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Shaikh Sir’s
Reliance Academy, Ghorpade galli, Near Malabar Hotel,
Shahupuri, Station road, Kolhapur
Maths-II {Engg. Maths}
Question wise Question bank
Shaikh sir’s Reliance Academy,
Shahupuri, Kolhapur.
Subject : Maths – II (G-Scheme)
Question : 1 This question contains 12 questions from all chapters, of 2 marks each Q1] Solve any Ten . [ 20 Marks ] a] Complex number b] Complex Number c] Function d] Function e] Limits {algebraic} f] Limits {Exponential} g] Limits {Trigonometric} h] Derivative i] Derivative j] Derivative k] Numerical Method{algebraic equation} l] Numerical Method{Simultaneous equation}
Types of Problems No. Type of Problem Q.B.Checked
1 Algebra of Complex number
2 Even or odd function
3 Simple algebraic function
4 Simple trigonometric functions
5 Limit of simple algebraic function
6 Limit of simple trigonometric function
7 Exponential function limx→0 kx
a −1kx
8 Exponential function [1 ]limx→0
+ 1kx
kx
9 Derivative:Product and division rule
10 Derivative: Composite function
11 Derivatives : parametric function
12 Derivatives: (function)function
13 Numerical :Check Root lies between
14 Numerical:One iteration solution
Type 1: Algebra Of complex Numbers 1. If (3+i)x +(1i) y =1+7i Find the value of x and y. 2. If (4+i)x +(3+i) y =2+8i Find the value of x and y. 3. If (2i)x +(4i) y =45i Find the value of x and y. 4. If x(1i)+y(2+i)=0 Find the value of x and y. 5 b. If x(2i)+y(4+i)=12i Find the value of x and y. 6. Express in the form of a+ib where a,b ,1−i
2+i R ∈ i = √− 1 7. Express in the form of a+ib where a,b ,3−i
2−3i R ∈ i = √− 1 8. Express in the form of a+ib where a,b ,10
3+4i R ∈ i = √− 1 9. Express in the form of a+ib where a,b ,1−i
1+i R ∈ i = √− 1 10. Express in the form of a+ib where a,b ,1
2+3i R ∈ i = √− 1 11. Show that is a real number. 1 + i10 + i20 + i30 12. Find the value of i i 2 6 − 3 8 + i7 13. Simplify i i i 2 2 − 3 7 + 4 6 + 2 14. Simplify i49 + i68 + i89 + i110 15.If 1 2 i) and z2 4 i) z = ( + 3 = ( + 5 Find the values of ) z1 2 2) z2 1 3) z1.z2 4) z1/z2 1 + z − z 16. Given Find 1 − i and z2 i z = 3 + 4 = 5 − 3 1
z1 +1z2
17.Given Find 1 and z2 i z = 3 − i = 5 + 4 1z1 +
1z2
18. Express the following in the form a+ib, 12−3i +
5−i6+2i
19.Express as a+ib4i −3i +38 9
3i −4i −211 10
20. If z= .i show that z z 1 + √3 2 + 4 = 2 Type 2.Even and odd function 1. State the function is even or odd (x) sinxf = x x7 − 5 5 + 3 2. State the function is even or odd (x) sinxf = x x4 − 5 2 + x 3. State the function is even or odd (x) cos x sin x f = 3x2 + x2 + 5 − 3 + 2 2 4. State the function is even or odd (x) f = cosx
1+sin x2 5. State the function is even or odd 2
a +ax −x 6. State the function is even or odd cos x sin x 3x4 + x2 + 5 − 3 + 2 2 7. state the function is even or odd og(x l +√x )2 + 1 Type 3.Simple algebraic function 1. , Find f(0) +f(3)f f(x) x I = x3 − 3 2 + 5 2. f f(x) find f(1) (2) I = x3 + x + f 3. f f(x) og x find f(1/2) I = 16x + l 4 4. f f(x) x , t find f(t) I = x2 + 6 − 8 = z + 2 5. f f(x) find f(x ) (x ) I = x2 + 4 + 1 − f − 1 6. f f(x) x, find ( ) I = 16x + log2 4
1 7. f f(x) x , find f(x ) I = x2 + 2 + 1 − 1 8. f f(x) x show that f(− ) .f(1) I = 3x2 − 5 + 7 1 = 3 9. f f(x) x inx cosx show that f(x) (− ) I = x3 − 3 + s + x + f x = 0
