OP Malhotra Class 12 Maths Solutions Chapter 10 Mean Value Theorems Chapter Test

S Chand Class 12 ICSE Maths Solutions Chapter 10 Mean Value Theorems Chapter Test

Verify Rolle's Theorem for the function:

Question 1. (i) f(x) = x² -5x + 4 on [1, 4].
Answer: Given the function \( f(x) = x^2 - 5x + 4 \) on the interval \( [1, 4] \).
Since \( f(x) \) is a polynomial, it is continuous on the closed interval \( [1, 4] \).
To find if it's derivable, we calculate the first derivative: \( f'(x) = 2x - 5 \). This derivative exists for all real numbers, so \( f(x) \) is derivable on the open interval \( (1, 4) \).
Next, we check the function values at the endpoints of the interval:
\( f(1) = (1)^2 - 5(1) + 4 = 1 - 5 + 4 = 0 \)
\( f(4) = (4)^2 - 5(4) + 4 = 16 - 20 + 4 = 0 \)
Since \( f(1) = f(4) = 0 \), all three conditions of Rolle's Theorem are satisfied.
Therefore, there must exist at least one real number \( c \) in the open interval \( (1, 4) \) such that \( f'(c) = 0 \).
Set the derivative to zero: \( 2c - 5 = 0 \)
\( \implies \) \( 2c = 5 \)
\( \implies \) \( c = \frac{5}{2} \)
Since \( \frac{5}{2} = 2.5 \), which is indeed within the interval \( (1, 4) \), Rolle's Theorem is verified. The theorem guarantees a point where the tangent is horizontal.
In simple words: We checked if the function is smooth, unbroken, and starts and ends at the same height. It met all conditions. Then, we found the exact spot where its slope is flat, and that spot was inside the allowed range.

🎯 Exam Tip: Remember to explicitly state and verify all three conditions of Rolle's Theorem: continuity, differentiability, and \( f(a) = f(b) \). Showing that the value of \( c \) lies within the open interval \( (a, b) \) is crucial for full marks.

Question 2. f(x) = (x – 1)(x – 2)² on [0, 2π].
Answer: Given the function \( f(x) = (x - 1)(x - 2)^2 \). The problem statement implies an interval of \( [0, 2\pi] \), but the solution provided uses \( [1, 2] \). We will proceed with the interval \( [1, 2] \) as per the detailed steps in the solution.
First, expand the function: \( f(x) = (x - 1)(x^2 - 4x + 4) = x^3 - 4x^2 + 4x - x^2 + 4x - 4 = x^3 - 5x^2 + 8x - 4 \).
Since \( f(x) \) is a polynomial, it is continuous on the closed interval \( [1, 2] \).
Next, find the first derivative: \( f'(x) = 3x^2 - 10x + 8 \). This derivative exists for all real numbers, so \( f(x) \) is derivable on the open interval \( (1, 2) \).
Now, evaluate the function at the endpoints:
\( f(1) = (1 - 1)(1 - 2)^2 = 0 \cdot (-1)^2 = 0 \)
\( f(2) = (2 - 1)(2 - 2)^2 = 1 \cdot (0)^2 = 0 \)
Since \( f(1) = f(2) = 0 \), all three conditions of Rolle's Theorem are satisfied.
Therefore, there must exist at least one real number \( c \) in \( (1, 2) \) such that \( f'(c) = 0 \).
Set the derivative to zero: \( 3c^2 - 10c + 8 = 0 \)
This is a quadratic equation. We can solve it using the quadratic formula \( c = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \):
\( c = \frac{10 \pm \sqrt{(-10)^2 - 4(3)(8)}}{2(3)} \)
\( c = \frac{10 \pm \sqrt{100 - 96}}{6} \)
\( c = \frac{10 \pm \sqrt{4}}{6} \)
\( c = \frac{10 \pm 2}{6} \)
This gives two possible values for \( c \):
\( c_1 = \frac{10 + 2}{6} = \frac{12}{6} = 2 \)
\( c_2 = \frac{10 - 2}{6} = \frac{8}{6} = \frac{4}{3} \)
We need \( c \) to be strictly within the open interval \( (1, 2) \).
\( c_1 = 2 \) is not in \( (1, 2) \) because the interval is open (does not include the endpoints).
\( c_2 = \frac{4}{3} \approx 1.33 \) is in \( (1, 2) \).
Thus, Rolle's Theorem is verified for \( c = \frac{4}{3} \). Finding the derivative helps locate points of zero slope.
In simple words: We checked if the function was smooth, unbroken, and had the same value at its start and end points. It met all these rules. We then found a point where the slope was flat, which was inside the allowed range, proving the theorem.

