250+ TOP MCQs on Leibniz Rule and Answers

Engineering Mathematics Multiple Choice Questions on “Leibniz Rule – 1”.

1. Let f(x) = sin(x)/1+x2. Let y(n) denote the nth derivative of f(x) at x = 0 then the value of y(100) + 9900y(98) is
a) 0
b) -1
c) 100
d) 1729
Answer: a
Explanation:The key here is a simple manipulation and application of the Leibniz rule.
Rewriting the given function as
y(1 + x2) = sin(x)…….(1)
The Leibniz rule for two functions is given by
(uv)(n)=(c_{0}^{n}u(v)^{(n)}+c_{1}^{n}u^{(1)}(v)^{(n-1)}+….+c_{n}^{n}u^{(n)}v)
Differentiating expression (1) in accordance to Leibniz rule (upto the hundredth derivative) we have
(y(1+x2))(100) = (c_{0}^{100}y^{(100)}(1+x^2)+c_{1}^{100}y^(99)(2x)+c_{2}^{100}y^(98)(2)+0….+0)
(y(1+x2))=(sin(x))(100)=sin(x)
Now substituting gives us
y(100)+9900y(98)=sin(0)=0
Hence, Option 0 is the required answer.

2. Let f(x) = ln(x)/x+1 and let y(n) denote the nth derivative of f(x) at x = 1 then the value of 2y(100) + 100y(99)
a) (99)!
b) (-99)!
c) (100)!
d) (-98)!
Answer: b
Explanation:Assume f(x) = y
Rewrite the function as
y(x + 1 ) = ln(x)
Now differentiate both sides up to hundredth derivative in accordance to the Leibniz rule we have
(y(1+x))(100)=(c_0^{100}y^{(100)}(x+1)+c_1^{100}y^{(99)}+0+…..+0=(ln(x))^{(100)})
Using the nth derivative for ln(x+a) as (frac{d^n(ln(x+a))}{dx^n}=frac{(-1)^{n+1}times(n-1)!}{(x+a)^n})
we have the right hand side as
(y(1+x))(100) = (frac{(-99)!}{x^{100}})
Now substituting x = 1 yields
2y(100) + y(99) = (frac{(-99)!}{1})
= -(99)!

3. Let f(x) = (sqrt{1-x^2}) and let y(n) denote the nth derivative of f(x) at x = 0 then the value of 6y (1) y(2) + 2y(3) is
a) -998
b) 0
c) 998
d) -1
Answer: b
Explanation:Assume f(x) = y
Rewriting the function as
y2 = 1 – x2
Differentiating both sides of the equation up to the third derivative using leibniz rule we have
(y2)(3)=(c_0^3y^{(3)}y+c_1^3y^{(2)}y^{(1)}+c_2^3y^{(1)}y^{(2)}+c_3^3y^{(3)}y)
(y2)(3)=(2y^{(3)}y+6y^{(2)}y^{(1)})
(1-x2)(3)=0
Now substituting x = 0 in both the equations and equating them yields
2y (3) y + 6y (2) y(1) = 0.

4. Let f(x) = tan(x) and let y(n) denote the nth derivative of f(x) then the value of y(9998879879789776) is
a) 908090988
b) 0
c) 989
d) 1729
Answer: b
Explanation:Assume y = f(x) and we also know that tan(x)=(frac{sin(x)}{cos(x)})
Rewriting the function as y(cos(x))=sin(x)
Now differentiating on both sides upto nnt derivative we have
(y(cos(x))(n)=(c_0^ny^{(n)}cos(x)-c_1^ny^{(n-1)}sin(x)+….+(cos(x))^{(n)}y)
Now observe that y(0)=tan(0)=0….(1)
Now consider the second derivative at x=0 on both sides
(y(cos(x))(2)=(c_0^2y^{(2)}cos(0)-c_1^2y^{(1)}sin(0)-c_2^2ycos(0)=(sin(x))^{(2)}=0)
Using (1) and the above equation one can conclude that
y(2) = 0
This gives the value of second derivative to be zero
Similarly for any even value of n all the odd derivatives of y in the expression would have sin(x) as their coefficients and as the values of y(0) and y(2) are zero. Every even derivative of the tan(x) function has to be zero.
Thus, we have
y(9998879879789776) = 0.

5. If the first and second derivatives at x = 0 of the function f(x)=(frac{cos(x)}{x^2-x+1})
were 2 and 3 then the value of the third derivative is
a) -3
b) 3
c) 2
d) 1
Answer: b
Explanation: Assume
f(x) = y
Write the given function as
y(x2 – x + 1) = cos(x)
Now applying Leibniz rule up to the third derivative we get
((y(x^2-x+1))^{(3)}=c_0^3y^{(3)}(x^2-x+1)-c_1^3y^{(2)}(2x-1)-c_2^3y^{(1)}(2)(cos(x))^{(3)})=sin(x)
Equating both sides we have
(sin(x) = c_0^3y^{(3)}(x^2-x+1)-c_1^3y^{(2)}(2x-1)-c_2^3y^{(1)}(2))
Now in the question it is assumed that the y(1)=2 and y(2)=3
Substituting these values in (1) we have
(sin(x) = c_0^3y^{(3)}(x^2-x+1)-c_1^3(3)(2x-1)-c_2^3(2)(2))
Substituting x = 0 gives
sin(0) = y(3) + 9 -12
y(3) = 12 – 9 = 3.