Type 4. Simple trigonometric function 1. f f(x) os x f(3x) [f(x)] f(x) I = c = 4 3 − 3 2. f f(x) inx show that f(3) f(x) f (x) I = s = 3 − 4 3 3. f f(x) anx show that f(2x) I = t = 2f(x)
1−[f(x)]2
4. f f(x) anx show that f(x ) I = t + y = f(x)+f(y)1−f(x).f(y)
Type 5. Limit of Simple algebraic limit 1. 2. 3. lim
x→2x +x−62
x +2x−82 limx→1 x −4x+3
2x +3x−42 lim
x→1x −8x+72
7x −6x−12
4. 5. 6. limx→3
x −4x+32
2x −3x−92 limx→7 x +7x−98
2x −18x+772 lim
x→−1 x +13x −3x−42
7. 8. 9. limx→2 x−2
x −42 limx→1 x−1
x −13 limx→5 x −25
2x −1253
Ans 1).5/6 2).5/2 3).3/4 4).2/9 5).4/21 6).5/3 7). 4 8). 3 9). 15/2 Type 6. Limits of Simple Trigonometric limit 1. 2. 3. lim
x→0 xsin3x lim
x→0sin4xtan5x lim
x→0 x2sin3x tan4x
4. 5. 6. limx→0 x2
1−cos2x limx→0 x2
1−cos8x limx→0 1−cos4x
sin4x tan3x
Type 7. Limits of Exponential function lim
x→0 kxa −1kx
1. limx→0 x
2 −15x2. lim
x→0 2 −1
x
sin2x 3. limx→0
3xa −14x
4. lim
x→0 tan4x3 −12x
5. 6. 7. limx→0
x3 −5x x
limx→0
2x3 −1tanx
limx→0
5xe −14x
Type 8. Limits of Exponential function [1 ]lim
x→0+ 1kx
kx
1. [1 ]limx→0
+ 32x x
52. [1 ]lim
x→∞ + x
1 3x 3. [1 ]limx→0
+ 35x
2x
4. [1 ]lim
x→0 − 3
4x x5
5. [1 ] limx→∞
+ 15x
x 6. [1 ]limx→∞
+ x2 x
7. ( )limx→0 x
x+1 x 8. 9. ( )limx→0 x
x−1 x ( )limx→0
xx+1
x
Type 9. Derivatives: Product rule and division rule
1. inx. logx y = s 2. 3. anx. e y = t x . tanx y = √x 4. 5. 6. in x.x y = s −1 x. x y = tan−1 . logx y = x3
7. 8. 9. y = x .5 √x ecx. tanx. y = s y = xlogx
10. 11. 12. y = exsinx y = x
√x y = extanx 13.
14. 15. y = √xlogx y = x
sin x−1 y = x.2x
16. 17. y = 5xlogx .sinx.cosx y = ex
Type 10. Derivatives : Composite function {Chain rule} 1. og(sin x) y = l 2. in(log x) y = s 3. og(e ) y = l x 4. )in( y = s √x 5. an (sinx) y = t −1 6. x. y = sin3 7. ec y = s √x 8. 9. ot ( ) y = c −1 √x y = etanx 10. og(tan x) y = l 11. osec( ) y = c √x 12. y = 2logx 13. (sinx) y = cos−1 14. y = elogx 15. (secx) y = tan−1
16. 17. 18. (tanx) y = cosec−1 in (log (sec x)) y = s og (tan (e )) y = l x 19. og (tan (3 x)) y = l − 4 Type 11. Derivatives : Product & div rule with composite function
1. find dy/dx .tan2x e3x 2. Find dy/dx og3x.sin4x l 3. find dy/dx sin4x . e3x
Type 12. Derivatives : Parametric Functions 1) ind if x at and yF dx
dy = 3 2 = 2at3 2) ind if x at and y t F dx
dy = 2 = a 3 3) ind if x inθ and y osθF dx
dy = s = c 4) ind if x cosθ and y sinθF dx
dy = a = b 5) ind if x (θ inθ) and y (1 osθ)F dx
dy = a + s = a − c 6) ind if x secθ and y tanθF dx
dy = a = a Type 13. Derivatives : (function)function
1) ind if yF dxdy = xx
2) ind if yF dxdy = xsinx
3) ind if yF dxdy = xcosx
4) ind if yF dxdy = xlogx
5) ind if yF dxdy = xtanx
6) ind if yF dxdy = xsecx
Type 14. Numerical Method :To find root lies between
1) Show that the root of the eqn. lies between 0 & 1.e x x = 1 2) Shown that there exist a root of the equation in thex x3 − 4 + 1 = 0
interval (1,2) 3) Find first two real roots of equation, using bisectionx x3 − 2 − 5 = 0
method Type 15. Numerical method : Solve one iteration
1) Find the first iteration by using Jacobi’s method for the following equation. 0x z 7, 3x 0y − 8, 2x y 0z 5 2 + y − 2 = 1 + 2 − z = 1 − 3 + 2 = 2
2) Find the first iteration by using Jacobi’s method for the following system of equation x , x y − , y z 5 − y = 9 − 5 + z = 4 − 5 = 6
3) Find the first iteration by using Jacobi’s method for the following system of equation x 0, x y , x z − 5 − y + z = 1 + 2 = 6 + y + 5 = 1
Shaikh sir’s Reliance Academy,
Shahupuri, Kolhapur.
