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

4 21. d n dxn (xn ) Gi gvb †KvbwU? n! x 1 0 DËi: n! e ̈vL ̈v: y = x n  y1 = nxn–1  y2 = n(n – 1)xn–2  y3 = n(n – 1)(n – 2)xn–3  yn = n! 22. y = 4ex + e–x Gi jNygvb KZ? – 4 3 4 5 DËi: 4 e ̈vL ̈v: y = 4ex + e–x  y1 = 4ex – e –x  y2 = 4ex + e–x Pig gv‡bi Rb ̈, y1 = 0  4ex = e–x  4ex = 1 e x  e x = 1 2 [x ev ̄Íe msL ̈v n‡j, e x me©`v abvZ¥K] GLb, e x = 1 2 n‡j, y2 = 4 > 0 ; jNygvb we` ̈gvb|  e x = 1 2 Gi Rb ̈ jNygvb y = 4  1 2 + 2 = 4 23. hw` y = sin–1 x nq, Z‡e y1 y2 Gi gvb †KvbwU? 1 1 – x 2 x 1 – x 2 x 1 – x 2 1 – x 2 x DËi: 1 – x 2 x e ̈vL ̈v: y = sin–1 x y1 = 1 1 – x 2 = (1 – x 2 ) – 1 2 y2 = – 1 2 (1 – x 2 ) – 3 2 . (– 2x) = x ( 1 – x ) 2 3  y1 y2 = 1 1 – x 2  ( 1 – x ) 2 3 x = 1 – x 2 x 24. †Kvb dvskbwUi Rb ̈ (1 – x 2 ) d 2 y dx2 – x dy dx = 2 mZ ̈? y =cos–1 x y = (cos–1 x)2 y = sin–1 x y = tan–1 x DËi: y = (cos–1 x)2 e ̈vL ̈v: Option Test K‡i, y = (cos–1 x)2 dy dx = 2cos–1 x    –  1 1 – x 2  1 – x 2 y1 = – 2cos–1 x  (1 – x 2 )y1 2 = 4(cos–1 x)2  (1 – x 2 ) y1 2 = 4y  (1 – x 2 )2y1y2 – 2xy1 2 = 4y1  (1 – x 2 ) d 2 y dx2 – x dy dx = 2 25. y = sin2x n‡j y99 = KZ? 2 99cos2x 2 100sin2x – 2 99sin2x – 2 99cos2x DËi: – 2 99cos2x e ̈vL ̈v: y = sin2x y = sinax n‡j, yn = an sin    n 2 + ax y99 = 299sin    99 2 + 2x = – 2 99cos2x 26. k Gi gvb KZ n‡j y – k(x – 1)(x + 2) = 0 eμ‡iLvi x = 1 we›`y‡Z ̄úk©K x A‡ÿi mv‡_ 60 †KvY Drcbœ Ki‡e? 1 3 2 3 3 2 3 DËi: 1 3 e ̈vL ̈v: y = k(x2 + x – 2)  dy dx = tan = k(2x + 1)  tan60 = k(2 × 1 + 1) [∵ x = 1]  3 3 = k  k = 1 3 27. hw` y = x 3 lnx nq, Z‡e d 4 y dx4 = ? 4 x 1 x 6 x 6 x 2 DËi: 6 x e ̈vL ̈v: y = x 3 lnx y1 = 3x2 lnx + x2 y2 = 6x lnx + 3x + 2x = 6x lnx + 5x y3 = 6 lnx + 6 + 5 = 6 lnx + 11 y4 = 6 x 28. x Gi †Kvb gv‡bi Rb ̈ F(x) =  x 0 t – 4 9 – t 2 dt dvskbwUi gvb e„nËg n‡e? 4 9 3 – 4 DËi: 4 e ̈vL ̈v: F(x) = x – 4 9 – x 2 = 0  x = 4