This set of Engineering Mathematics Multiple Choice Questions & Answers (MCQs) focuses on “Rolle’s Theorem – 1”.
1. For y = -x2 + 2x there exist a c in the interval [ – 19765, 19767] Such that f'(c) = 0
Explanation: The key here is to rewrite the function as y = -(x – 1)2 + 1
Observe here that on substituting – 19765 and 19767 in the equation we get
(- 19766)2 + 1 and (- 19766)2 respectively.
As we are dealing with their squared values they have to be equal
f(- 19765) = f(19767) Polynomial functions are continuous and differentiable over the whole domain and hence by Rolles Theorem we must have a c such that f'(c) = 0 in the interval [-19765, 19767] Hence, the claim is true.
2. For the function f(x) = sin(x)⁄x2 How many points exist in the interval [0, 7π] Such that f’ (c) = 0
Explanation: We know that sine is a periodic function and it is divided by x2.
Observe that the sine takes the value of zero at integral arguments, hence at every interval of the form
We have f(nπ) = f((n + 1) π)
The sine and the polynomial combination is continuous and differentiable at every point except x = 0
Every such interval has a point such that f’ (c) = 0
Hence, by Rolles theorem, in every interval of the form [nπ, (n + 1)π] we must have a point such that
Leaving the interval [0, π] we are left with six such intervals from 0 to 7π.
3. f(x) = sin(x)⁄x, How many points exist such that f’ (c) = 0 in the interval [0, 18π]
Explanation: We have the sine function that takes the value of zero at
Integral multiples, But for sin(x)⁄x we have the exceptional value of limx→0 sin(x)⁄x reaching one.
So leaving the first interval [0, π], for every other interval of the form [nπ (n + 1)π ] we must have f(nπ) = f((n + 1)π)
By Rolles theorem we have
f’ (c) = 0 For every interval of the form [nπ (n + 1)π ] There are 17 such intervals.
4. Let f(x) = x + sin(x) Every point on the graph is rotated by 45 degree with respect to the origin along the radius equal to the radius vector at that point. How many c that belong to [0, 11π] exist Such that f'(c) = 0
Explanation: The function has the beautiful property that the sine function is built over the line y = x
In other words the axis for the sine graph is no more on the x axis but is the line y = x .
When this graph is rotated by 45 degrees which Is equal to the slope of the line y = x we must get the original graph sin(x) again. The number of c in the interval [0, 11π] is then obviously equal to eleven.
5. A Function f(x) has the property f(a) = f(b) for ∀a,b…∊….I and a + b = 20 then which of the following even degree polynomials could be f(x)
a) x4⁄4 – 10x3 + 150x2 – 1000x + 10131729
b) x2 + 5x + 6
c) x2 + x + 1
d) Polynomial functions are inadequate representations
Explanation: Given that f(a) = f(b) for all integral values of a, b, one additional constraint is that a + b⁄2 = 10
This means that the graph is symmetric about the line x = 10
This means that the even degree polynomial has a Rolle point(the only Rolle point) at x = 10
We need to differentiate the functions given in the options and observe which of these has a single(OR repeated root), which is equal to 10.
The first Option when differentiated yields
f'(x) = x3 – 30x2 + 300x – 1000
Equating to zero, we see that x = 10
satisfies the equation
(10)3 – 30(10)2 + 300(10) – 1000 = 0.
6. For some function f(x) we have f(a) = f(b) for ∀a,b…∊….I and a + b = 2 then which of the following even degree polynomials could f(x) be
a) x2 + 3x +1
b) 5x2⁄2 – 5x + 101
c) x2 + 2x + 1
d) Even degree polynomials of such kind cannot exist
Explanation: Given that f(a) = f(b) for all integral values of a, b with the condition a + b ⁄ 2 = 1
We know that the function is symmetric about the line x = 1
Thus we must have a Rolle point at 1 for any function that satisfies these conditions.
Differentiating the given functions in the options we have for a
2x + 3 = 0
x = -3⁄2 ≠ we get
5x – 5 = 0
7. For all second degree polynomials with y = ax2 + bx + k, it is seen that the Rolles’ point is at c = 0 . Also the value of k is zero. Then what is the value of b
Explanation: Given that k = 0 we must rewrite the function to get
y = ax2 + bx
Now differentiating the function yields
y’ = 2ax + b = 0
Equating it to zero we get the Rolle point which is also zero
2a(0) + b = 0
b = 0 .
8. For second degree polynomial it is seen that the roots are equal. Then what is the relation between the Rolles point c and the root x
a) c = x
b) c = x2
c) They are independent
d) c = sin(x)
Explanation: For a polynomial with equal roots we can write the polynomial as f(x) = (x – a)2
Where a is the repeated root.
Differentiating it and equating the function to zero we get the Rolles point,
f'(x) = 2(x – a) = 0
x = a = c.
9. For any second degree polynomial with two real unequal roots. The relation between Rolles point r1 and the two roots r2 is
a) They are independent
b) c = r1 – r2
c) c = r1 * 1⁄r2
d) c = r1 + r2⁄2
Explanation: Given the polynomial has two real and unequal roots, we can write the polynomial in the following form
y = (x – a)(x – b)
Where a ≠ b are the two, unequal, real roots.
Now differentiation and equating to zero yields
y = (x – b) + (x -a ) = 0
Put x = c (because is the Rolles point here)
c = a + b / 2 = r1 + r2⁄2.
10. If the domain of a function can be broken into infinite number of disjoint subsets such that every subset has a Rolles point then the function cannot be in a
Explanation: The function could be an infinite degree polynomial (similar to Taylor series), such that after differentiating once we get another polynomial with infinite degree.
Hence, we may conclude that it has infinite distinct roots (with proper choice of polynomial) and hence, the domain can be broken into epsilon (very small) intervals
around the roots of the differentiated polynomial to get infinite intervals containing the Rolles point.
Sanfoundry Global Education & Learning Series – Engineering Mathematics.
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