1. For the function f(x) = x3 + x + 1. We do not have any Rolles point. The coordinate axes are transformed by rotating them by 60 degrees in anti-clockwise sense. The new Rolles point is?
a) \(\frac{\sqrt{3}}{2}\)
b) The function can never have a Rolles point
c) \(3^{\frac{1}{2}}\)
d) \(\sqrt{\frac{\sqrt{3}-1}{3}}\)
Explanation: The question is simply asking us to find if there is some open interval in the original function f(x)
where we have f'(x) = tan(60)
We have
f'(x) = 3x2 + 1 = tan(60)
3x2 = √3 – 1
x=\(\sqrt{\frac{\sqrt{3}-1}{3}}\)
2. What is the minimum angle by which the coordinate axes have to be rotated in anticlockwise sense (in Degrees), such that the function f(x) = 3x3 + 5x + 1016 has at least one Rolles point
a) π⁄180 tan-1(5)
b) tan-1(5)
c) 180⁄π tan-1(5)
d) -tan-1(5)
Explanation: For the transformed function to have a Rolles point is equivalent to the existing function having a Lagrange point somewhere in the real number domain, we are finding
the point in the domain of the original function where we have f'(x) = tan(α)
Let the angle to be rotated be α
We have
f'(x) = 9x2 + 5 = tan(α)
9x2 = tan(α) – 5
For the given function to have a Lagrange point we must have the right hand side be greater than zero, so
tan(α) – 5 > 0
tan(α) > 5
α > tan-1(5)
In degrees we must have,
αdeg > 180⁄π tan-1(5).
3. For a third degree monic polynomial, it is seen that the sum of roots are zero. What is the relation between the minimum angle to be rotated to have a Rolles point (α in Radians) and the cyclic sum of the roots taken two at a time c
a) α = π⁄180 * tan-1(c)
b) Can never have a Rolles point
c) α = 180⁄π tan-1(c)
d) α = tan-1(c)
Explanation: From Vietas formulas we can deduce that the x2 coefficient of the monic polynomial is zero (Sum of roots = zero). Hence, we can rewrite our third degree polynomial as
y = x3 + (0) * x2 + c * x + d
Now the question asks us to relate α and c
Where c is indeed the cyclic sum of two roots taken at a time by Vietas formulae
As usual, Rolles point in the rotated domain equals the Lagrange point in the existing domain. Hence, we must have
y ‘ = tan(α)
3x2 + c = tan(α)
To find the minimum angle, we have to find the minimum value of α
such that the equation formed above has real roots when solved for x.
So, we can write
tan(α) – c > 0
tan(α) > c
α > tan-1(c)
Thus, the minimum required angle is
α = tan-1(c).
4. For the infinitely defined discontinuous function
\(\begin{cases}x+sin(2x)& :x\in [0,\pi] \\ x+sin(4x)& :x\in (\pi,2\pi] \\ x+sin(6x)& :x\in (2\pi,3\pi] \\.\\.\\x+sin(2nx)& :x\in [(n-1)\pi,n\pi)\\.\\.\end{cases}\)
How many points c∈[0,16x] exist, such that f'(c) = 1
a) 256
b) 512
c) 16
d) 0
Explanation: To find points such that f'(c) = 1
We need to check points on graph where slope remains the same (45 degrees)
In every interval of the form [(n – 1)π, nπ] we must have 2n – 1 points
Because sine curve there has frequency 2n and the graph is going to meet the graph y = x at 2n points.
Hence, in the interval [0, 16π] we have
= 1 + 3 + 5…….(16terms)
=(16)2 = 256.
5. Let g(x) be periodic and non-constant, with period τ. Also we have g(nτ) = 0 : n ∈ N. The function f(x) is defined as
f(x)=\(\begin{cases}2x+g(2x)& :x\in [0,\tau] \\ 2x+g(4x)& :x\in (\tau,2\tau] \\2x+g(6x)& :x\in (2\tau,3\tau]\\.\\.\\2x+g(2nx)& :x\in [(n-1)\tau,n\tau)\\.\\.