🎯 Exam Tip: Always pay attention to whether the interval is open or closed when checking the value of \( c \). For Rolle's Theorem, \( c \) must be in the open interval \( (a, b) \), meaning \( a < c < b \).

Question 3. f (x) = cos x + sin x in [0, 2π].
Answer: Given the function \( f(x) = \cos x + \sin x \) in the interval \( [0, 2\pi] \).
Since both \( \sin x \) and \( \cos x \) are continuous functions everywhere, their sum \( f(x) \) is also continuous on the closed interval \( [0, 2\pi] \).
Next, find the first derivative: \( f'(x) = -\sin x + \cos x \). This derivative exists for all real numbers, so \( f(x) \) is derivable on the open interval \( (0, 2\pi) \).
Now, evaluate the function at the endpoints:
\( f(0) = \cos(0) + \sin(0) = 1 + 0 = 1 \)
\( f(2\pi) = \cos(2\pi) + \sin(2\pi) = 1 + 0 = 1 \)
Since \( f(0) = f(2\pi) = 1 \), all three conditions of Rolle's Theorem are satisfied.
Therefore, there must exist at least one real number \( c \) in \( (0, 2\pi) \) such that \( f'(c) = 0 \).
Set the derivative to zero: \( -\sin c + \cos c = 0 \)
\( \implies \) \( \cos c = \sin c \)
If \( \cos c \neq 0 \), we can divide by \( \cos c \):
\( \implies \) \( \tan c = 1 \)
In the interval \( (0, 2\pi) \), the values of \( c \) for which \( \tan c = 1 \) are \( c = \frac{\pi}{4} \) and \( c = \pi + \frac{\pi}{4} = \frac{5\pi}{4} \).
Both \( c = \frac{\pi}{4} \) and \( c = \frac{5\pi}{4} \) are in the open interval \( (0, 2\pi) \).
Thus, Rolle's Theorem is verified. The tangent to the curve is horizontal at these points.
In simple words: The function starts and ends at the same value in the given range, and it's smooth everywhere. We found two specific angles where the function's slope becomes completely flat, which proves the theorem is true for this function.

🎯 Exam Tip: When solving trigonometric equations for \( c \), always consider all possible solutions within the specified open interval \( (a, b) \). Make sure to include all relevant quadrants.

