Title Differentiation-2 & Plant Physiology- Daily-12 MCQ (Set-A)-With Solve.pdf ✅

2 8. x = t, y = t2 , y2 = ? 4x2 4x3 12x 12x2 DËi: 12x2 e ̈vL ̈v: x 2 = t  x 4 = t2  x 4 = y  y = x 4  y1 = 4x3  y2 = 12x2 9. y = 5x – x 2 eμ‡iLvwU †h mKj we›`y‡Z Aÿ؇qi mv‡_ mgvb †KvY Drcbœ K‡i, Zvi ̄’vbv1⁄4 KZ? (1, 6) I (2, 6) (2, 4) I (2, 6) (2, 6) I (3, 6) (4, 6) I (5, 6) DËi: (2, 6) I (3, 6) e ̈vL ̈v: y = 5x – x 2  dy dx = 5 – 2x †h‡nZz ̄úk©K Aÿ؇qi mv‡_ mgvb †KvY Drcbœ K‡i ZvB, dy dx =  1  5 – 2x =  1  x = 2, 3 x = 2 n‡j, y = 10 – 4 = 6 x = 3 n‡j, y = 15 – 9 = 6  we›`y ̧‡jv (2, 6) I (3, 6) 10. †MvjK R mg‡qi mv‡_ †MvjKwUi AvqZb I e ̈vmva© cwieZ©‡bi nv‡ii AbycvZ R n‡j, †Mvj‡Ki e ̈vmva© KZ? 2 1 2 wbY©q m¤¢e bq 1 4 DËi: 1 4 e ̈vL ̈v: V = 4 3 R 3  dV dt = 4 3    3R2 dR dt  dV dt dR dt = 4R 2  R = 4R 2  R = 1 4 11. y = x 3 – 2x2 + 4 eμ‡iLvi (2, 4) we›`y‡Z Awfj‡¤^i mgxKiY †KvbwU? y – 2 = – 1 4 (x + 4) y – 4 = 4(x – 2) y – 4 = – 1 4 (x – 2) y – 2 = – 1 4 (x + 4) DËi: y – 4 = – 1 4 (x – 2) e ̈vL ̈v: y = x 3 – 2x2 + 4 dy dx = 3x2 – 4x (2, 4) we›`y‡Z dy dx = 3.22 – 4  2 = 4 Awfj‡¤^i Xvj = – 1 4  Awfj‡¤^i mgxKiY: y – y1 = m(x – x1)  y – 4 = – 1 4 (x – 2) 12. f(x) = x – x 2 2 + x 3 3 – x 4 4 + .....  n‡j, f(x) = ? 1 1 + x2 – 1 (1 + x) 2 1 (1 + x) 2 – 1 1 + x2 DËi: – 1 (1 + x) 2 e ̈vL ̈v: Avgiv Rvwb, ln(1 + x) = x – x 2 2 + x 3 3 – x 4 4 + ....   f(x) = ln(1 + x) f(x) = 1 1 + x  f(x) = – 1 (1 + x) 2 13. f(x) = |x| x dvskbwU x = 0 †ZÑ Amxg Awe‡”Q` ̈ we‡”Q` ̈ †Kv‡bvwUB bq DËi: we‡”Q` ̈ e ̈vL ̈v: f(x) = |x| x LHL = lim x  0 – – x x = – 1; RHL = lim x  0 + x x = 1; LHL  RHL  dvskbwU we‡”Q` ̈| 14. y = xlnx eμ‡iLvi †h we›`y‡Z ̄úk©K x A‡ÿi mgvšÍivj Zvi fzR KZ? e – e 1 e – 1 e DËi: 1 e
3 e ̈vL ̈v: y = xlnx y = x . 1 x + lnx = 1 + lnx ̄úk©K x A‡ÿi mgvšÍivj n‡j, Xvj = 0 kZ©g‡Z, 1 + lnx = 0 lnx = – 1 x = e–1 = 1 e 15. x + y = 20 n‡j, xy Gi m‡e©v”P gvb KZ? 10 20 102 wbY©q Am¤¢e DËi: 102 e ̈vL ̈v: x + y = 20 ........ (i) awi, dvskb A = xy; = x(20 – x) dA dx = d dx (20x – x 2 ) = 20 – 2x m‡e©v”P ev me©wb¤œ gv‡bi Rb ̈ dA dx = 0  20 – 2x = 0  x = 10  y = 10  Amax = (xy)max = 102 16. d dx (xx 2 ) = ? x x 2 2(1 + lnx) x x 2 2x(1 + lnx) x x 2 (x + 2xlnx) x x 2–1 . 2x DËi: x x 2 (x + 2xlnx) e ̈vL ̈v: awi, y = x x 2  lny = x2 lnx  1 y dy dx = x 2  1 x + lnx  2x  dy dx = y (x + 2x lnx)  d dx (xx 2 ) = xx 2 (x + 2x lnx) 17. hw` y = secx nq, Z‡e y2 + y Gi gvb †KvbwU? 2 2y 2y2 2y3 DËi: 2y3 e ̈vL ̈v: y = secx  y1 = secx.tanx    ⸪  dy dx = y1  y2 = secx.sec2 x + tanx.secx.tanx = sec3 x + tan2 x.secx = sec3 x + secx(sec2 x – 1) = y 3 + y(y2 – 1)  y2 = 2y3 – y  y2 + y = 2y3 18. y = e–x n‡j y5 †KvbwU? – e –x e –x – 5e–x 5e–x DËi: – e –x e ̈vL ̈v: y = e–x  y1 = – e –x  y2 = e–x  y3 = – e –x  y4 = e–x  y5 = – e –x 19. x y = y x n‡j dy dx = ? x(y lny – y) y(x lny – x) y(x lny – y) x(y lnx – x) y(x lny + y) x(y lnx + x) x(y lnx – y) y(x lny – x) DËi: y(x lny – y) x(y lnx – x) e ̈vL ̈v: x y = y x  lnxy = lnyx  ylnx = xlny  y  1 x + lnx  dy dx = x. 1 y dy dx + lny  dy dx     lnx – x y = lny – y x  dy dx    ylnx – x y = xlny – y x  dy dx = xlny – y x  y ylnx – x = y(x lny – y) x(y lnx – x) 20. y = 2x2 + 3x + 5 eμ‡iLvi (0, 1) we›`y‡Z Awfj‡¤^i Xvj KZ? – 3 – 1 3 1 3 3 DËi: – 1 3 e ̈vL ̈v: dy dx = 4x + 3 (0, 1) we›`y‡Z ̄úk©‡Ki Xvj dy dx = 4  0 + 3 = 3  Awfj‡¤^i Xvj = – 1 3