Subject : Maths – II (G-Scheme)
Question : 2 This question contains 6 questions from following chapters Q2] Solve any Four . [ 16 Marks ] a] Complex Number {Algebra of Complex number} b] Complex Number {Simplify De-Moivre Theorem} c] Complex Number {Problem on Euler's Formula} d] Complex Number {Proof using De-Moivre's theorem/Find Roots} e] Function { Function of function type problem } f] Function {Logarithmic functions}
Types of Problems
No. Type of Problem Q.B.Checked Revision
1 Simplify Using D.M.Theorem [ ] [ ] [ ] [ ] [ ]
2 Function of function type problem [ ] [ ] [ ] [ ] [ ]
3 Logarithmic functions [ ] [ ] [ ] [ ] [ ]
4 Prove Using Euler's Formula [ ] [ ] [ ] [ ] [ ]
5 Algebra of complex no. [ ] [ ] [ ] [ ] [ ]
6 Express in Polar Form [ ] [ ] [ ] [ ] [ ]
7 Proofs using D.M.theorem [ ] [ ] [ ] [ ] [ ]
8 Find roots of complex number [ ] [ ] [ ] [ ] [ ]
Type 1: Simplify Using D.M. Theorem 1. Simplify (cos 2θ+ i sin 2θ) (cosθ − isinθ)3
(cos 3θ + isinθ) (cos 5θ − isin 5θ)2 4
2. Simplify (cosθ − isinθ) (cos 2θ + isin 2θ)3 3
(cos 3θ − isin3θ) (cos 5θ − isin 5θ) 2
3. Simplify (cos θ +isin θ) (cos θ − isin θ) 41
41 4
32
32 3
(cos θ − isin θ) (cos θ + isin θ)52
52 5
72
72
7
4. Simplify (cos θ +isin θ) (cos θ +isin θ) 53
53 5
54
54 10
(cos 3θ +isin 3θ) (cos 5θ − isin 5θ) 4 54
5. Simplify (cos 4θ +isin 4θ) (cos 5θ+isin 5θ)3 −4
(cos 3θ + isin 3θ) (cos 4θ − isin 4θ)4 5
6. Simplify (cos 4θ + isin 4θ) (cosθ − isinθ) 7 8(cos 2θ − isin 2θ) (cos 3θ+isin 3θ)5 6
7. Simplify (cos 5θ − isin5θ ) (cos θ + isin θ) 52
72
72 7
(cos 4θ +isin 4θ) (cos θ − isin θ) 41
32
32 3
Type 2.Function of function type problem 1. f f (x) & t , f ind value of f (t). I = 3x−4
2x+5 = 3x−25+4x
2. f f (x) & t , show that f (t) . I = x+24x−3 = 4x−1
2+3x = x 3. f y (x) , show f (y) . I = f = 3x−2
2x−3 = x 4. f y (x) , show f (y) . I = f = x−5
5x−1 = x 5. f f (x) , show that f{f (x)} . I = 5x−3
3x+4 = x 6. f ϕ(x) , f ind ϕ [ϕ (x)]I = x−2
2x−3
Type 3.Logarithmic functions 1. f f (x) log[ ], show that f (a) (b) [ ] I = 1−x
1+x + f = f a+b1+ab
2. f f (x) log[ ], show that f (y) (z) [ ]I = 1−x1+x − f = f y−z
1−yz 3. f f (x) log[ ], show that f [ ] .f (x) I = 1−x
1+x 2x1+x2 = 2
4. f f (x) log[ ], show that f (y) (− ) (y ) I = xx−1 + f y = f 2
5. f f (x) log[ ], show that f (y) (z) [ ]I = 1−x1+x + f = f y+z
1+yz 6. f f (x) log[x ], show that, f (x) (− ) I = + √x2 + 1 + f x = 0 7. f f (x) log[ ], show that f (a ) (a) log[ ] I = x
x−1 + 1 + f = a−1a+1
Type 4.Prove Using Eulers Formula
1. sing Euler s formula prove that, sin θ os θ u ′ 2 + c2= 1