Question 4. Using Rolle's Theorem, find points on the curve y = 16 − x², x∈[-1, 1], where tangent is parallel to x-axis.
Answer: Given the function \( f(x) = 16 - x^2 \) on the interval \( [-1, 1] \).
Since \( f(x) \) is a polynomial, it is continuous on the closed interval \( [-1, 1] \).
Next, find the first derivative: \( f'(x) = -2x \). This derivative exists for all real numbers, so \( f(x) \) is derivable on the open interval \( (-1, 1) \).
Now, evaluate the function at the endpoints:
\( f(-1) = 16 - (-1)^2 = 16 - 1 = 15 \)
\( f(1) = 16 - (1)^2 = 16 - 1 = 15 \)
Since \( f(-1) = f(1) = 15 \), all three conditions of Rolle's Theorem are satisfied.
Therefore, there must exist at least one real number \( c \) in \( (-1, 1) \) such that \( f'(c) = 0 \). This means the tangent at that point is parallel to the x-axis.
Set the derivative to zero: \( -2c = 0 \)
\( \implies \) \( c = 0 \)
Since \( c = 0 \) is in the open interval \( (-1, 1) \), Rolle's Theorem is verified.
To find the point on the curve, substitute \( c = 0 \) back into the original function \( f(x) = 16 - x^2 \):
\( f(0) = 16 - (0)^2 = 16 \)
So, the required point on the parabola where the tangent is parallel to the x-axis is \( (0, 16) \). This point is the vertex of the parabola.
In simple words: The function is a smooth curve that has the same height at its two end points. Rolle's Theorem tells us there's a point in between where the curve is perfectly flat. We found this point to be \( (0, 16) \).

🎯 Exam Tip: When a question asks for a "point," make sure to provide both the x-coordinate (the value of \( c \)) and the y-coordinate (f(c)) by plugging \( c \) back into the original function.

Question 5. Verify Lagrange's Mean Value Theorem for the function f(x) = 2sinx + sin 2x on [0, π].
Answer: Given the function \( f(x) = 2\sin x + \sin 2x \) in the interval \( [0, \pi] \).
Since both \( \sin x \) and \( \sin 2x \) are continuous functions everywhere, their sum \( f(x) \) is also continuous on the closed interval \( [0, \pi] \).
Next, find the first derivative: \( f'(x) = 2\cos x + 2\cos 2x \). This derivative exists for all real numbers, so \( f(x) \) is derivable on the open interval \( (0, \pi) \).
Both conditions for Lagrange's Mean Value Theorem (LMVT) are satisfied.
According to LMVT, there must exist at least one real number \( c \) in \( (0, \pi) \) such that \( f'(c) = \frac{f(\pi) - f(0)}{\pi - 0} \).
First, calculate the function values at the endpoints:
\( f(0) = 2\sin(0) + \sin(2 \cdot 0) = 2(0) + 0 = 0 \)
\( f(\pi) = 2\sin(\pi) + \sin(2\pi) = 2(0) + 0 = 0 \)
Now, calculate the slope of the chord:
\( \frac{f(\pi) - f(0)}{\pi - 0} = \frac{0 - 0}{\pi} = 0 \)
Set \( f'(c) = 0 \):
\( 2\cos c + 2\cos 2c = 0 \)
Divide by 2: \( \cos c + \cos 2c = 0 \)
Using the identity \( \cos 2c = 2\cos^2 c - 1 \):
\( \cos c + (2\cos^2 c - 1) = 0 \)
\( 2\cos^2 c + \cos c - 1 = 0 \)
This is a quadratic equation in \( \cos c \). Let \( y = \cos c \):
\( 2y^2 + y - 1 = 0 \)
Factorizing the quadratic equation:
\( (2y - 1)(y + 1) = 0 \)
So, \( 2\cos c - 1 = 0 \implies \cos c = \frac{1}{2} \)
Or, \( \cos c + 1 = 0 \implies \cos c = -1 \)
For \( \cos c = \frac{1}{2} \) in the interval \( (0, \pi) \), the value of \( c \) is \( \frac{\pi}{3} \).
For \( \cos c = -1 \), the value of \( c \) is \( \pi \). However, \( c \) must be in the open interval \( (0, \pi) \), so \( c = \pi \) is not included.
Since \( c = \frac{\pi}{3} \) is in \( (0, \pi) \), Lagrange's Mean Value Theorem is verified. This means there's a point where the tangent is parallel to the chord.
In simple words: We confirmed that the function is continuous and smooth in the given range. Then, we found a point where the slope of the curve is exactly the same as the slope of the straight line connecting its start and end points. This showed that Lagrange's theorem holds true.

