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

REVIEW TEST-1 Class : XI (PQRS) 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. [Sol. logx9 – log3x2 = 3 2[logx3 + 2log3x] = 3 logx3 + log3x = 3/2 [Ans. No solution] [3] log3x + 1 3 log3 x = 2 let log3x = a m a + 1 = 3 a 2 1 1 which is not possible as a + a > 2 if a > 0 and a + a < – 2 if a < 0; hence no solution ] Q.2 Simplify: cos x · sin(y – z) + cos y · sin(z – x) + cos z · sin (x – y) where x, y, z ϵ R. [3] [Ans. 0] 1 [Sol. 2 [sin(y – z + x) + sin(y – z – x) + sin(z – x + y) + sin(z – x – y) + sin(x – y + z) + sin(x – y – z)] = 0 ] Q.3 If logx–3(2x – 3) is a meaningful quantity then find the interval in which x must lie. [3] [Ans. (3, 4) u (4, )] [Sol. x – 3 > 0, x – 3 s 1 and 2x – 3 > 0 x > 3 ; x s 4 and x > 3/2 ] 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] [Ans. 5] [Sol. a + b + c = – 1  c = – 2 and 8 + 4a + 2b + c = 0  4a + 2b = – 6  2a + b = – 3  a = – 4, b = 5 hence a = – 4; b = 5; c = – 2 ] Q.5 tan θ = 1 where θ ϵ (0, 2π), find the possible value of θ. [3] 2 + 2 + 1 2 +O  1 [Ans. tan θ = 1 –1  θ = π 9π 8 or 8 ] [Sol. Let tan θ = x = 2 + 1 = 2 + x 2 + 1 2 +O  x2 + 2x – 1 = 0  x = – 2  2 = ( –1) m tan θ = –1  θ = π 9π 8 or 8 ] Q.6 Find the exact value of 2 sin 42°sin 30° cos12° – cos 72° sin 72° + sin12° . [3] [Sol. 2 sin 42°cos 30° = tan 30° = Ans. ] Q.713 ph-1 If the tangent of 3DAB is expressed as a ratio of positive integers a b in lowest term, then find the value of (a + b). [3] [Ans. 37] [Hint: tan (3DAB) = tan(θ1 + θ2) = m a + b = 37 Ans. ] 23 = 14 12 A Q.8 If sin A = 13 . Find the value of tan 2 . [3] 12 [Sol. sinA = 13 ⇒ A is in I quadrant or II quadrant m cos A can be 5/13 or – 5/13 Case-I: A is in I quadrant m 0 < A/2 < π/4 m cos A = 5/13 m tan A = 2 1– cos A 1+ cos A = = 2 = = 3 Case-II: A is in II quadrant m π/4 < A/2 < π/2 m cos A = – 5/13 m tan A = 2 = = 3 = = 2 m tan A can be 2 2 3 3 or 2 ] 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. ( 15 ⎞ [Sol. log x = log(0.15)20 = 20 log|100 | [Ans. – 17; 0.52] [3] ⎝ ⎠ = 20[log 15 – 2] = 20[log 3 + log 5 – 2] = 20[log 3 + 1 – log 2 – 2] = 20[– 1 + log 3 – log 2] = 20[– 1 + 0.477 – 0.301] = – 20 × 0.824 = – 16.48 = 17.52 hence characteristic = – 17 and mantissa = 0.52 Ans. ] 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] [Sol. θ = 22.5° tan 22.5° = x · 2 [Ans. –1] 2 1 x = tan 22.5° = –1 ] PART-B cos3( π + θ⎞cot(3π + θ) sec(θ – 3π) cosec( 3π – θ⎞ Q.11 Simplify | | ⎝ ⎠ tan2 (θ – π) sin(θ – 2π) | | ⎝ ⎠ . [4] [Sol. (– sin3 θ)(cotθ)(– secθ)(– secθ) tan2 θsin θ sin3 θ·cos2 θ·cos θ = – sin 2 θ·sin θ·cos2 θ·sin θ cos θ = – sin θ = – cot θ Ans. ] Q.12 Find the sum of the solutions of the equation 2e2x – 5ex + 4 = 0. [4] [Ans. ln 2] [Sol. 2e2x – 5ex + 4 = 0 let roots be x1 and x2 product of the roots = ex1 ex2 = 4 2 = 2 ex1+x2 = 2 m x1 + x2 = ln 2 Ans. ] Q.13 Prove the