2. sing Euler s formula prove, cosh θ inh θ u ′ 2 − s2= 1
3. rove using Euler s formula sin2θ sinθ.cosθ P ′ = 2 4. rove using Euler s formula cos2θ θ P ′ = cos θ in2 − s 2 5. rove by Euler s formula sinh2θ sinhθ.coshθ P ′ = 2 6. rove by Euler s formula cosh2x cosh x P ′ = 2 2 − 1 7. sing Euler s formula show that, u ′ & . cos3x cos x cosx 1 = 4 3 − 3 . sin3x sinx sin x 2 = 3 − 4 3 8. rove that 1 θ ec θ. using Euler s formula. P + tan2 = s 2 ′ 9. sing Euler s formula prove, u ′ . cosA osB cos( ). cos( ) 1 + c = 2 2
A+B2A−B
. sinα inβ cos( ).sin( ) 2 − s = 2 2α+β
2α−β
10. f x y in(A B) prove that, I + 1 = s + i . 1 x2
cosh B2 + y2sinh B2 = 1
. 2 x2sin A2 − y2
cos A2 = 1 11. os(x y) B prove that, c − 1 = A + i . 1 A2
cosh y2 + B2sinh y2 = 1
. 2 A2cos x2 − B2
sin x2 = 1 Type 5. Algebra of complex no. 1. Find modulus and argument of (3−i)
(2+2) (2−2) 2. f z − i, z i, f ind I 1 = 3 + 4 2 = 5 − 3 1
z1+ 1z2
3. Find real and imaginary part of where ,z + z−1 z = 1−i3+4i
4. Separate into real and imaginary part 3+23−5i + 3+5i3−2i
5. Find real and imaginary part, (2+i)2
(2+3i) 6. xpress in x y f rom E + 2 3i −2i +24 5 8
2−3i +i +4i5 7 3
7. xpress in a b f rom E + i 4i−3i +39
3i −4i −211 10 Type 6.Express in Polar Form
.Find modulus and arguments of i, and express in polar form 1 2−1 + 2
√3 . Express in polar form z i 2 = 2
1 − 2√3
. Express in polar form 3 1−3i1+2i
. Express in polar form 2 i 4 + 2√3
. Express in polar form [ ] 5 3−i2+i 2
. Express in polar form 1 osα sinα 6 − c + 2 Type 7. Proofs using D.M.theorem 1. rove that, (1 ) 1 ) 2. Using De Moivre s theorem P + i 8 + ( − i 8 = 3 ′ 2. how using D.M theorem. (1 ) 1 ) − 28 S + i 12 + ( − i 12 = 1 3. how that (1 ) 1 ) S + i√3 15 + ( − i√3 15 = − 216
4. rove that, ( ) P √3 + i14+ ( )√3 − i
14= 214
Type 8. To find roots of Complex Number 1. Find the cube root of unity. 2. Find the cube root of 1 3. Solve the equation x3 + 1 = 0 4. Solve the equation x4 + 1 = 0
Shaikh sir’s Reliance Academy,
Shahupuri, Kolhapur.
Subject : Maths – II (G-Scheme)
Question : 3 This question contains 6 questions from following chapters Q3] Solve any Four . [ 16 Marks ] a] Functions {Algebraic} b] Function {trigonometric} c] Limits {Algebraic limits by Factorization} d] Limits {Rationalization & limit tends to infinity} e] Limits { Trigonometric limits } f] Limits {Exponential Limits}
Types of Problems
No. Type of Problem Q.B.Checked Revision
1 Algebraic functions [ ] [ ] [ ] [ ] [ ]
2 Trigonometric functions [ ] [ ] [ ] [ ] [ ]
3 Limits by factorization [ ] [ ] [ ] [ ] [ ]
4 Limits by rationalization [ ] [ ] [ ] [ ] [ ]