🎯 Exam Tip: For LMVT problems, carefully calculate \( \frac{f(b) - f(a)}{b - a} \) as this value is crucial for finding \( c \). Also, remember to factorize trigonometric equations or use identities to solve them efficiently.

Question 6. Verify Lagrange's Mean Value Theorem for the function f(x) = x³ – 5x² – 3x in the interval [1, 3]. Find c ∈ (1, 3) for which f'(c) = \frac{f(3)-f(1)}{3-1}.
Answer: Given the function \( f(x) = x^3 - 5x^2 - 3x \) in the interval \( [1, 3] \).
Since \( f(x) \) is a polynomial, it is continuous on the closed interval \( [1, 3] \).
Next, find the first derivative: \( f'(x) = 3x^2 - 10x - 3 \). This derivative exists for all real numbers, so \( f(x) \) is derivable on the open interval \( (1, 3) \).
Both conditions for Lagrange's Mean Value Theorem (LMVT) are satisfied.
According to LMVT, there must exist at least one real number \( c \) in \( (1, 3) \) such that \( f'(c) = \frac{f(3) - f(1)}{3 - 1} \).
First, calculate the function values at the endpoints:
\( f(1) = (1)^3 - 5(1)^2 - 3(1) = 1 - 5 - 3 = -7 \)
\( f(3) = (3)^3 - 5(3)^2 - 3(3) = 27 - 5(9) - 9 = 27 - 45 - 9 = -27 \)
Now, calculate the slope of the chord:
\( \frac{f(3) - f(1)}{3 - 1} = \frac{-27 - (-7)}{2} = \frac{-27 + 7}{2} = \frac{-20}{2} = -10 \)
Set \( f'(c) = -10 \):
\( 3c^2 - 10c - 3 = -10 \)
\( 3c^2 - 10c + 7 = 0 \)
This is a quadratic equation. We can factorize it:
\( 3c^2 - 3c - 7c + 7 = 0 \)
\( 3c(c - 1) - 7(c - 1) = 0 \)
\( (3c - 7)(c - 1) = 0 \)
This gives two possible values for \( c \):
\( 3c - 7 = 0 \implies c = \frac{7}{3} \)
\( c - 1 = 0 \implies c = 1 \)
We need \( c \) to be strictly within the open interval \( (1, 3) \).
\( c = 1 \) is not in \( (1, 3) \).
\( c = \frac{7}{3} \approx 2.33 \) is in \( (1, 3) \).
Thus, Lagrange's Mean Value Theorem is verified for \( c = \frac{7}{3} \). This means there's a point on the curve where the tangent slope matches the average slope of the whole interval.
In simple words: The function is smooth and unbroken. We found a point on the curve where its slope is exactly the same as the slope of the imaginary straight line connecting the function's start and end points in the given interval.

🎯 Exam Tip: Always double-check your factorization or quadratic formula calculations, as errors here will lead to incorrect values of \( c \) and potential loss of marks.

Question 7. Verify Lagrange's Mean Value Theorem for the function f (x) = x (x – 2) on [1, 3].
Answer: Given the function \( f(x) = x(x - 2) \) on the interval \( [1, 3] \).
First, expand the function: \( f(x) = x^2 - 2x \).
Since \( f(x) \) is a polynomial, it is continuous on the closed interval \( [1, 3] \).
Next, find the first derivative: \( f'(x) = 2x - 2 \). This derivative exists for all real numbers, so \( f(x) \) is derivable on the open interval \( (1, 3) \).
Both conditions for Lagrange's Mean Value Theorem (LMVT) are satisfied.
According to LMVT, there must exist at least one real number \( c \) in \( (1, 3) \) such that \( f'(c) = \frac{f(3) - f(1)}{3 - 1} \).
First, calculate the function values at the endpoints:
\( f(1) = 1(1 - 2) = 1(-1) = -1 \)
\( f(3) = 3(3 - 2) = 3(1) = 3 \)
Now, calculate the slope of the chord:
\( \frac{f(3) - f(1)}{3 - 1} = \frac{3 - (-1)}{2} = \frac{3 + 1}{2} = \frac{4}{2} = 2 \)
Set \( f'(c) = 2 \):
\( 2c - 2 = 2 \)
\( \implies \) \( 2c = 4 \)
\( \implies \) \( c = 2 \)
Since \( c = 2 \) is in the open interval \( (1, 3) \), Lagrange's Mean Value Theorem is verified. This means the curve's instantaneous rate of change at \( c=2 \) matches its average rate of change over the interval.
In simple words: We checked that the function is continuous and smooth. We then found a point where the curve's slope matches the average slope between its two end points, proving the theorem.

