Engineering Questions with Answers - Multiple Choice Questions
Engineering Mathematics MCQ – Leibniz Rule – 1
1 - Question
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
View Answer
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)=cn0u(v)(n)+cn1u(1)(v)(n−1)+….+cnnu(n)v
Differentiating expression (1) in accordance to Leibniz rule (upto the hundredth derivative) we have
(y(1+x2))(100) = c1000y(100)(1+x2)+c1001y(99)(2x)+c1002y(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 - Question
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)!
View Answer
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)=c1000y(100)(x+1)+c1001y(99)+0+…..+0=(ln(x))(100)
Using the nth derivative for ln(x+a) as dn(ln(x+a))dxn=(−1)n+1×(n−1)!(x+a)n
we have the right hand side as
(y(1+x))(100) = (−99)!x100
Now substituting x = 1 yields
2y(100) + y(99) = (−99)!1
= -(99)!
3 - Question
Let f(x) = 1−x2−−−−−√ 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
View Answer
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)=c30y(3)y+c31y(2)y(1)+c32y(1)y(2)+c33y(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 - Question
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
View Answer
Answer: b
Explanation:Assume y = f(x) and we also know that tan(x)=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)=cn0y(n)cos(x)−cn1y(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)=c20y(2)cos(0)−c21y(1)sin(0)−c22ycos(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 - Question
If the first and second derivatives at x = 0 of the function f(x)=cos(x)x2−x+1 were 2 and 3 then the value of the third derivative is
a) -3
b) 3
c) 2
d) 1
View Answer
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(x2−x+1))(3)=c30y(3)(x2−x+1)−c31y(2)(2x−1)−c32y(1)(2)(cos(x))(3)=sin(x)
Equating both sides we have
sin(x)=c30y(3)(x2−x+1)−c31y(2)(2x−1)−c32y(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)=c30y(3)(x2−x+1)−c31(3)(2x−1)−c32(2)(2)
Substituting x = 0 gives
sin(0) = y(3) + 9 -12
y(3) = 12 – 9 = 3.
6 - Question
For the given function f(x)=x3+x7−−−−−−√ 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
View Answer
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)
c30y(3)y+c31y(2)y(1)+c32y(1)y(2)+c33yy(3)=3!+(7.6.5)x4….(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 - Question
Let f(x)=ex×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) πn4
b) ∑i<=ni=0(−1)icn2i2i+1
c) πn
d) π2n
View Answer
Answer: b
Explanation: Expanding sin(x)/x into Taylor series we have
sin(x)x=11!−x23!+x45!−…..∞
Now Taking the nth derivative of function using Leibniz rule we have
(ex(sin(x)x))(n)=cn0ex(11!−x23!+x45!−…∞)+cn1ex(−2x3!−4x35!−…∞)+….
Now substituting x=0 we have
[(ex(sin(x)x))(n)]x=0=cn0(11!)+cn2(−2!3!)+c4n(4!5!)…∞
[(ex(sin(x)x))(n)]x=0=∑i<=ni=0(−1)icn2i2i+1
Hence, Option ∑i<=ni=0(−1)icn2i2i+1 is the right answer.
Explanation: Expanding sin(x)/x into Taylor series we have
sin(x)x=11!−x23!+x45!−…..∞
Now Taking the nth derivative of function using Leibniz rule we have
(ex(sin(x)x))(n)=cn0ex(11!−x23!+x45!−…∞)+cn1ex(−2x3!−4x35!−…∞)+….
Now substituting x=0 we have
[(ex(sin(x)x))(n)]x=0=cn0(11!)+cn2(−2!3!)+c4n(4!5!)…∞
[(ex(sin(x)x))(n)]x=0=∑i<=ni=0(−1)icn2i2i+1
Hence, Option ∑i<=ni=0(−1)icn2i2i+1 is the right answer.
8 - Question
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) ∑ni=0cn2i2i+1
b) ∑ni=012i+1
c) 1
d) Has no general form
View Answer
Answer: a
Explanation: The key here is to find a separate Taylor expansion for sinh(x) / x which is
sinh(x)x=x1!+x33!+x55!…∞x
sinh(x)x=x1!+x23!+x45!…∞
Now consider ((ex)(sinh(x)x)) applying the Leibniz rule for nth derivative we have
(ex(sin(x)x))(n)=cn0ex(11!−x23!+x45!−…∞)+cn1ex(2x3!+4x35!…∞)
+cn2(23!+12x25!….∞)+…
Now substituting x=0 yields
(ex(sin(x)x))(n)=cn01+cn23+cn45
∑ni=0cn2i2i+1