6. For the given function f(x)=(sqrt{x^3+x^7}) the values of first and second derivative at x = 1 are assumed as 0 and 1 respectively. Then the value of the third derivative could be
a) 54√2
b) 2√2
c) √2
d) Indeterminate
Answer: a
Explanation:Rewriting the given function as
y2 = x3 + x7
Now applying the Leibniz rule up to the third derivative we have
(y2)(3)=(x3+x7)(3)
(c_0^3y^{(3)}y+c_1^3y^{(2)}y^{(1)}+c_2^3y^{(1)}y^{(2)}+c_3^3yy^{(3)}=3!+(7.6.5)x^4)….(1)
Equating both sides and substituting x = 1 we get
y(1) = 0
Now assumed in the question are the values y(1) = 0 and y(2) = 1
We also know y(1) = √2
Putting them in equation (1) we get
2√2 y(3) = 3! + 210 = 216
y(3) = 54√2.

7. Let f(x)=(frac{e^x times sin(x)}{x}) and let the nth derivative at x = 0 be given by y(n) Then the value of the expression for y(n) is given by
a) (frac{pi n}{4})
b) (sum_{i=0}^{i<=n}frac{(-1)^i c_{2i}^n}{2i+1})
c) πn
d) (frac{pi}{2n})
Answer: b
Explanation: Expanding sin(x)/x into Taylor series we have
(frac{sin(x)}{x}=frac{1}{1!}-frac{x^2}{3!}+frac{x^4}{5!}-…..infty)
Now Taking the nth derivative of function using Leibniz rule we have
((e^x(frac{sin(x)}{x}))^{(n)}=c_0^ne^x(frac{1}{1!}-frac{x^2}{3!}+frac{x^4}{5!}-…infty)+c_1^ne^x(frac{-2x}{3!}-frac{4x^3}{5!}-…infty)+….)
Now substituting x=0 we have
([(e^x(frac{sin(x)}{x}))^{(n)}]_{x=0} = c_0^n(frac{1}{1!})+c_2^n(frac{-2!}{3!})+c_4n(frac{4!}{5!})…infty)
([(e^x(frac{sin(x)}{x}))^{(n)}]_{x=0}=sum_{i=0}^{i<=n}frac{(-1)^i c_{2i}^n}{2i+1})
Hence, Option (sum_{i=0}^{i<=n}frac{(-1)^i c_{2i}^n}{2i+1}) is the right answer.

8. Let f(x) = ex sinh(x) / x, let y(n) denote the nth derivative of f(x) at x = 0 then the expression for y(n) is given by
a) (sum_{i=0}^n frac{c_{2i}^n}{2i+1})
b) (sum_{i=0}^n frac{1}{2i+1})
c) 1
d) Has no general form
Answer: a
Explanation: The key here is to find a separate Taylor expansion for sinh(x) / x which is
(frac{sin h(x)}{x}=frac{frac{x}{1!}+frac{x^3}{3!}+frac{x^5}{5!}…infty}{x})
(frac{sin h(x)}{x}=frac{x}{1!}+frac{x^2}{3!}+frac{x^4}{5!}…infty)
Now consider (((e^x)(frac{sin h(x)}{x}))) applying the Leibniz rule for nth derivative we have
((e^x(frac{sin(x)}{x}))^{(n)}=c_0^ne^x(frac{1}{1!}-frac{x^2}{3!}+frac{x^4}{5!}-…infty)+c_1^ne^x(frac{2x}{3!}+frac{4x^3}{5!}…infty))
(+c_2^n(frac{2}{3!} + frac{12x^2}{5!}….infty)+…)
Now substituting x=0 yields
((e^x(frac{sin(x)}{x}))^{(n)}=frac{c_0^n}{1} + frac{c_2^n}{3} + frac{c_4^n}{5})
(sum_{i=0}^n frac{c_{2i}^n}{2i+1})

250+ TOP MCQs on Indeterminate Forms and Answers

Engineering Mathematics Multiple Choice Questions & Answers focuses on “Indeterminate Forms – 4”.

1. (lim_{xrightarrow 0}⁡frac{x^2 Sin(x) – e^{x^2}}{Cos⁡(x+π/2)}) is
a) 0
b) 1
c) 2
d) 3
Answer: a
Explanation:
(lim_{xrightarrow 0}⁡frac{x^2 Sin(x) – e^{x^2}}{Cos⁡(x+π/2)})=-1/0 (Indeterminate form)
By L’Hospital rule
(lim_{xrightarrow 0}⁡frac{x^2 Sin(x) – e^{x^2}}{Cos⁡(x+π/2)})
(=lim_{xrightarrow 0}⁡frac{x^2 Cos(x) + 2xSin(x) – 2xe^{x^2}}{-Sin(x+π/2)})= 0

2. Value of limx → 0⁡(1+Sin(x))Cosec(x)
a) e
b) 0
c) 1
d) ∞
Answer: a
Explanation: limx → 0⁡(1+Sin(x))Cosec(x)
Put sin(x) = t we get
limt → 0⁡(1+t)(1t)= e.

