Content text Inverse Engineering Practice Sheet Solution.pdf
wecixZ w·KvYwgwZK dvskb I w·KvYwgwZK mgxKiY Engineering Practice Sheet Solution 1 07 wecixZ w·KvYwgwZK dvskb I w·KvYwgwZK mgxKiY Inverse Trigonometric Functions and Trigonometric Equations WRITTEN weMZ mv‡j BUET-G Avmv cÖkœvejx 1. r x wP‡Î, f() = x r mgvavb Ki: 1 f(2) – 1 f() = 2; – [BUET 23-24] mgvavb: GLv‡b, f() = x r = cos GLb, 1 f(2) – 1 f() = 2 1 cos2 – 1 cos = 2 cos – cos2 cos2.cos = 2 cos – cos2 = 2cos2 cos cos– cos2 = cos3 + cos cos3 + cos2 = 0 2cos 5 2 cos 2 = 0 nq, cos 5 2 = 0 5 2 = (2n + 1) 2 = (2n + 1) 5 cos 2 = 0 2 = (2n + 1) 2 = (2n + 1) – e ̈ewa‡Z = 5 , 3 5 , (Ans.) 2. mgvavb Ki: cos3 x – 1 2 sin2x = 1 + sin3 x [BUET 22-23] mgvavb: cos3 x – sin3 x = 1 + 1 2 sin2x (cosx – sinx)(cos2 x + sin2 x + sinxcosx) = 1 + sinxcosx (cosx – sinx)(1 + sinxcosx) – (1 + sinxcosx) = 0 (1 + sinxcosx)(cosx – sinx – 1) = 0 nq, cosx – sinx = 1 1 2 cosx – 1 2 sinx = 1 2 cos x + 4 = cos 4 x + 4 = 2n 4 x = 2n, 2n – 2 ; n Z (Ans.) A_ev, 1 + sinxcosx = 0 sinxcosx = – 1 2sinxcosx = – 2 sin2x = – 2 wKš‘, sin2x – 2 3. log4 (sinx – cos2x)3 = 2 tan–1– 5 2 + tan–13 7 n‡j, x Gi gvb wbY©q Ki| hLb [– 2 x 2] [BUET 21-22] mgvavb: †`Iqv Av‡Q, log4 (sinx – cos2x)3 = 2 tan–1– 5 2 + tan–13 7 3 log4 (sinx – cos2x) = 2tan–1 (– 1) 3 log4 (sinx – cos2x) = 2 × 3 4 log4 (sinx – cos2x) = 1 2 sinx – cos2x = 4 1 2 sinx – cos2x = 2 sinx + 2sin2 x – 1 = 2 [∵ 1 – cos2x = 2sin2 x] 2sin2 x + sinx – 3 = 0 2sin2 x + 3sinx – 2sinx – 3 = 0 sinx(2sinx + 3) – 1(2sinx + 3) = 0 (sinx – 1)(2sinx + 3) = 0 nq, sinx – 1 = 0 A_ev, 2sinx + 3 = 0 sinx = 1 sinx = – 3 2 wKš‘ sinx – 3 2 x = (4n + 1) 2 ; n Z [– 2 x 2] mxgvi g‡a ̈ MÖnY‡hvM ̈ gvb ̧‡jv n‡jv: x = 2 , – 3 2 (Ans.)
2 Higher Math 2nd Paper Chapter-7 4. sin(x) + sin x 2 = 0 n‡j, x = ? [0 x 2] [BUET 20-21] mgvavb: sin(x) + sin x 2 = 0 2sinx 2 cos x 2 + sinx 2 = 0 sinx 2 2cos x 2 + 1 = 0 nq, cos x 2 = – 1 2 cos x 2 = cos 2 3 x 2 = 2n 2 3 x = 4n 4 3 ; n Z n = 0 n‡j, x = 4 3 ∵ – 4 3 MÖnY‡hvM ̈ bq Avevi, sinx 2 = 0 x 2 = n x = 2n ; n Z n = 0 n‡j, x = 0 n = 1 n‡j, x = 2 wb‡Y©q gvb, x = 0 , 4 3 , 2 (Ans.) 5. mgvavb Ki: sin–1 2x + sin–1 x = 3 [BUET 18-19] mgvavb: sin–1 2x + sin–1 x = 3 sin–1 2x = 3 – sin–1 x 2x = sin 3 – sin–1 x 2x = sin 3 cos(sin–1 x) – cos 3 sin(sin–1 x) 2x = 3 2 1 – x 2 – 1 2 x 5x 2 = 3 2 1 – x 2 25x2 = 3 – 3x2 28x2 = 3 x = 3 28 ïw× cixÿv K‡i cvB, x = 3 28 (Ans.) 6. mgvavb Ki: cos–1 x – sin–1 x = sin–1 (1 – x) [BUET 17-18; MIST 22-23] mgvavb: †`Iqv Av‡Q, cos–1 x – sin–1 x = sin–1 (1 – x) sin–1 ( 1 – x ) 2 – sin–1 (x) = sin–1 (1 – x) sin–1 ( 1 – x ) 2 . 