This set of Engineering Mathematics Multiple Choice Questions & Answers (MCQs) focuses on “Lagrange’s Mean Value Theorem – 1”.
1. For the function f(x) = x2 – 2x + 1. We have Rolles point at x = 1. The coordinate axes are then rotated by 45 degrees in anticlockwise sense. What is the position of new Rolles point with respect to the transformed coordinate axes?
a) 3⁄2
b) 1⁄2
c) 5⁄2
d) 1
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Explanation:The coordinate axes are rotated by 45 degree then the problem transforms into that of Lagrange mean value theorem where the point in some interval has the slope of tan(45).
Hence differentiating the function and equating to tan(45).
We have
f'(x) = tan(45) = 2x – 2
2x – 2 = 1
x = 3⁄2.
2. 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) \(\sqrt{\frac{\sqrt{3}-1}{3}}\)
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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=\(\sqrt{\frac{\sqrt{3}-1}{3}}\)
3. 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)
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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).
4. 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)
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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).
5. 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
view answer
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.
6. 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
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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.
7. 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
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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.
8. A function f(x) with n roots should have n – 1 unique Lagrange points.
a) True
b) False
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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.
9. 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
view answer
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.
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