5 Algebraic Limit limx→∞
[ ] [ ] [ ] [ ] [ ]
6 Trigonometric limx→0
[ ] [ ] [ ] [ ] [ ]
7 Trigonometric limx→π, ,2
π4π [ ] [ ] [ ] [ ] [ ]
8 Exponential limits type A [ ] [ ] [ ] [ ] [ ]
Type 1: Solution of Algebraic functions 1. f f (x) x . Find f (x ) & (x ) I = x2 + 4 + 1 − 1 + 1 2. f f (x) x x . f ind f (1 ) & f (2x ) I = 2 − 3 + 4 − x + 1 3. f f (x) . Show that f (x) (x ) (x ) I = x
1 − f + 1 = f 2 + x 4. f f (x) . Show that [f (x)] (x ) .f ( ) I = x − x
1 3 = f 3 − 3 x1
5. f f (x) . Find f{f (x)} I = 11−x
6. f f (x) . Show that, f{f{f (x)}}I = 11−x = x
Type 2.Solution of trigonometric functions 1. f f (x) an x . Show that f (2x)I = t = 2.f(x)
1−[f(x)]2
2. f f (t) 0 sin (100 π t .04). Show that, f [ ] (t) I = 5 + 0 2100 + t = f
3. f f (x) in x & ϕ(x) os x . Show that, ϕ(x ) (x).ϕ(y) (x).f (y) I = s = c + y = ϕ − f 4. f f (t) 0 sin 100 π t. Find f [ ] I = 1 1
200 − t 5. f f (x) an x, Show that, f (2x)I = t = 2.f(x)
1−[f(x)]2
6. f f (x) og(1 an x). Show that, f ( ) og2 (x) I = l + t 4π − x = l − f
Type 3.Limits by Factorization 1. 2. 3. lim
x→1 x −4x+32x +3x−42 lim
x→1x −8x+72
7x −6x−12 limx→7 x +7x−982
x −18x+772
4. 5. 6. lim
x→2x −83
x −3x+22 limx→−1 x +13
x −3x−42 limx→−4
x +3x−42
x +7x+122
7. 8. 9. lim
x→2x −5x+62
x −2x +x−23 2 limx→3
x −5x+62
x +3x −18x3 2 limx→2
] [ 1x−2 −
6x8−x3
10. 11. lim
x→2] [ 1
x−2 −2
x −2x2 limx→3
] [ 1x−3 −
3x(x −5x+6)2
Type 4.Limits by rationalization
1. 2. limx→1 (x−1)
−√x+4 √5 limx→2
2−x−√2x+5 √5x−1
3. 4. lim
x→0 x−√a+x √a−x lim
x→4 1−√5−x3−√5+x
5. lim
x→13−√4+5x5−√12+13x
Type 5. Algebraic Limits x→ ∞
1. [ ] 2. limx→∞
√x2 + x − x limx→∞
√x2 + 1 − x 3. 4. [ ]limx→∞
] [√x x2 + 5 − x limx→∞
√x2 + x + 1 − x 5. 6. [ ]x [ ]limx→∞
3 √x2 + 1 − √x2 − 1 limx→∞
x √x2 + 1 − √x2 − 1 Type 6. Trigonometric limits x→ 0 1. 2. 3. limx→0
x tan x1−cos x lim
x→0 3x+ 2tan3x3sin2x + 2x lim
θ→02θ+3sinθ3θ+5tanθ
4. 5. 6. limx→0 x2
cos3x−cos x limx→0 x. tan x
cos 6x− cos 8x limx→0 x2
cos 6x − cos 4x
7. 8. 9. limx→0
cos 8x − cos 2xcos 12x − cos4x lim
x→0 x3tan x − sin x lim
x→0 x tan xx + 1−cos x2
10. lim
x→0 x32sinx−sin2x
Type 7. Trigonometric limits , , ...x→ π 2
π4π
1. 2. 3. limθ→ 2
π cos θ21−sinθ lim
θ→π sin θ21+cos θ3 lim
x→ 2π cos x2
1−sin x3
4. 5. limx→π sin x2
1+cos x limx→ 4
π 1−tan x2−sec x2
Type 8. Exponential limits on formula lim
x→0 kxa −1kx
1. 2. limx→0 x2
6 −3 −2 +1x 3 x
limx→0 x.sin x
10 −2 −5 +1x x x
3. 4. limx→0 tan x2
18 −6 −3 +1x x xlimx→0
5 −4x x
tan2 x
5. 6. limx→0 x2
3 +3 −2x −xlimx→0 x2
4 +4 −2x −x
Shaikh sir’s Reliance Academy,
Shahupuri, Kolhapur.