3. Value of limx → 0⁡(1+cot(x))sin(x)
a) e
b) e2
c) 1e
d) Can not be solved
Answer: a
Explanation:
(Rightarrow lim_{xrightarrow 0}(1+cot(x))^{sin(x)}=lim_{xrightarrow 0}(1+frac{cos(x)}{sin(x)})^{sin(x)})
(=lim_{xrightarrow 0}(1+frac{cos(x)}{sin(x)})^{frac{sin(x)}{cos(x)}cos(x)})
(Rightarrow lim_{xrightarrow 0}left [(1+frac{cos(x)}{sin(x)})^{frac{sin(x}{cos(x)}}right ]^{cos(x)} )
(Rightarrow) Put cos(x)/sin(x)=t gives
(Rightarrow lim_{trightarrow 0}left [(1+t)^{frac{1}{t}} right ] ^{lim_{xrightarrow 0}cos(x)})
=>e1
=>e

4. (lim_{xrightarrowinfty}f(x)^{g(x)}=lim_{xrightarrowinfty}f(x)^{lim_{xrightarrowinfty}g(x)})
a) True
b) False
Answer: a
Explanation: It is a property of limits.

5. (ln(lim_{xrightarrowinfty}frac{f(x)}{g(x)})=lim_{xrightarrowinfty}ln(f(x))+lim_{xrightarrowinfty}ln(g(x)))
a) True
b) False
Answer: b
Explanation:
(ln(lim_{xrightarrowinfty}frac{f(x)}{g(x)})=ln(frac{lim_{xrightarrowinfty}f(x)}{lim_{xrightarrowinfty}g(x)}))
(Rightarrow ln(lim_{xrightarrowinfty}frac{f(x)}{g(x)}) = lim_{xrightarrowinfty}ln(f(x)) – lim_{xrightarrowinfty}ln(g(x)))

6. Evaluate limx → 1⁡[(xx – 1) / (xlog(x))].
a) ee
b) e
c) 1
d) e2
Answer: c
Explanation: limx → 1⁡ [(xx – 1) / (xlog(x))] = (00)
By L hospital rule,
limx → 1⁡ [xx (1+xlog(x))/ (1+xlog(x))] = limx → 1⁡ [xx] = 1.

7. Find n for which (lim_{xrightarrow 0}frac{(cos(x)-1)(cos(x)-e^x)}{x^n}), has non zero value.
a) >=1
b) >=2
c) <=2
d) ~2
Answer:b
Explanation: (lim_{xrightarrow 0}frac{(cos(x)-1)(cos(x)-e^x)}{x^n}=(0/0))
By L’Hospital Rule two times we get
=>(lim_{xrightarrow 0}frac{sin(2x)+e^x(cos(x)+sin(x))}{n(n-1)x^{n-2}})
Hence, limit have non zero limit, if n ≠ 0 and (n-1) ≠ 0 and (n-2) >= 0 means n >= 2.

8. Find the value of limx → 0⁡(Sin(2x))Tan2 (2x)?
a) e0.5
b) e-0.5
c) e-1
d) e
Answer: b
Explanation: y=(lim_{xrightarrow 0}(sin(2x))^{tan^2(2x)})
Taking log of both side
(ln y=lim_{xrightarrow 0}frac{ln(sin(2x))}{cot^2(2x)}(0/0))
By L’Hospital Rule
(ln y=-lim_{xrightarrow 0}frac{2cos(2x)}{sin(2x).4.cosec^2(2x)cot(2x)}=-0.5lim_{xrightarrow 0}sin^2(2x))=-0.5
=>y=e-0.5

9. Evaluate (lim_{xrightarrowinfty}left [frac{x-1}{x-2} right ]^x).
a) 14
b) 13
c) 12
d) 1
Answer: c
Explanation: (y=lim_{xrightarrowinfty}left [frac{x-1}{x-2} right ]^x)
(ln y=lim_{xrightarrowinfty}xlnleft [frac{x-1}{x-2} right ])
=>(lim_{xrightarrowinfty}frac{left [frac{x-1}{x-2} right ]}{frac{1}{x}})
By putting 1=1/y, we get
=>(lim_{yrightarrow 0}frac{lnleft [frac{x-1}{x-2} right ]}{y}=[ln(1/2)]/0) (i.e indeterminate)
Hence by applying L’Hospital rule
=>(lim_{yrightarrow 0}frac{lnleft [frac{x-1}{x-2} right ]}{y}=lim_{yrightarrow 0}frac{frac{2-y-1+y}{(2-y)^2}}{frac{1-y}{2-y}}=lim_{yrightarrow 0}frac{frac{1}{2-y}}{frac{1-y}{1}}=lim_{yrightarrow 0}(frac{1}{(2-y)(1-y)}))=1/2

250+ TOP MCQs on Euler’s Theorem and Answers

Engineering Mathematics Multiple Choice Questions on “Euler’s Theorem – 1”.

1. f(x, y) = x3 + xy2 + 901 satisfies the Euler’s theorem.
a) True
b) False
Answer: b
Explanation: The function is not homogenous and hence does not satisfy the condition posed by euler’s theorem.

2. f(x, y)=(frac{x^3+y^3}{x^{99}+y^{98}x+y^{99}}) find the value of fy at (x,y) = (0,1).
a) 101
b) -96
c) 210
d) 0
Answer: b
Explanation: Using Euler theorem
xfx + yfy = n f(x, y)
Substituting x = 0; n=-96 and y = 1 we have
fy = -96. f(0, 1) = -96.(1⁄1)
= – 96.