🎯 Exam Tip: Simplify the function and its derivative carefully before proceeding. A common mistake is to incorrectly calculate \( f(a) \), \( f(b) \), or the slope of the chord.

Question 8. Using Lagrange's Mean Value Theorem, find a point on the parabola y = (x + 3)², where the tangent is parallel to the chord joining (-3, 0) and (-4, 1).
Answer: Given the function \( f(x) = (x + 3)^2 \). The chord joins the points \( (-4, 1) \) and \( (-3, 0) \). This implies we are working on the interval \( [-4, -3] \).
Since \( f(x) \) is a polynomial, it is continuous on the closed interval \( [-4, -3] \).
Next, find the first derivative: \( f'(x) = 2(x + 3) \). This derivative exists for all real numbers, so \( f(x) \) is derivable on the open interval \( (-4, -3) \).
Both conditions for Lagrange's Mean Value Theorem (LMVT) are satisfied.
According to LMVT, there must exist at least one real number \( c \) in \( (-4, -3) \) such that \( f'(c) = \frac{f(-3) - f(-4)}{-3 - (-4)} \).
First, calculate the function values at the endpoints:
\( f(-4) = (-4 + 3)^2 = (-1)^2 = 1 \) (This matches the point \( (-4, 1) \))
\( f(-3) = (-3 + 3)^2 = (0)^2 = 0 \) (This matches the point \( (-3, 0) \))
Now, calculate the slope of the chord:
\( \frac{f(-3) - f(-4)}{-3 - (-4)} = \frac{0 - 1}{-3 + 4} = \frac{-1}{1} = -1 \)
Set \( f'(c) = -1 \):
\( 2(c + 3) = -1 \)
\( \implies \) \( c + 3 = -\frac{1}{2} \)
\( \implies \) \( c = -\frac{1}{2} - 3 \)
\( \implies \) \( c = -\frac{1}{2} - \frac{6}{2} \)
\( \implies \) \( c = -\frac{7}{2} \)
Since \( c = -\frac{7}{2} = -3.5 \), which is in the open interval \( (-4, -3) \), Lagrange's Mean Value Theorem is verified.
To find the required point on the curve, substitute \( c = -\frac{7}{2} \) back into the original function \( f(x) = (x + 3)^2 \):
\( y = \left(-\frac{7}{2} + 3\right)^2 = \left(-\frac{7}{2} + \frac{6}{2}\right)^2 = \left(-\frac{1}{2}\right)^2 = \frac{1}{4} \)
So, the required point on the parabola where the tangent is parallel to the chord joining \( (-3, 0) \) and \( (-4, 1) \) is \( \left(-\frac{7}{2}, \frac{1}{4}\right) \). This point represents where the instantaneous slope equals the average slope.
In simple words: We found a point on the parabola where the curve's slope is exactly the same as the slope of the straight line connecting the two given points. This point on the curve is \( \left(-\frac{7}{2}, \frac{1}{4}\right) \).

🎯 Exam Tip: When given two points, the interval for the theorem is derived from their x-coordinates. Always plug the \( c \) value back into the *original function* to find the y-coordinate of the point.