1. The pth derivative of a qth degree monic polynomial, where p, q are positive integers and 2p4 + 3pq3⁄2 = 3q3⁄2 + 2qp3 is given by?
a) Cannot be generally determined
b) (q – 1)!
c) (q)!
d) (q – 1)! * pq
Explanation: First consider the equation
2p4 + 3pq3⁄2 = 3q3⁄2 + 2qp3
After simplification, we get
(2p3 + 3q1⁄2) (p – q) = 0
This gives us two possibilities
2p3 = – 3q1⁄2
OR
p = q
The first possibility can’t be true as we are dealing with positive integers
Hence, we get
p = q
Thus the pth derivative of any monic polynomial of degree p(p = q) is
p! =q!.
2. The first and second derivatives of a quadratic Polynomial at x = 1 are 1 and 2 respectively. Then the value of f(1) – f(0) Is given by?
a) 3⁄2
b) 1⁄2
c) 1
d) 0
Explanation: Let the quadratic polynomial be
f(x) = ax2 + bx + c
The first derivative at x = 1 is given by
2a + b = 1
Now consider the second derivative at x = 1 which is given by
2a = 2
Solving for the coefficients using equations, we get the values as a = 1 and b = -1
Putting these values back in the polynomial yields
f(x) = x2 – x + c
Now the required value can be computed as
f(1) – f(0) = (12 – 1 + c) – (02 – 0 + c)
= (0 + c) – (0 + c) = 0.
3. Let f(x) = \(\frac{sin(x)}{x – 54}\), then the value of f(100)(54) is given by?
a) Undefined
b) 100
c) 10
d) 0
Explanation: The key here is to expand the numerator into a taylor series centered at 54
Doing this gives us the following
Sin(x)=Sin(54)+\(\frac{(x-54)\times Cos(54)}{1!}-\frac{(x-54)^2\times Sin(54)}{2!}…\infty \)
Now the function transforms into f(x)=\(\frac{Sin(54)}{(x-54)}+\frac{Cos(54)}{1!}-\frac{(x-54)\times Sin(54)}{2!}…\infty \)
Observe the first term in the infinite series there is always (x – 54) in the denominator which goes to 0 when we substitute x = 54
Every derivative also will have this term
Hence any derivative of the given function goes to ∞ as x = 54
Hence, the answer is Undefined.
4. Let f(x) = ln(x2 + 5x + 6) then the value of f(30)(1) is given by?
a) (29!)\((\frac{1}{3^{30}}+\frac{1}{4^{30}})\)
b) (-29!)\((\frac{1}{3^{30}}+\frac{1}{4^{30}})\)
c) (30!)\((\frac{1}{3^{30}}+\frac{1}{4^{30}})\)
d) (-30!)\((\frac{1}{3^{30}}+\frac{1}{4^{30}})\)
Explanation: Given function
f(x) = ln(x2 + 5x + 6)
Factorising the inner polynomial we get
f(x) = ln((x + 3) (x + 2))
Now using the rule of logarithms ln (m * n) = ln(m) + ln(n)
we get
f(x) = ln(x + 3) + ln(x + 2)
Now using the nth derivative of logarithmic function \(\frac{d^n(ln(x+a))}{dx^n} = \frac{(-1)^{n+1}\times(n-1)!}{(x+a)^n}\)
We have
\(\frac{d^n(f(x))}{dx^n}=\frac{d^n(ln(x+3))}{dx^n}+\frac{d^n(ln(x+2))}{dx^n}\)
Which simplifies to \(\frac{d^n(f(x))}{dx^n}=\frac{(-1)^{n+1}\times(n-1)!}{(x+3)^n}+\frac{(-1)^{n+1}\times(n-1)!}{(x+2)^n}\)
Substituting x=1 and n=30
Gives us the answer as (-29!)\((\frac{1}{3^{30}}+\frac{1}{4^{30}})\)
5. f(x) = \(\int_{0}^{\pi/2}sin(ax)da\) then the value of f(100)(0) is?
a) a(100) sin(a)
b) -a(100) sin(a)
c) a(100) cos(a)
d) 0
Explanation: First solve the integral f(x)=\(\int_{0}^{\pi/2}sin(ax)da\)
Which gives =\([\frac{-cos(ax)}{x}]_{0}^{\pi/2}=\frac{1-cos(\frac{\pi x}{2})}{x}\)
Now expanding into Taylor series yields
\(=(\frac{(\frac{\pi x}{2})^2}{x2!}-\frac{(\frac{\pi x}{2})^4}{x4!}….\infty)\)
Observe that every term in the expansion is odd powered
Hence even derivative at x = 0 has to be 0.
6. Let f(x) = x9 ex then the ninth derivative of f(x) at x = 0 is given by?
a) 9!
b) 9! * e9
c) 10!
d) 21!
Explanation: The key here is to expand ex as a mclaurin series and then multiply it by x9
We have ex=\(\frac{x^0}{0!}+\frac{x^1}{1!}+\frac{x^2}{2!}…\infty\)
Now multiplying it by x9 we get
f(x)=\(\frac{x^9}{0!}+\frac{x^{10}}{1!}+\frac{x^{11}}{2!}…\infty\)
This simplifies into polynomial differentiation and the ninth derivative is as follows
f(9)(x)=\(9!+\frac{10!\times x}{1!}+\frac{p_9^{11}\times x^2}{2!}…\infty\)
Substituting x = 0 gives us the result
f(9)(0) = 9!.