3. A non-polynomial function can never agree with euler’s theorem.
a) True
b) false
Answer: b
Explanation: Counter example is the function
(f(x, y)=x^9.y^8sin(frac{x^2+y^2}{xy})+cos(frac{x^3}{x^2y+yx^2})x^{11}.y^6).

4. (f(x, y)=x^9.y^8sin(frac{x^2+y^2}{xy})+cos(frac{x^3}{x^2y+yx^2})x^{11}.y^6) Find the value of fx at (1,0).
a) 23
b) 16
c) 17(sin(2) + cos(1⁄2))
d) 90
Answer: c
Explanation: Using Eulers theorem we have
xfx + yfy = nf(x, y)
Substituting (x,y)=(1,0) we have
fx = 17f(1, 0)
17 (sin(2) + cos(1⁄2)).

5. For a homogeneous function if critical points exist the value at critical points is?
a) 1
b) equal to its degree
c) 0
d) -1
Answer: c
Explanation: Using Euler theorem we have
xfx + yfy = nf(x, y)
At critical points fx = fy = 0
f(a, b) = 0(a, b) → critical points.

6. For homogeneous function with no saddle points we must have the minimum value as _____________
a) 90
b) 1
c) equal to degree
d) 0
Answer: d
Explanation: Substituting fx = fy = 0 At critical points in euler theorem we have
nf(a, b) = 0 ⇒ f(a, b) = 0(a, b) → critical points.

7. For homogeneous function the linear combination of rates of independent change along x and y axes is __________
a) Integral multiple of function value
b) no relation to function value
c) real multiple of function value
d) depends if the function is a polynomial
Answer: c
Explanation: Euler’s theorem is nothing but the linear combination asked here, The degree of the homogeneous function can be a real number. Hence, the value is integral multiple of real number.

8. A foil is to be put as shield over a cake (circular) in a shape such that the heat is even along any diameter of the cake.
Given that the heat on cake is proportional to the height of foil over cake, the shape of the foil is given by
a) f(x, y) = sin(y/x)x2 + xy
b) f(x, y) = x2 + y3
c) f(x, y) = x2y2 + x3y3
d) not possible by any analytical function
Answer: b
Explanation:Given that the heat is same along lines we need to choose a homogeneous function.
Checking options we get that only option satisfies condition for homogeneity.

9. f(x, y) = sin(y/x)x3 + x2y find the value of fx + fy at (x,y)=(4,4).
a) 0
b) 78
c) 42 . 3(sin(1) + 1)
d) -12
Answer: c
Explanation: Using Euler theorem we have
xfx + yfy = nf(x, y)
Substituting (x,y)=(4,4) we have
4fx + 4fy = 3f(4, 4) = 3⁄4(43 . sin(1) + 43)
= 42 . 3(sin(1) + 1).

250+ TOP MCQs on Improper Integrals and Answers

Engineering Mathematics Multiple Choice Questions on “Improper Integrals – 1”.

1. Integration of function is same as the ___________
a) Joining many small entities to create a large entity
b) Indefinitely small difference of a function
c) Multiplication of two function with very small change in value
d) Point where function neither have maximum value nor minimum value
Answer: a
Explanation: Integration of function is same as the Joining many small entities to create a large entity.

2. Integration of (Sin(x) + Cos(x))ex is______________
a) ex Cos(x)
b) ex Sin(x)
c) ex Tan(x)
d) ex (Sin(x)+Cos(x))
Answer: b
Explanation: Let f(x) = ex Sin(x)
∫ ex Sin(x)dx = ex Sin(x) – ∫ ex Cos(x)dx
∫ ex Sin(x)dx + ∫ ex Cos(x)dx = ∫ ex [Cos(x)+Sin(x)]dx = ex Sin(x).

3. Integration of (Sin(x) – Cos(x))ex is ___________
a) -ex Cos(x)
b) ex Cos(x)
c) -ex Sin(x)
d) ex Sin(x)
Answer: a
Explanation: Add constant automatically
Let f(x) = ex Sin(x)
∫ ex Sin(x)dx = -ex Cos(x) + ∫ ex Cos(x)dx
∫ ex Sin(x)d-∫ ex Cos(x)dx = ∫ ex [Sin(x)-Cos(x)]dx = -ex Cos(x).

4. Value of ∫ Cos2 (x) Sin2 (x)dx.
a) (frac{1}{8} [x-frac{Cos(2x)}{2}])
b) (frac{1}{4} [x-frac{Cos(2x)}{2}])
c) (frac{1}{8} [x-frac{Sin(2x)}{2}])
d) (frac{1}{4} [x-frac{Sin(2x)}{2}])
Answer: c
Explanation: Add constant automatically
Given,f(x)=(int Cos^2 (x) Sin^2 (x)dx=frac{1}{4} int Sin^2 (2x) dx=frac{1}{4} int frac{[1-Cos(2x)]}{2} dx=frac{1}{8} [x-frac{Sin(2x)}{2}])

5. If differentiation of any function is zero at any point and constant at other points then it means?
a) Function is parallel to x-axis at that point
b) Function is parallel to y-axis at that point
c) Function is constant
d) Function is discontinuous at that point
Answer: a
Explanation: Since slope of a function is given by dydx at that point. Hence, when dydx = 0 means slope of a function is zero i.e, parallel to x axis.
Function is not a constant function since it has finite value at other points.