Subject : Maths – II (G-Scheme)
Question : 4 This question contains 6 questions from following chapters Q2] Solve any Four . [ 16 Marks ] a] Derivatives : Proof based on first principle b] Derivatives : Proof of Rules of derivatives c] Derivatives : Problems on function)( function
d] Derivatives : Problems on Implicit function e] Derivatives : Problems on Parametric function f] Derivatives : Problems on Higher order derivatives
Types of Problems
No. Type of Problem
1 Derivatives :Derivatives : Problems based on first principle
2 Derivatives :Proof of Rules of derivatives
3 Derivatives : Problems on function)( function
4 Derivatives :Problems on Implicit function
5 Derivatives :Problems on Parametric function
6 Derivatives :Problems on Higher order derivatives
Type1: Derivatives : Proof of addition/subtraction, product rule and division rule
1) If u and v are differential function of x and y = u - v then prove
that dxdy = dx
du − dxdv
2) If u and v are differntial function of x and y = u + v then prove
that dxdy = dx
du + dxdv
3) If u and v are differentiable functions of x and y = u,v, than prove
that . .dxdy = u dx
du + v dxdv
4) If u and v are differentiable function of x and y = , then provevu
that dxdy = v2
v. −u.dxdu
dxdv
Type 2: Derivatives : Problems on FIRST PRINCIPLE
A) Find derivatives using first principle, 1. (x)f = √x 2. (x) inxf = s 3. (x) osxf = c 4. (x) anxf = t 5. (x)f = ex 6. (x) ogxf = l 7. (x)f = ax Type 3: Derivatives : Problems on function) ( function 1. Find of dx
dy xsinx 2.Find dx
dy (sinx)x 3.Find dx
dy (logx)x 4.Find dx
dy (sinx)logx 5.Find dx
dy (tanx)logx 6. ind , f y F dx
dy i = xsinx + sinxx
Type 4. Problems on Implicit function
1. Find x y dxdy 3 + 2 2 = 0
2. Find x y dxdy 3 + y3 = x
3. Find x xy dxdy 2 + 2 + y2 = 0
4. Find x xy dxdy 2 + 3 + y2 = 0
5. Find x xy dxdy 2 + y2 = 4
6. Find x xy dxdy 3 − 2 + y3 = 0
Proofs on implicit function 1. f x.siny .sinx Prove that −I + y = 0 dx
dy = (x cosy+sinx)(y cosx+sinx)
2. Show that ey = yx dxdy = (log)2
(log y−1) 3. f y .siny Show that I = x dx
dy = yx(1−x.cosy)
4. .logy Prove that y = x dxdy = y2
x(y−x) 5. f x Prove that I y = ex−y dx
dy = logx(1+logx)2
Type 5. Problems on parametric functions 1. f x .cosθ .sinθ, y .sinθ .cosθ Show that −I = a + b = a − b dx
dy = yx
2. f x .cos t., y .sin t Find I = a 3 = a 3dxdy
3. f x in(logt), y os(logt) Find I = s = c dxdy
4. f x , y Prove that −I = ecos2t = esin2t dxdy = x logy
y logx
5. f x .log(cost), y .log(sint) Find I = a = a dxdy
Type 6. Higher order derivative 1. f xy , Prove that x .I = a dx2
d y2 + 2 dxdy = 0
2. f y in5x cos5x Show that 5.yI = s − 3 dx2d y2 + 2 = 0
3. f y .e .e Prove that yI = A 3x + B 3xdx2d y2 − 9 = 0
4. Prove that (1 ). 2x ).y = etan x−1 + x2 dx2d y2 + ( − 1 dx
dy = 0
5. in(logx) Prove that x . .y = s 2dx2d y2 + x dx
dy = 0
Shaikh sir’s Reliance Academy,
Shahupuri, Kolhapur.
Subject : Maths – II (G-Scheme)
Question : 5 This question contains 6 questions from following chapters Q2] Solve any Four . [ 16 Marks ] a] Limits-{logarithmic limits} b] Limits {Combined algebraic,trigonometric and exponential limits} c] Numerical method- Bisection method d] Numerical Method - Regula Falsi Method e] Numerical Method- Newton-Raphson Method f] Numerical method- Newton Raphson Method
Types of Problems
No. Type of Problem Revision
1 Bisection method [ ] [ ] [ ] [ ] [ ]
2 Regula Falsi Method [ ] [ ] [ ] [ ] [ ]
3 NewtonRaphson Method [ ] [ ] [ ] [ ] [ ]
4 Limits by Logarithmic methods [ ] [ ] [ ] [ ] [ ]
5 Limits combined algebraic,trignometric and exponential method
[ ] [ ] [ ] [ ] [ ]
Type 1: Bisection Methods Problems Ex. 1: Use the bisection method to find the root of which lies between 2 x 1x3 − 5 − 1 = 0 and 3. {Three iterations only} Ex.2: Using bisection method find the approximate root of the equation xx3 − 5 + 1 = 0 Ex.3: Using bisection method find the root between 1 and 2 of the equation using three iterations.0x4 − x − 1 = 0
Ex.4: Find a root of using bisection method {Three iterations only}xx3 − 4 − 9 = 0
Ex.5: Find a root of using bisection method {Three iterations only}x3 − x − 4 = 0