7. The following moves are performed on g(x).
(i) Pick (x0, y0) on g(x) and travel toward the left/right to reach the y = x line. Now travel above/below to reach g(x). Call this point on g(x) as (x1, y1)
(ii) Let the new position of (x0, y0)be (x0, y1)
This is performed for all points on g(x) and a new function is got. Again these steps may be repeated on new function and another function is obtained. It is observed that, of all the functions got, at a certain point (i.e. after finite number of moves) the nth derivatives of the intermediate function are constant, and the curve passes through the origin. Then which of the following functions could be g(x)
a) y = \(\sqrt{1 – x^2}\)
b) xy3⁄2 + y = constant
c) x9 y3⁄2 + y6 x3⁄2 = constant
d) x7 y 8 + 4y = constant
Explanation: The phrasing is a bit convoluted. It is asking us to convert a given function f(x) into its own composition f(f(x))
Now the set of finite number of moves asks us to find the composition of composition i.e. f(f(x)) and f(f(x)) and so on.
It also says that the intermediate composition function has the property of all nth derivatives being constant. This is possible only if the intermediate function is of the form k(x) = ax + b (Linear Function)
Given another condition that it has to pass through origin leads to the conclusion that b = 0
So the intermediate function has the form
k(x) = ax
The first possibility gives rise to
k(x) = x
Now one has to select from the options as to which curve would give k(x) = f(f(….(x)…)) = x when composed with itself a finite number of times.
This can be done by simply interchanging the position of y and x
in the options and check whether it preserves its structure.
For y = \(\sqrt{1 – x^2}\) we have after changing the position
This has the same structure.
8. The first, second and third derivatives of a cubic polynomial f(x) at x = 1, are 13, 23 and 33 respectively. Then the value of f(0) + f(1) – 2f(-1) is?
a) 76
b) 86
c) 126
d) 41.5
Explanation: Assume the polynomial to be of the form f(x) = ax3 + bx2 + cx + d
Now the first derivative at x = 1 yields the following equation
13 = 1 = 3a + 2b + c
The second derivative at x = 1 yields the following expression
23 = 8 = 6a + 2b
The third derivative at x = 1 yields the following equation
33 = 27 = 6a
Solving for a, b and c simultaneously yields
(a, b, c) = (9⁄2, -19⁄2, 13⁄2)
Hence the assumed polynomial is f(x) = 9x3 – 19x2 + 13x ⁄ 2 + d
Now the given expression can be evaluated as
f(0) + f(1) – 2f(-1) = (d) + (3⁄2 + d) – 2(-20 + d)
= 40 + 3⁄2
= 41.5.
9. Let g(x) = ln(x) ⁄ x – 1 Then the hundredth derivative at x = 1 is?
a) 100!⁄101
b) 99!⁄101
c) 101⁄100!
d) 1⁄99!
Explanation: The key here is to again expand the numerator as a Taylor series centered at x = 1
Hence we have the Taylor series as
ln(x)=\(\frac{(x-1)}{1}-\frac{(x-1)^2}{2}+\frac{(x-1)^3}{3}…..\infty\)
Hence our function g(x) transforms into
g(x)=\(1-\frac{(x-1)}{2}-\frac{(x-1)^2}{3}+\frac{(x-1)^3}{4}…..\infty\)
This now simplifies into polynomial differentiation and the hundredth derivative can be written as
g(100)(x)=\(\frac{100!}{101}-\frac{101!\times(x-1)}{102}…..\infty\)
Substituting x = 1 yields
100!⁄101.
10. Let f(x) = ln(x3 – 3x2 – 16x – 12) , then the 1729th derivative at x = 234 is?
a) (1728!)\(\times(\frac{1}{228^{1729}}+\frac{1}{236^{1729}}+\frac{1}{235^{1729}})\)
b) (-1728!)\(\times(\frac{1}{228^{1729}}+\frac{1}{236^{1729}}+\frac{1}{235^{1729}})\)
c) (1728!)\(\times(\frac{1}{228^{1729}}+\frac{1}{236^{1728}}+\frac{1}{235^{1729}})\)
d) (-1729!)\(\times(\frac{1}{228^{1729}}+\frac{1}{236^{1729}}+\frac{1}{235^{1729}})\)
Explanation: The function can be written as f(x) = ln((x – 6)(x + 1)(x + 2))
Using property of logarithms we have
f(x) = ln(x + 1) + ln(x + 2) + ln(x – 6)
Using the nth derivative of logarithmic functions
\(\frac{d^n(ln(x+a))}{dx^n}=\frac{(-1)^{n+1}\times(n-1)!}{(x+a)^n}\)
We have the 1729th derivative of f(x) as
\(f^{(1729)} (x) = \frac{1728!}{(x-6)^{1729}} + \frac{1728!}{(x+2)^{1729}} + \frac{1728!}{(x+1)^{1729}}\)
Now substituting x=234 we get our answer as
= (1728!)\(\times(\frac{1}{228^{1729}}+\frac{1}{236^{1729}}+\frac{1}{235^{1729}})\).