6. If differentiation of any function is infinite at any point and constant at other points then it means ___________
a) Function is parallel to x-axis at that point
b) Function is parallel to y-axis at that point
c) Function is constant
d) Function is discontinuous at that point
Answer: a
Explanation: Since slope of a function is given by dydx at that point.Hence,when dydx = ∞ means slope of a function is 90 degree i.e,parallel to y axis.

7. Integration of function y = f(x) from limit x1 < x < x2 , y1 < y < y2, gives ___________
a) Area of f(x) within x1 < x < x2
b) Volume of f(x) within x1 < x < x2
c) Slope of f(x) within x1 < x < x2
d) Maximum value of f(x) within x1 < x < x2
Answer: a
Explanation: Integration of function y=f(x) from limit x1 < x < x2 , y1 < y < y2, gives area of f(x) within x1 < x < x2.

8. Find the value of ∫ ln⁡(x)x dx.
a) 3a2
b) a2
c) a
d) 1
Answer: a
Explanation: Add constant automatically
Given, f(x)=(int frac{ln⁡(x)}{x} dx)
Let, z=ln⁡(x)=>dz=(frac{dx}{x}=>f(x)=int zdz=z^2/2=frac{ln^2⁡(x)}{2})

9. Find the value of ∫t(t+3)(t+2) dt, is?
a) 2 ln⁡(t+3)-3 ln⁡(t+2)
b) 2 ln⁡(t+3)+3 ln⁡(t+2)
c) 3 ln⁡(t+3)-2 ln⁡(t+2)
d) 3 ln⁡(t+3)+2ln⁡(t+2)
Answer: c
Explanation: Add constant automatically
Given, et = x => dx = et dt,
Given, f(x)=(int frac{ln⁡(x)}{x} dx)
Let, z=ln⁡(x)=>dz=(frac{dx}{x}=>f(x)=int zdz=frac{z^2}{2}=frac{ln^2⁡(x)}{2})

10. Find the value of ∫ cot3(x) cosec4 (x).
a) –([frac{cot^4⁡(x)}{4}+frac{cosec^6⁡(x)}{6}])
b) –([frac{cosec^4⁡(x)}{4}+frac{cosec^6⁡(x)}{6}])
c) –([frac{cot^4⁡(x)}{4}+frac{cot^6⁡(x)}{6}])
d) –([frac{cosec^4⁡(x)}{4}+frac{cot^6⁡(x)}{6}])
Answer: c
Explanation: Add constant automatically
Given, (int cot^3⁡(x)cosec^4 (x)dx=-int cot^3⁡(x)cosec^2 (x)dcot(x))
=-(int t^3 (1+t^2)dt=-[frac{t^4}{4}+frac{t^6}{6}]=-[frac{cot^4⁡(x)}{4}+frac{cot^6⁡(x)}{6}])

11. Find the value of (int frac{sec^4⁡(x)}{sqrt{tan⁡(x)}} dx).
a) (frac{2}{5}sqrt{tan⁡(x)}[5+sec^2⁡(x)])
b) (frac{2}{5}sqrt{sec⁡(x)}[5+tan^2⁡(x)])
c) (frac{2}{5}sqrt{tan⁡(x)}[6+tan^2⁡(x)])
d) (frac{2}{5}sqrt{tan⁡(x)}[5+tan^2⁡(x)])
Answer: d
Explanation: Add constant automatically
Given, (int frac{sec^4⁡(x)}{sqrt{tan⁡(x)}} dx)
=(int frac{sec^2⁡(x) sec^2⁡(x)}{sqrt{tan⁡(x)}} dx)
=(int frac{1+t^2}{sqrt{t}} dt)
=(int [frac{1}{sqrt{t}}+t^{3/2}]dt)
=(2sqrt{t}+frac{2}{5} t^{5/2})
=(frac{2}{5}sqrt{tan⁡(x)}[5+tan^2⁡(x)])

12. Find the value of (int frac{1}{4x^2+4x+5} dx).
a) 18 sin(-1)⁡(x + 12)
b)14 tan(-1)⁡(x + 12)
c) 18 sec(-1)⁡(x + 12)
d) 14 cos(-1)⁡(x + 12)
Answer: b
Explanation: Add constant automatically
Given, (int frac{1}{4x^2+4x+5} dx)
=(int frac{1}{4 (x^2+x+frac{5}{4}+frac{1}{4}+frac{1}{4})} dx =int frac{1}{4[(x+frac{1}{2})^2+1^2])}dx=frac{1}{4} tan^{-1}(x+frac{1}{2}))