\end{cases}\)
How many points c ∈ [0, 18τ] exist such that f'(c) = tan-1(2)
a) 325
b) 323
c) 324
d) 162
Explanation: To find points such that f ‘(c) = f ‘(c) = tan-1(2)
We need to check points on the graph such that the slope remains the same ( tan-1(2) radians)
In every interval of the form [(n – 1)τ, nτ] we must have 2n points on the graph because the frequency of periodic function in that interval is 2n and we have g(nτ) = 0
And we have 2n – 1
Lagrange points in the interval [(n – 1)τ, nτ]
The total number of such points in the interval [0, 18τ] is given by
= 1 + 3 + 5…….(18 terms)
= (18)2 = 324.
6. Let f(x)=\(x-\frac{x^3}{3^2.2!}+\frac{x^5}{5^2.4!}-\frac{x^7}{7^2.6!}+…\infty\). Find a point nearest to c such that f'(c) = 1
a) 1
b) 0
c) 2.3445 * 10-9
d) 458328.33 * 10-3
Explanation: First find f'(x)
f'(x)=1-\(\frac{x^2}{3!}+\frac{x^4}{5!}-\frac{x^6}{7!}+…\infty\)
Multiplying and dividing by x We have the well known series expansion of \(\frac{sin(x)}{x}\)
We get
f'(x)=\(\frac{1}{x}\times (x-\frac{x^3}{3!}+\frac{x^5}{5!}-\frac{x^7}{7!}+…\infty)\)
f'(x)=\(\frac{sin(x)}{x}\)
Equating this to 1 We have
\(\frac{sin(x)}{x}\) = 1
We know the well known limit \(lim_{x \rightarrow 0} \frac{sin(x)}{x}\) = 1
Thus we have to choose a point nearer to 0 as our answer which is,
458328.33 * 10-3.
7. A function f(x) with n roots should have n – 1 unique Lagrange points.
a) True
b) False
Explanation: If the roots of the polynomial are equal then we will have less than n – 1 Lagrange points that are unique. Hence, the right Option is False.
8. Let f(x)=\(\frac{x^{100}}{100}+\frac{x^{101}}{101}+…\infty\). Find a point c ∈ (- ∞, ∞) such that f'(c) = 0
a) 1
b) 2
c) 0
d) -1
Explanation: f'(x) = x99+x100+x101+…∞
Using geometric series we have
f'(x)=\(\frac{x^{99}}{1-x}\)
Equating to 0 we have
x99 = 0
Observe that x = 0 satisfies the equation. Neglect other roots as they are complex roots.
9. Cauchy’s Mean Value Theorem can be reduced to Lagrange’s Mean Value Theorem.
a) True
b) False
Explanation: Cauchy’s Mean Value Theorem is the generalized form of Lagrange’s Mean Value Theorem and can be given by,
\(\frac{f'(a+θh)}{g'(a+θh)} = \frac{f(a+h)-f(a)}{g(a+h)-g(a)}, 0 < θ < 1\)
Hence, if g(x) = x, then CMV reduces to LMV.
10. Which of the following is not a necessary condition for Cauchy’s Mean Value Theorem?
a) The functions, f(x) and g(x) be continuous in [a, b]
b) The derivation of g'(x) be equal to 0
c) The functions f(x) and g(x) be derivable in (a, b)
d) There exists a value c Є (a, b) such that, \(\frac{f(b)-f(a)}{g(b)-g(a)} = \frac{f'(c)}{g'(c)}\)
Explanation: Cauchy’s Mean Value theorem is given by, \(\frac{f(b)-f(a)}{g(b)-g(a)} = \frac{f'(c)}{g'(c)}\), where f(x) and g(x) be two functions which are derivable in [a, b] and g'(x)≠0 for any value of x in [a, b] and where c Є (a, b).