1 – x 2 – x 1 – 1 + x2 = sin–1 (1 – x) sin–1 (1 – x 2 – x 2 ) = sin–1 (1 – x) sin–1 (1 – 2x2 ) = sin–1 (1 – x) 1 – 2x2 = 1 – x 2x2 = x x(2x – 1) = 0 x = 0 A_ev, x = 1 2 wb‡Y©q mgvavb, x = 0 , 1 2 (Ans.) 7. mgvavb Ki: 2(sinx cosx + 3) = 3cosx + 4sinx, 0 < x < [BUET 17-18; MIST 16-17] mgvavb: 2sinx cosx + 2 3 – 3cosx – 4sinx = 0 2sinx(cosx – 2) – 3(cosx – 2) = 0 (2sinx – 3)(cosx – 2) = 0 2sinx – 3 = 0 A_ev, cosx – 2 = 0 wKš‘, cosx 2 sinx = 3 2 = sin 3 x = n + (– 1)n 3 ; n Z n 0 1 2 x 3 2 3 7 3 cÖ`Ë e ̈ewa‡Z wb‡Y©q mgvavb, x = 3 2 3 (Ans.) 8. mgvavb Ki: cotx + cot2x + cot3x = cotx cot2x cot3x [BUET 14-15] mgvavb: cotx + cot2x + cot3x = cotx cot2x cot3x cot2x + cotx = cot3x(cotx cot2x – 1) cot2x + cotx cotx cot2x – 1 = cot3x 1 cot(x + 2x) = cot3x tan3x = 1 tan3x tan 2 3x = 1 tan3x = 1 = tan 4 3x = n 4 x = n 3 12 ; n Z (Ans.) 9. mgvavb Ki: tan–1 x + 2cot–1 x = 2 3 [BUET 10-11] mgvavb: tan–1 x + cot–1 x + cot–1 x = 2 3 2 + cot–1 x = 2 3 cot–1 x = 6 = cot–1 3 x = 3 (Ans.)
wecixZ w·KvYwgwZK dvskb I w·KvYwgwZK mgxKiY Engineering Practice Sheet Solution 3 10. mgvavb Ki: sin + sin2 + sin3 = 1 + cos + cos2; 0 < < [BUET 08-09; RUET 15-16] mgvavb: sin + sin2 + sin3 = 1 + cos + cos2 2sin2cos + sin2 = 2cos2 + cos sin2(2cos + 1) = cos(1 + 2cos) (1 + 2cos)(sin2 – cos) = 0 nq, 1 + 2cos = 0 cos = – 1 2 cos = cos 2 3 = 2n 2 3 A_ev, sin2 – cos = 0 2sincos – cos = 0 cos(2sin – 1) = 0 cos = 0 Avevi, sin = 1 2 = (2n + 1) 2 sin = sin 6 = n + (– 1)n 6 = (2n + 1) 2 , n + (– 1)n 6 , 2n 2 3 ; n Z 0 < < e ̈ewa‡Z Gi MÖnY‡hvM ̈ gvb: 2 , 6 , 2 3 , 5 6 (Ans.) 11. mgvavb Ki: 4cosx cos2x cos3x = 1; 0 < x < [BUET 08-09] mgvavb: 4cosx cos2x cos3x = 1 2cos2x(cos4x + cos2x) = 1 2cos2x cos4x + 2cos2 2x – 1 = 0 2cos2x cos4x + cos4x = 0 cos4x (2cos2x + 1) = 0 nq, cos4x = 0 4x = (2n + 1) 2 x = (2n + 1) 8 ; n Z A_ev, 2cos2x + 1 = 0 cos2x = – 1 2 = cos 2 3 2x = 2n 2 3 x = n 3 ; n Z x = 8 , 3 , 3 8 , 2 3 , 5 8 , 7 8 (Ans.) 12. mgvavb Ki: 3sinx – cosx = 2, 0 < x < 2 [BUET 07-08] mgvavb: 3sinx – cosx = 2 3 2 sinx – 1 2 cosx = 1 [Dfqcÿ‡K 2 Øviv fvM K‡i] sinx cos 6 – cosx sin 6 = 1 sin x – 6 = 1 x – 6 = (4n + 1) 2 x = (4n + 1) 2 + 6 ; n Z wb‡Y©q mgvavb : x = 2 3 (Ans.) 13. mgvavb Ki: 3 sin2x – 1 cos2x = 4 [BUET 06-07; MIST 17-18] mgvavb: 3 sin2x – 1 cos2x = 4 3cos2x – sin2x sin2x.cos2x = 4 3 2 cos2x – 1 2 sin2x = 2sin2x.cos2x sin 3 cos2x – cos 3 sin2x = 2sin2x.cos2x sin 3 – 2x = sin4x – sin 2x – 3 = sin4x sin4x + sin 2x – 3 = 0 2sin 3x – 6 cos x + 6 = 0 nq, sin 3x – 6 = 0 3x – 6 = n x = (6n + 1) 18 A_ev, cos x + 6 = 0 x + 6 = (2n + 1) 2 x = (3n + 1) 3 wb‡Y©q mgvavb: x = (6n + 1) 18 , (3n + 1) 3 ; n Z (Ans.) 14. mgvavb Ki: tan–1 1 – x 1 + x = 1 2 tan–1 x [BUET 06-07] mgvavb: tan–1 1 – x 1 + x = 1 2 tan–1 x tan–1 1 – tan–1 x = 1 2 tan–1 x tan–1 1 = 3 2 tan–1 x = 4 tan–1 x = 2 12 x = tan 6 = 1 3 (Ans.)