13. Find the value of (int sqrt{4x^2+4x+5} dx).
a) (2left [frac{1}{2} (x+frac{1}{2}) sqrt{{(x+frac{1}{2})^2+1)}}right ]+ln⁡left [(x+frac{1}{2})+sqrt{(x+frac{1}{2})^2+1} right ])
b) (2left [frac{1}{2} sqrt{(x+frac{1}{2})^2+1)}right ]+frac{1}{2} ln⁡left [(x+frac{1}{2})+sqrt{(x+frac{1}{2})^2+1} right ])
c) (2left [frac{1}{2} (x+frac{1}{2}) sqrt{(x+frac{1}{2})^2+1)}right ]+frac{1}{2} ln⁡left [(x+frac{1}{2})+sqrt{(x+frac{1}{2})^2+1} right ])
d) (2left [(x+frac{1}{2}) sqrt{{(x+frac{1}{2})^2+1)}}right ]+frac{1}{2} ln⁡left [(x+frac{1}{2})+sqrt{(x+frac{1}{2})^2+1} right ])
Answer: c
Explanation: Add constant automatically
Given, (int sqrt{4x^2+4x+5} dx=int 2sqrt{(x+frac{1}{2})^2+1^2} dx)
=(int 2sqrt{t^2+1^2} dt=2left [frac{1}{2} tsqrt{t^2+1}right ]+frac{1}{2} ln⁡[t+sqrt{t^2+1}])
=(2left [frac{1}{2} (x+frac{1}{2}) sqrt{(x+frac{1}{2})^2+1)} right ]+frac{1}{2} ln⁡left [(x+frac{1}{2})+sqrt{(x+1/2)^2+1}right ])

250+ TOP MCQs on Clairaut’s and Lagrange Equations and Answers

Ordinary Differential Equations Multiple Choice Questions on “Clairaut’s and Lagrange Equations”.

1. Singular solution for the Clairaut’s equation (y = y’x+frac{a}{y’}) is given by _______
a) (frac{x^2}{a^2} + frac{y^2}{a^2} = 1)
b) y2=-4ax
c) y2=4ax
d) x2=-2ay
Answer: c
Explanation: Let ( p = y’ = frac{dy}{dx})
Given equation is of the form y = px + f(p), whose general solution is y = cx + f(c)
thus the general solution is (y = cx + frac{a}{c}) …………..(1), to find the value of c
we differentiate (1) partially w.r.t ‘c’ i.e ( 0 = x-frac{a}{c^2} rightarrow c^2 = frac{a}{x} rightarrow c = sqrt{frac{a}{x}})
hence (1) becomes (y = sqrt{frac{a}{x}} x + a sqrt{frac{x}{a}} rightarrow y=2sqrt{ax})
y2=4ax is the singular solution.

2. Obtain the general solution for the equation xp2+px-py+1-y=0 where p=(frac{dy}{dx}).
a) y=cx+(frac{1}{c+1})
b) x=cy-(c+1)
c) x=cy-(frac{1}{c+1})
d) y=cx+(c+1)
Answer: a
Explanation: xp2+px-py+1-y=0
xp2+px+1=y(p+1)
(y=frac{xp(p+1)+1}{p+1} ,or, y=px+frac{1}{p+1})……(1) thus (1) is in the Clairaut’s equation form y=px+f(p),
thus general solution is y=cx+(frac{1}{c+1}).

3. Find the general solution for the equation (px-py)(py+x)=2p by reducing into Clairaut’s form by using the substitution X=x2, Y=y2 where p=(frac{dy}{dx}).
a) (y^2 = x + frac{c}{c+1})
b) (y^2 = cx^2 – frac{2c}{c+1})
c) (x^2 = cy^2 – frac{1}{2c+1})
d) (x^2 = y^2 + frac{c}{2c+2})
Answer: b
Explanation: (X=x^2 rightarrow frac{dX}{dx} = 2x)
(Y=y^2 rightarrow frac{dY}{dy} = 2y)
now (p = frac{dy}{dx} = frac{dy}{dY} frac{dY}{dX} frac{dX}{dx} ,and, ,let, P=frac{dY}{dx})
(p=frac{1}{2y} * P * 2x ,or, p=frac{x}{y} ,P ,i.e, p=sqrt{frac{X}{Y}} P)
now consider (px-py)(py+x)=2p substituting the value of p we get
(left(sqrt{frac{X}{Y}} P sqrt{X} – sqrt{Y}right)left(sqrt{frac{X}{Y}} P sqrt{Y} + sqrt{X}right) = 2sqrt{frac{X}{Y}} P)
(frac{(PX-Y)}{sqrt{Y}} (P+1)sqrt{X} = 2sqrt{frac{X}{Y}} P rightarrow (PX-Y)(P+1)=2P ,or, Y(P)=PX-frac{2P}{P+1}) is in the Clairaut’s form
hence general solution is (y^2 = cx^2 – frac{2c}{c+1}).

4. Find the general solution of the D.E 2y-4xy’-log y’=0.
a) (y(p) = frac{2c}{p} – 1 + frac{log⁡p}{2} )
b) (y(p) = frac{c}{2p} – 2 + log⁡p)
c) (x(p) = frac{-1}{p} + frac{c}{p^2} )
d) (x(p) = frac{1}{2p} + frac{c}{p^{1/2}} )
Answer: a
Explanation: Let y’=p and hence given equation is in Lagrange equation form
i.e 2y=4xp+log p …..(1) differentiating both sides of the equation
2dy=4xdp+4pdx+(frac{dp}{p}) and dy=pdx
–> 2pdx=4xdp+4pdx+(frac{dp}{p}) –> -2pdx=4pdx+(frac{dp}{p})
-2p(frac{dx}{dp} = 4x + frac{1}{p} rightarrow frac{dx}{dp} + frac{2}{p} x = frac{-1}{2p^2}) (p≠0)…….this is a linear D.E for the function x(p)
I.F is (e^{int frac{2}{p} ,dp} = e^{log⁡p^2} = p^2) and solution is x(p) (p^2 = int p^2 *frac{-1}{2p^2} ,dp + c)
(x(p) = frac{-1}{2p} + frac{c}{p^2}) substituting back in (1) we get (2y=4pleft(frac{-1}{2p} + frac{c}{p^2}right) + log p)
(y(p) = frac{2c}{p} – 1 + frac{log⁡p}{2} ).