identity cot A + cosec A –1 = cot A . [4] [Sol. LHS = cot A – cosec A +1 2 cot A + cosec A – (cosec2 A – cot 2 A) cot A – cosec A +1 = (cot A + cosec A)(1– cosec A + cot A) (1– cosec A + cot A) 1+ cos A 2 cos2 A A = cot A + cosec A = sin A = 2 sin 2 A cos A 2 2 = cot 2 = RHS. Hence proved ] Q.14 If log (log ( log x)) = log (log ( log y)) = 0 then find the value of (x + y). [5] [Ans. 17] [Sol. log (log ( log x)) = 0 ⇒ log (log x) = 1 ⇒ log x = 2 2 2 3 2 3 3 ⇒ x = 9 |||ly log (log ( log y)) y = 8 ⇒ log (log y) = 1 ⇒ log y = 3 ⇒ y = 8 m x + y = 17 Ans. ] (( 1 ⎞2 ⎞ ( 1 ⎞ Q.154 log 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] [Ans. 2a – 3b – 1] [Sol. log 25 = a; log 225 = b 2 log 5 = a ; log(25 · 9) = b or log 25 + 2 log 3 = b ⇒ 2 log 3 = b – a (1) ( 1 ⎞2 ( 1 ⎞ now log| 9 | + log| 2250 | ⎝ ⎠ ⎝ ⎠ = – 2 log 9 – log 2250 ⇒ – 4 log 3 – [log 225 + log 10] = – 2 (b – a) – [b + 1] = – 2b + 2a – b – 1 = 2a – 3b – 1 Ans. ] Q.16 Prove that the expression sin2θ + sin2(120° + θ) + sin2(120° – θ) remains constant 6 θ ϵ R. Find also the value of the constant. [5] [Sol. LHS = sin2θ + 1 – [cos2(120° + θ) – sin2(120° – θ)] = sin2θ + 1 – [cos240° – cos2θ] = sin2θ + 1 + = sin2θ + 1 + 1 cos2θ 2 1 (1 – 2 sin2θ) = 3/2 Ans. ] 2 Q.1722 log Suppose that x and y are positive numbers for which log9x = log12y = log16(x + y). If the value y of x = 2 cos θ, where θ ϵ (0, π 2) find θ. [6] [Ans. π/5] [Hint: Given log9x = log12y = log16(x + y) = k (say) x = 9k ; y = 12k ; x + y = 16k y we have to find x ( 4 ⎞k = | 3 | ⎝ ⎠ now 9k + 12k = 16k ( 4 ⎞k divide by 9k, 1 + | 3 | ( 16 ⎞k = | 9 | ( 4 ⎞2k = | 3 | ; put ( 4 ⎞k | 3 | y = z = x > 0 ⎝ ⎠ ⎝ ⎠ ⎝ ⎠ ⎝ ⎠ m z2 – z – 1 = 0 ⇒ z = 1 5 2 ⇒ z = 5 +1 2 (as 1– 5 2 is rejected); ( 4 ⎞k y y ( 5 +1⎞ π m | 3 | = x = ; m x = | | · 2 = 2 cos . 5 ⎝ ⎠ 2 ⎝ ⎠ Hence θ = π/5 Ans. ] tan θ 1 cot θ Q.18 If tan θ – tan 3θ = 3 , find the value of cot θ – cot 3θ . [6] tan θ 1 [Sol. tan θ – tan 3θ = 3 ⇒ 3 tan θ = tan θ – tan 3θ ⇒ 2 tan θ + tan 3θ = 0 2 tan θ + 3 tan θ – tan3 θ 1– 3 tan2 θ = 0 2(1 – 3 tan2θ) + 3 – tan2θ = 0 (cancelling tan θ throughout) 7 tan2θ = 5 ⇒ tan2θ = 5 7 (1) now, cot θ cot θ – cot 3θ tan 3θ = tan 3θ – tan θ = 3 tan θ – tan3 θ 2 ( 3 tan θ – tan3 θ ⎞ – tan θ| tan θ(3 – tan 2 θ)(1 – 3 tan 2 θ) = tan θ(1 – 3 tan 2 θ)(3 – tan 2 θ –1 + 3 tan 2 θ) | 1– 3 tan2 θ | 3 – tan2 θ = 2(1+ tan2 θ) = tan θ 16 2 = 2·12 = 3 cot θ Ans. Alternatively: Prove that tan θ – tan 3θ + cot θ – cot 3θ = 1 now proceed ] π 2π 3π 4π Q.19 Let S = sec 0 + sec + sec + sec + sec and P = tan π · tan 2π · tan 4π then prove 5 5 5 5 9 9 9 S + P = 2 2 sin 7π [6] 12 π [Sol. S = 1 + 0 ( sec 5 4π + sec 5 2π = 0 and sec 5 3π + sec 5 = 0) now P = tan 20° · tan 40° · tan 80° now sin 20° · sin 40° · sin 80° = 3 and cos 20° · cos 40° · cos 80° = 1 8 8 dividing tan 20° · tan 40° · tan 80° = m P = m S + P = 1 + = (2 2 ) = 2 2 sin 7π 12 Hence proved. ]