of flow of charge through a cross sectional area is called Current. if q charge flows in time interval t then average current is given by q Iav = t and Instantaneous current i =. lim q  dq t0 t dt (b) Direction of current is along the direction of flow of positive charge or opposite to the direction of flow of negative charge. But the current is a scalar quantity. i i q+ velocity q– velocity SI unit of current is ampere and 1 Ampere = 1 coloumb/sec 1 coloumb/sec = 1A 2 . CONDUCTOR In some materials, the outer electrons of each atom or molecule are only weakly bound to it. These electrons are almost free to move throughout the body of the material and are called free electrons. They are also known as conduction electrons. When such a material is placed in an electric field, the free electrons drift in a direction opposite to the field. Such materials are called conductors. 3 . INSULATOR Another class o
Read more