Ex.6: Find a root of using bisection method {Three iterations only}xx3 + 2 − 1 = 0
Ex.7: Find a root of using bisection method {Three iterations only}xx3 − 2 − 5 = 0
Answers Ex.1 [X = 2.875] Ex.2 [X= 0.125] Ex.3 – [X = 1.875] Ex.4 – [X = 2.625] Ex.5 – [X = 1.875] Ex.6 – [X = 0.375] Ex.7 – [X = 2.125]
Type 2 : RegulaFalsi Method
Problems Ex.1: Find the approximate value of the root of the equation lying between x x3 − 2 2 − 5 = 0 2 and 3 by false position method. Ex.2: Find the root of the equation which lies between 2 and 3 by the xx3 − 5 − 7 = 0 method of false position. Ex.3: Find the approximate value of the root of the equation lying between xx3 − 9 + 1 = 0 2 and 3 by false position method. Ex.4: Find the root of which lies between 3.5 and 4 by false position equation. Ex.5: Find a real root of by regula falsi method.xx3 − 3 + 4 = 0
Ex.6: Find a real root of by regula falsi method.x2 − x − 1 = 0
Ex.7: Find a real root of by regula falsi method.x3 − x − 1 = 0
Ex.8: Find a real root of by regula falsi method. x2 + x − 3 = 0
Answers Ex.1 [ 2.687 ] Ex.2 – [ 2.746 ] Ex. 3– [ 2.943 ] Ex.4 [ 3.789 ] Ex.5 – [ 2.187 ] Ex.6 – [ 1.615 ] Ex.7 – [ 1.3112 ] Ex.8 – [ 1.3000 ]
Type 3: Newton Raphson Method Problems Ex.1 : Find the root of using NewtonRaphson method.0x4 − x − 1 = 0
Ex.2 : Find the real root of upto three iterations using NewtonRaphsonxx3 − 2 − 5 = 0 method. Ex.3 : Find the root of the equation lying between 2 and 1 up to threex2x3 − 3 + 4 iterations using NewtonRaphson method. Ex. 4 : Find a root of by NewtonRaphson method.x 8x3 − 9 2 − 1 = 0
Ex.5 : Find the root of which is near to 2 correct to three places of 0x4 − x − 1 = 0 decimals using NewtonRaphson method. Ex.6 : Find the value of using NewtonRaphson method. √10 Ex.7 : Find the value of by Newton – Raphson method. √3 20 Ex.8: Find the value of by Newton – Raphson method. √3 100
Answers 1] [ X = 1.855] 2] [ X = 2.0945] 3] [ X = 1.719] 4] [ X = 9.212] 5] [ X = 1.856} 6] [ X = 3.162] 7] [ X = 2.714] 8] [ X = 4.642]
Type 4.Logarithmic limits
1. 2. limx→0 sin x
log (1+4x) limx→0 x
log 10 +log(x+0.1)
3. 4. limx→0 x
log (5+x)− log(5−x) limx→0 x−5
log x−log 5
5. 6. limx→0 x−3
log x− log 3 limx→0 x−2
log x − log 2
7. limx→0
log [ ] x1
1+3x1+2x
Type 5. Combined Algebraic,trigonometric and exponential limits
1. 2. 3. limx→0 −2√x +42
tan 3x (3 −1)x
limx→0 −2√x +42
sin 4x (5 −1)x
limx→0 −1√x +12
(2 −1). sin 3xx
4. 5. 6.limx→0
5 −3x x
−2√x+4 limx→0 x4
sin 2x.tan4x.log(1+x )2
7. limx→0 x. log (1+5x)
(5 −1) (e −1)2x 3x
limx→0
3 −2x x
sin π x
Shaikh Sir's Reliance Academy,
Coaching Classes for Diploma EngineeringShahupuri Kolhapur
Subject : Maths – II (G-Scheme)
Question : 6This question contains 6 questions from following chaptersQ6] Solve any Four . [ 16 Marks ]
a] Derivative {Higher order derivative}
b] Derivative{Higher order derivative}
c] Numerical method-Gauss elimination method
d] Numerical method-Jacobi’s Method
e] Numerical method- Jacobi’s /Gauss-Seidel Method/Gauss elimation
f] Numerical method- Gauss-Seidel method
Types of Problems
No. Type of Problem
1 Gauss Elimination Method
2 Jacobi’s Iterative Method
3 Gauss-Seidal Method
4 Higher order derivatives
Type 1: Gauss Elimination Method
Ex.1: Apply Gauss elimination method to solve the equations. x + 3y – 2z = 5, 2x + y -3z = 1, 3x + 2y – z = 6.Ex.2: Solve by Gauss elimination method. 2x + y + z = 10, 3x + 2y + 3z = 18 , x + 4y + 9z = 16.Ex.3: Solve the following equations by using Gauss elimination method. 2x + 3y + z = 13, x – y – 2z = - 1, 3x + y + 4z = 15.Ex.4: Solve the following equations by using Gauss elimination method. 2x + 2y + 3z = 15, 3x + y + 2z = 11, 2x + 3y + z = 11.Ex.5: Solve the following equations by using Gauss elimination method x + y + z = 2, 2x + 2y – z = 1, 3x + 4y + z = 9. Ex.6: Solve the following equations by using Gauss elimination method 2 x + 3y - z = 7, x + 2y + z = 6, x + y - z = 2. Ex.7: Solve the following equations by using Gauss elimination method x+y+z = 6,2x + y + 3z = 13, 3x + 3y +4z = 20. Ex.8: Solve the following equations by using Gauss elimination method x+y+z = 6,3x - y + 3z = 10, 5x + 5y -4z = 3.