4 Higher Math 2nd Paper Chapter-7 15. mgvavb Ki: 1 + sin2 + sin2 = cos(2 + 2) [0 , 90] [BUET 05-06] mgvavb: 1 + sin2 + sin2 = cos(2 + 2) 2sin( + )cos( – ) + 1 – cos2( + ) = 0 2sin( + )cos( – ) + 2sin2 ( + ) = 0 2sin( + ){cos( – ) + sin( + )} = 0 2sin( + ) = 0 + = n ⸪ 0 , 90 + = 0, 180 = = 0 (Ans.) A_ev, = = 90 (Ans.) cos( – ) + sin( + ) 0 [0 , 90 Gi g‡a ̈ I Gi †Kv‡bv gv‡bi R‡b ̈B cos( – ) = 0 n‡Z cv‡i bv] 16. mgvavb wbY©q Ki: 1 – 2sin = cos [BUET 05-06] mgvavb: cos + 2sin = 1 1 5 cos + 2 5 sin = 1 5 coscos + sinsin = cos cos( – ) = cos – = 2n 5 1 2 awi, 1 5 = cos sin = 2 5 = 2n + 2, 2n ; n Z †hLv‡b, = cos–1 1 5 (Ans.) 17. mgvavb Ki: cos + sin = 2 [BUET 02-03] mgvavb: cos + sin = 2 1 2 cos + 1 2 sin = 2 2 sin 4 cos + cos 4 sin = 1 sin 4 + = 1 4 + = (4n + 1) 2 = (4n + 1) 2 – 4 = 2n + 4 ; n Z (Ans.) weMZ mv‡j KUET-G Avmv cÖkœvejx 18. cÖgvY Ki †h, costan–1 cotsin–1 x = x (wPÎ Avek ̈K) [KUET 19-20] mgvavb: L.H.S = costan–1 cotsin–1 x = costan–1 cotcot–1 1 – x 2 x = costan–1 1 – x 2 x = cos cos–1 x 1 = x = R.H.S (Proved) x 1 – x 2 1 x 1 – x 2 1 19. cÖgvY Ki: cos–1 x = 2sin–1 1 – x 2 = 2cos–1 1 + x 2 [KUET 03-04] mgvavb: awi, cos–1 x = cos = x GLb, sin 2 = 1 – cos 2 = 1 – x 2 2 = sin–1 1 – x 2 = 2sin–1 1 – x 2 cos–1 x = 2sin–1 1 – x 2 Avevi, cos 2 = 1+ cos 2 = 1 + x 2 2 = cos–1 1 + x 2 = 2cos–1 1 + x 2 cos–1 x = 2cos–1 1 + x 2 cos–1 x = 2sin–1 1 – x 2 = 2cos–1 1 + x 2 (Proved) Shortcut: 1 2 tan–1 x = tan–1 1 + x2 – 1 x 1 2 sin–1 x = tan–1 1 – 1 – x 2 x 1 2 cos–1 x = cos–1 1 + x 2 = sin–1 1 – x 2 = tan–1 1 – x 1 + x 20. mgvavb Ki: tan2 = 3cosec2 – 1 for 0 2 [KUET 03-04] mgvavb: tan2 = 3cosec2 – 1 1 + tan2 = 3cosec2 sec 2 = 3cosec2 sin2 cos 2 = 3 tan2 = 3 tan = 3 GLb, tan = 3 n‡j, tan = tan 3 = n + 3 ; n Z Avevi, tan = – 3 n‡j, tan = – tan 3 = n – 3 ; n Z cÖ`Ë e ̈ewa‡Z = 3 , 4 3 , 2 3 , 5 3 (Ans.)