if , passing through (1, 2), satisfies the differential equation , then&#8230;.
if , passing through (1, 2), satisfies the differential equation , then
if , passing through (1, 2), satisfies the differential equation , then
y = ex + 1 differentiating both sides of this equation with respect to x, we get: thus, the given function is the solution of the corresponding differential equation. a=c and b=d
b=d and `anec`
`ab=1 " and "cdne1`
`cd=1" and "ab=ne1`
answer: a
solution: using `a.m.gtg.m,` we have   
 `a+(1)/(b)gt2sqrt((a)/(b))b+(1)/( c )gt2sqrt((b)/( c )),c+(1)/(d)gt2sqrt(( c )/(d))" and "d+(1)/(a)gt2sqrt((d)/(a))`   
 `implies(a+(1)/(b))(b+(1)/( c ))(c+(1)/(d))(d+(1)/(a))gt16`   
 but,`(a+(1)/(b))(b+(1)/( c ))(c+(1)/(d))(d+(1)/(a))=4xx1xx4xx1=16`   
 `:." "a=(1)/(b),b=(1)/( c ),c=(1)/(d)" and "d=(1)/(a)`   
 `implies" "a=(1)/(b)=2,b=(1)/( c )=(1)/(2),c=(1)/(d)=2" and "d=(1)/(a)=(1)/(2)`   
 `implies" "a=2,b=(1)/(2),c=2" and "d=(1)/(2)`   
 `implies" "a=c" and "b=d` 1) p(x) = 1/x2 + 1 if x ≤ 3 it is clear that x= 3 is included, so you have to use this function.2) loo k at table d, and compare the given information with the data written inside the table. to find: limitsnote: first check the form of imit. used this method if the limit is satisfying any one from 7 indeterminate form.in this case, indeterminate form is formula used: = 1so, by using the above formula, we have = = therefore, = given;let ⇒ p(k+1) is true when p(k) is true.∴ by mathematical induction is true for all natural number n≥2. it is given in the question that,
from class a students who passed in first class = = 

also,
from class b students who passed in first class = = 

therefore, from above results it is clear that in the both the classes equal fraction of students passed in the first class solution: given that, `sinx+cosx=a`   
 on squaring both sides, we get   
 `" "(sinx+cosx)^(2)=(a)^(2)`   
 `rarr" "sin^(2)x+cos^(2)x+2sinxcosx=a^(2)`   
 `rarr" "sinx*cosx=(1)/(2)(a^(2)-1)`   
 `(i) sin^(6)x+cos^(6)x=(sin^(2)x)^(3)+(cos^(2)x)^(3)`   
 `" "=(sin^(2)x+cos^(2)x)(sin^(4)x-sin^(2)xcos^(2)x+cos^(4)x)` ltbrgt `" "=sin^(4)x+cos^(4)x-(1)/(4)(a^(2)-1)^(2)`   
 `" "=(sin^(2)x+cos^(2)x)^(2)-2sin^(2)xcos^(2)x-(1)/(4)(a^(2)-1)^(2)`   
 `" "=1-2*(1)/(4)(a^(2)-1)^(2)-(1)/(4)(a^(2)-1)^(2)=(1)/(4)[4-3(a^(2)-1)^(2)]`   
 `(ii) abs(sinx-cosx)=sqrt((sinx-cosx)^(2))`   
 `" "=sqrt(sin^(2)x+cos^(2)x-2sinxcosx)`   
 `" "=sqrt(1-2(1)/(2)(a^(2)-1))=sqrt(1-a^(2)+1)=sqrt(2-a^(2))` formula :- therefore ,put -2x dx = dt several ways to do this, here's one:

to go from (-6,4) to (-2,5) you need to go 4 in the x direction and 1 in the y direction. if you move again 4 in the x direction and 1 in the y direction you end up at (2, 6). that's p2 let a (x, y) be the point on the curve y = f(x). then slope of the tangent to the curve at point a is `"dy"/"dx"`. according to the given condition, `"dy"/"dx" = "x" + 3"y" - 1` ∴ `"dy"/"dx" - 3"y" = "x - 1"` ....(1) this is the linear differential equation of the form `"dy"/"dx" + "py" = "q"`, where p = - 3 and q = x - 1 ∴ i.f. = `"e"^(int "p dx") = "e"^(int - 3"dx") = "e"^(- 3"x")` ∴ the solution of (1) is given by `"y" * ("i.f.") = int "q" * ("i.f.") "dx" + "c"` ∴ `"y" * "e"^(- 3"x") = int ("x - 1") * "e"^(-3"x") "dx" + "c"` ∴ `"e"^(- 3"x") * "y" = ("x - 1") int "e"^(- 3"x") - int ["d"/"dx" ("x - 1") * int "e"^(- 3"x") "dx"] "dx" + "c"\_1` ∴ `"e"^(- 3"x") * "y" = ("x - 1") * "e"^(- 3"x")/-3 - int 1 * "e"^(- 3"x") * "y"/-3 "dx" + "c"\_1` ∴ `"e"^(- 3"x") * "y" = - 1/3 ("x - 1") * "e"^(- 3"x") + 1/3 int "e"^(- 3"x") "dx" + "c"\_1` ∴ `"e"^(- 3"x") * "y" = - 1/3 ("x - 1") "e"^(- 3"x") + 1/3 * "e"^(- 3"x")/-3 + "c"\_1` ∴ `"e"^(- 3"x") * "y" = - 1/3 ("x - 1")"e"^(- 3"x") - 1/9 "e"^(- 3"x") + "c"\_1` ∴ 9y = - 3(x - 1) - 1 + 9`"c"\_1 * e^("3x")` ∴ 9y + 3(x - 1) + 1 = `9"c"\_1 * e^("3x")` ∴ 9y + 3x - 3 + 1 = `9"c"\_1 * e^("3x")` ∴ 3(x + 3y) = 2 + `9"c"\_1 * e^("3x")` ∴ 3(x + 3y) = 2 + `"c" * e^("3x")` where c = 9c1 ....(2) this is the general equation of the curve. but the required curve is passing through the origin (0, 0). ∴ by putting x = 0 and y= 0 in (2), we get 0 = 2 + c ∴ c = - 2 ∴ from (2), the equation of the required curve is 3(x + 3y) = 2 - 2`e^("3x")` i.e. 3(x + 3y) = 2(1 - `e^("3x")`).