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Answers
1] [ X = 1, Y = 2, Z = 1] 2] [ X = 7, Y = -9, Z = 5] 3] [ Z = 1, Y = 2, X = 3] 4] [ X = 1, Y= 2, Z = 3] 5] [ X = -4, Y= 5, Z= 1] 6] [ X = 1, Y= 2, Z = 1]7] [ X = 3, Y= 1, Z= 2] 8] [ X = 1, Y= 2, Z = 3]
Type 2. Jacobi’s Iterative methodEx. 1 : Solve the following equations by Jacobi’s iteration method. 15x + 2y + z = 18, 2x + 20y – 3z = 19, 3x – 6y + 25z = 22.Ex. 2 : Solve by Jacobi’s method. (carry two iterations only) 4x + y + 3z = 17, x + 5y + z = 14, 2x – y + 8z = 12.Ex. 3 : Solve the following equations by Jacobi’s method. 10x + y + 2z = 13, 3x + 10y + z = 14, 2x + 3y + 10z = 15.Ex. 4 : Solve the following equations by Jacobi’s method. 15x + y - z = 14, x + 20y + z = 23, 2x - 3y + 18z = 37Ex. 5 : Solve by Jacobi’s method. {only two iterations} 12x + 2y + z = 27, 2x + 15y - 3z = 16, 2x - 3y + 25z = 23.Ex. 6 : Solve the following equations by Jacobi’s method. 8x - y + 2z = 13, x - 10y + 3z = 17, 3x + 2y + 12z = 25.Ex. 7 : Solve the following equations by Jacobi’s method. 10x + y + z = 12, 2x + 10y + z = 13, 2x + 2y + 10z = 14.Ex. 8 : Solve the following equations by Jacobi’s method. 20x + y -2z = 17, 3x + 20y - z = -18, 2x - 3y + 20z = 25.-------------------------------------------------------------------------------------------------------------------------------
Answers1] [ X = 1.200,1.015,1.007 Y =0.950,0.962,0.993 Z =0.880,0.964,0.989] ] 2] [ X = 4.250,2.425, Y = 2.8,1.650, Z = 1.5,0.787] 3] [ X = 1.300,0.860,1.050 Y =1.400,0.860,1.060 Z =1.500 ,0.820,1.070] ] 4] [ X =0.933,1.066 , Y =1.050, ,1.001 Z = 2.055 ,2.143 , ]5] [ X = , , Y = , , Z = , , ] 6] [ X = 1.625 ,0.892 ,1.021 Y =-1.700,-0.913,-1.023 Z =2.083,1.960,2.012 ] 7] [ X = , , Y = , , Z = , , ] 8][ X = , , Y = , , Z = , , ]
Type 3.Gauss-Seidel Method
Ex.1 : Solve the following equations by Gauss-Seidel method 27x + 6y – z = 85, 6x + 15y + 2z = 72, x + y + 54z = 110.Ex.2 : Solve the following equations by Gauss-Seidel method 10x1 + x2 + x3 = 12, 2x1 + 10x2 + x3 = 13, 2x1 + 2x2 + 10x3 = 14.Ex.3 : Solve the following equations by Gauss-Seidel method 23x + 4y – z = 32, 2x + 17y +4z = 35, x + 3y + 10z = 24.Ex.4 : Solve the following equations by Gauss-Seidel method ,upto two iterations. 4x - 2y – z = 40, x - 6y +2z = - 28, x - 2y + 12z = - 86. Ex.5 : 15x + y + z = 17, 2x + 15y + z = 18, x + 2y + 15z = 18.Ex.6 : 10x + 2y + z = 9, 2x + 20y - 2z = - 44, -2x + 3y + 10z = 22Ex.7 : 82x - 3y + z = 75, x + 75y - 2z = 153, 3x - 2y + 85z = - 86.Ex.8: 20x + y - z = 40, 2x + 18y + z = 21, x + 2y + 25z = - 21--------------------------------------------------------------------------------------------------------------------------------
Answers 1] [ X = , , Y = , , Z = , , ] 2] [ X = 4.250,2.425, Y = 2.8,1.65, Z = 1.5,0.787] 3] [ X = , , Y = , , Z = , , ] 4] [ X = , , Y = , , Z = , , ] 5] [ X = , , Y = , , Z = , , ] 6] [ X = , , Y = , , Z = , , ] 7] [ X = , , Y = , , Z = , , ]
Type 4. Higher order derivative