5. Find the general solution of the D.E y = 2xy’ – 3(y’)2.
a) (y(p) = p^{1/2} + frac{c}{2p})
b) (y(p) = p^2 + frac{2c}{p})
c) (x(p) = -cp + frac{c}{p^2})
d) (x(p) = 2p + frac{2c}{p^2} )
Answer: b
Explanation: Let y’=p –> y = 2xp – 3p2 ….(1) is in the Lagrange equation form
now differentiating we get dy=2xdp+2pdx-6pdp and dy=pdx
thus pdx=2xdp+2pdx-6pdp→-pdx=2xdp-6pdp –> (frac{dx}{dp} + frac{2}{p} x – 6=0)…(2)
(2) is a linear D.E whose I.F=(e^{int frac{2}{p} ,dp} = p^2) hence its solution is
(p^2 x(p) = int 6p^2 ,dp + c rightarrow x(p) = 2p + frac{c}{p^2}) ….substituting in (1) we get
(y(p) = 2(2p+frac{c}{p^2})p-3p^2 = p^2 + frac{2c}{p}.)

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250+ TOP MCQs on Laplace Transform by Properties and Answers

Engineering Mathematics Multiple Choice Questions on “Laplace Transform by Properties – 3”.

1. Time domain function of (frac{s}{a^2+s^2}) is given by?
a) Cos(at)
b) Sin(at)
c) Cos(at)Sin(at)
d) Sin(t)
Answer: a
Explanation: L[Cos(at)] = (frac{s}{a^2+s^2})
L-1 ([frac{s}{a^2+s^2}]) = Cos(at).

2. Inverse Laplace transform of (frac{1}{(s+1)(s-1)(s+2)}) is?
a) –12 et + 16 e-t + 13 e2t
b) –12 e-t + 16 et + 13 e-2t
c) 12 e-t16 et13 e-2
d) –12 e-t + 16 e-t + 13 e-2
Answer: b
Explanation:
Given, (F(s)=frac{1}{(s+1)(s-1)(s+2)}=frac{-1}{2(s+1)} +frac{1}{6(s-1)}+frac{1}{3(s+2)})
Hence, inverse laplace transform is (f(t)=-frac{1}{2} e^{-t}+frac{1}{6} e^t+frac{1}{3} e^{-2})

3. Inverse laplace transform of (frac{1}{(s-1)^2 (s+5)}) is?
a) 16 e – t136 et + 136 e-5t
b) 16 ett – 136 et + 136 e-5t
c) 16 e-tt2136 e-t + 136 e5t
d) 16 e-t t-136 e-t + 136 e5t
Answer: a
Explanation:
Given, (F(s)=frac{1}{(s-1)^2 (s+5)}=frac{1}{(s-1)} left [frac{1}{(s-1)(s+5)}right ])
=(frac{1}{(s-1)} left [frac{1}{6(s-1)}-frac{1}{6(s+5)}right ])
=(frac{1}{6} left [frac{1}{(s-1)^2}-frac{1}{(s-1)(s+5)}right ])
=(frac{1}{6} left [
frac{1}{(s-1)^2} – frac{1}{6} left [frac{1}{(s-1)} – frac{1}{(s+5)}right ]right ])
=(frac{1}{6(s-1)^2}-frac{1}{36(s-1)}+frac{1}{36(s+5)})
Inverse Laplace transform is (f(t)=frac{1}{6} e^t t-frac{1}{36} e^t+frac{1}{36} e^{-5t})

4. Find the inverse laplace transform of (frac{1}{(s^2+1)(s – 1)(s + 5)}).
a) 112 et113 Cos(-t) – 112 Sin(-t) – 1156 e-5t
b) 112 e-t113 Cos(t) – 112 Sin(t) – 1156 e5t
c) 112 et113 Cos(t) – 112 Sin(t) – 1156 e-5t
d) 112 et + 113 Cos(t) + 112 Sin(t) + 1156 e-5t
Answer: c
Explanation:
Given , F(s)=(frac{1}{(s^2+1)(s-1)(s+5)})
F(s)=(frac{1}{6(s^2+1)}left [frac{1}{s-1}-frac{1}{s+5}right ]=frac{1}{6(s^2+1)(s-1)}-frac{1}{6(s^2+1)(s+5)})
=(frac{1}{6} left [frac{1}{2*(s – 1)}-frac{1}{2} frac{s+1}{(s^2+ 1)}right ]-frac{1}{6}left [frac{1}{26*(s + 5)}-frac{1}{26} frac{s-5}{(s^2+1)}right ])
=(frac{1}{12(s – 1)}-frac{1}{26} frac{2s+3}{(s^2+ 1)}-frac{1}{156(s + 5)})
=(frac{1}{12(s – 1)}-frac{1}{13} frac{s}{(s^2+ 1)}-frac{1}{12} frac{1}{(s^2+ 1)}-frac{1}{156(s + 5)})
=(frac{1}{12} e^t-frac{1}{13} Cos(t)-frac{1}{12} Sin(t)-frac{1}{156}e^{-5t})

5. Find the inverse laplace transform of (frac{s}{(s^2+ 4)^2}).
a) 14 sin(2t)
b) t24 sin(2t)
c) t4 sin(2t)
d) t4 sin(2t2)
Answer: c
Explanation:
Given, (Y(s)=frac{s}{(s^2+ 4)^2})
Inverse Laplace transform of (frac{1}{s^2+4})=sin⁡(2t)
Now, (frac{d}{ds} (frac{1}{s^2+4}))=-tsin(2t)
Inverse lapalce of (frac{-2s}{(s^2+4)^2}=-frac{t}{2} sin(2t))
Inverse lapalce of (frac{s}{(s^2+4)^2}=frac{t}{4} sin(2t))

6. Final value theorem states that _________
a) x(0)=(lim_{xrightarrow ∞} sX(s))
b) x(∞)=(lim_{xrightarrow ∞} sX(s))
c) x(0)=(lim_{xrightarrow 0} sX(s))
d) x(∞)=(lim_{xrightarrow 0} ⁡sX(s))
Answer: d
Explanation: Final value theorem states that
x(∞)=(lim_{xrightarrow 0} ⁡sX(s))

7. Initial value theorem states that ___________
a) x(0)=(lim_{xrightarrow ∞} sX(s))
b) x(∞)=(lim_{xrightarrow ∞} sX(s))
c) x(0)=(lim_{xrightarrow 0} sX(s))
d) x(∞)=(lim_{xrightarrow 0} ⁡sX(s))
Answer: a
Explanation: Initial value theorem states that
x(0)=(lim_{xrightarrow ∞} sX(s))

8. Find the value of x(∞) if (X(s)=frac{2s^2+5s+12/s}{s^3+4s^2+14s+20}).
a) 5
b) 4
c) 1220
d) 2
Answer: c
Explanation:
Given, (X(s)=frac{2s^2+5s+12/s}{s^3+4s^2+14s+20})
Hence, (sX(s)=frac{2s^3+5s^2+12}{s^3+4s^2+14s+20})
Hence, by final value theorem,
(x(∞)=lim_{xrightarrow 0} ⁡sX(s)=frac{12}{20})

9. Find the value of x(0) if (X(s)=frac{2s^2+5s+12/s}{s^3+4s^2+14s+20}).
a) 5
b) 4
c) 12
d) 2
Answer: d
Explanation:
Given, (X(s)=frac{2s^2+5s+12/s}{s^3+4s^2+14s+20})
Hence, (sX(s)=frac{2s^3+5s^2+12}{s^3+4s^2+14s+20})
Hence, by initial value theorem,
(x(0)=lim_{xrightarrow infty} ⁡sX(s)=2)

10. Find the inverse lapace of (frac{(s+1)}{[(s+1)^2+4][(s+1)^2+1]}).
a) 13 et [Cos(t) – Cos(2t)].
b) 13 e-t [Cos(t) + Cos(2t)].
c) 13 et [Cos(t) + Cos(2t)].
d) 13 e-t [Cos(t) – Cos(2t)].
Answer: d
Explanation:
Given, (Y(s)=frac{(s+1)}{[(s+1)^2+4][(s+1)^2+1]})
=(frac{s+1}{3(s^2+ 2*s + 2)}-frac{s+1}{3(s^2+ 2*s + 5)})
=(frac{s+1}{3[(s+1)^2+1]}-frac{s+1}{3[(s+1^2+4)]})
=(frac{1}{3} [e^{-t} Cos(t)]-frac{1}{3}[e^{-t} Cos(2t)])
=(frac{1}{3} e^{-t} [Cos(t)-Cos(2t)])

11. Find the inverse laplace transform of (Y(s)=frac{2s}{1-s^2}e^{-s}).
a) -e-t + 1 + et – 1
b) -e-t + 1 – et + 1
c) -e-t + 1 + et + 1
d) -e-t + 1 – et – 1
Answer: d
Explanation: Given,
Y(s)=(frac{2s}{1-s^2}e^{-s})
Let,G(s)=(frac{2s}{1-s^2}=-frac{1}{s – 1}-frac{1}{s + 1})
hence,g(t)=(-e^{-t} – e^t)
Since,Y(s)=(e^{-s} G(s)=>y(t)=g(t-1))
hence,y(t)=(-e^{-t+1}-e^{t-1})

12. Find the inverse laplace transform of (frac{1}{s(s-1)(s^2+1)}).
a) 12 e-t + 12 Sin(-t) – 12 Cos(-t)
b) 12 et + 12 Sin(t) – 12 Cos(t)
c) 12 et + 12 Sin(t) + 12 Cos(t)
d) 12 et12 Sin(t) – 12 Cos(t)
Answer: b
Explanation: We know that,
Given, Y(s)=(frac{1}{s(s-1)(s^2+1)})
Let, G(s)=(frac{1}{(s-1)(s^2+1)}=frac{1}{2(s^2-1)}-frac{s+1}{2(s^2+1)}=frac{1}{2*(s-1)}-frac{s}{2(s^2+1)}-frac{1}{2(s^2+1)})
Now, g(t)=(frac{1}{2}e^t-frac{1}{2}cos(t)-frac{1}{2}cos(t))
Now, Y(s)=(frac{1}{2}G(s)=>y(t)=int_0^t g(t)dt=frac{1}{2}e^t+frac{1}{2}sin(t)-frac{1}{2}cos(t))