OP Malhotra Class 12 Maths Solutions Chapter 10 Mean Value Theorems Exercise 10 (A)

Question 1. \( f(x) = x^2 - x - 6 \) on \([-2, 3]\).
Answer: We are given the function \( f(x) = x^2 - x - 6 \) on the interval \([-2, 3]\). Since \( f(x) \) is a polynomial, it is continuous and differentiable everywhere on the real number line, R. Therefore, \( f(x) \) is continuous on the closed interval \([-2, 3]\) and differentiable on the open interval \( (-2, 3) \). Next, we evaluate the function at the endpoints of the interval: \( f(-2) = (-2)^2 - (-2) - 6 = 4 + 2 - 6 = 0 \) \( f(3) = (3)^2 - 3 - 6 = 9 - 3 - 6 = 0 \) So, we have \( f(-2) = f(3) = 0 \). This means the function has the same value at both ends of the interval. All three conditions for Rolle's Theorem are satisfied. This guarantees that there is at least one real number \( c \) in the open interval \( (-2, 3) \) such that \( f'(c) = 0 \). Let's find the derivative of \( f(x) \): \( f'(x) = 2x - 1 \) Now, we set \( f'(c) = 0 \): \( 2c - 1 = 0 \)
\( \implies \) \( 2c = 1 \)
\( \implies \) \( c = \frac{1}{2} \) Since \( c = \frac{1}{2} \) is indeed within the interval \( (-2, 3) \), Rolle's Theorem is verified. This value of \( c \) is where the tangent line to the curve is horizontal.In simple words: This problem checks if Rolle's Theorem works for a given function. We showed that the function is smooth, starts and ends at the same height, and found a spot where its slope is zero, all confirming the theorem.

🎯 Exam Tip: Remember the three key conditions for Rolle's Theorem: continuity, differentiability, and \( f(a) = f(b) \). Always show each step clearly, from checking the conditions to finding the value of \( c \).

Question 2. \( f(x) = x^2 - 6x + 5 \), in the interval \([1, 5]\).
Answer: We are given the function \( f(x) = x^2 - 6x + 5 \) on the interval \([1, 5]\). Since \( f(x) \) is a polynomial function, it is continuous on the closed interval \([1, 5]\) and differentiable on the open interval \( (1, 5) \). Every polynomial function is infinitely differentiable. Let's find the derivative of \( f(x) \): \( f'(x) = 2x - 6 \) This derivative exists for all real numbers, so it exists for all \( x \in (1, 5) \). Now, we evaluate the function at the endpoints of the interval: \( f(1) = (1)^2 - 6(1) + 5 = 1 - 6 + 5 = 0 \) \( f(5) = (5)^2 - 6(5) + 5 = 25 - 30 + 5 = 0 \) Thus, \( f(1) = f(5) = 0 \). The function values at the interval's boundaries are equal. All three conditions of Rolle's Theorem are met. This means there is at least one real number \( c \) in the open interval \( (1, 5) \) such that \( f'(c) = 0 \). Set the derivative equal to zero: \( 2c - 6 = 0 \)
\( \implies \) \( 2c = 6 \)
\( \implies \) \( c = 3 \) Since \( c = 3 \) is in the interval \( (1, 5) \), Rolle's Theorem is verified for this function. This point \( c \) is where the graph has a flat slope.In simple words: We checked if Rolle's Theorem works for this function. Since it is smooth, has the same value at both ends, we found a point where its slope is zero, exactly as the theorem predicts.

🎯 Exam Tip: When dealing with polynomials, stating that they are inherently continuous and differentiable is sufficient for the first two conditions of Rolle's Theorem.

Question 3. \( f(x) = x^2 - 5x + 6 \), \( 2 \le x \le 3 \).
Answer: We are given the function \( f(x) = x^2 - 5x + 6 \) on the interval \([2, 3]\). Since \( f(x) \) is a polynomial, it is continuous on \([2, 3]\) and differentiable on \( (2, 3) \). Polynomials are well-behaved functions that are continuous and differentiable everywhere. Next, we find the function's values at the endpoints of the interval: \( f(2) = (2)^2 - 5(2) + 6 = 4 - 10 + 6 = 0 \) \( f(3) = (3)^2 - 5(3) + 6 = 9 - 15 + 6 = 0 \) So, we see that \( f(2) = f(3) = 0 \). All three conditions of Rolle's Theorem are satisfied. Therefore, there must exist at least one real number \( c \) in the open interval \( (2, 3) \) such that \( f'(c) = 0 \). Let's find the derivative: \( f'(x) = 2x - 5 \) Now, set \( f'(c) = 0 \): \( 2c - 5 = 0 \)
\( \implies \) \( 2c = 5 \)
\( \implies \) \( c = \frac{5}{2} \) Since \( c = \frac{5}{2} = 2.5 \) is indeed in the interval \( (2, 3) \), Rolle's Theorem is verified. This point represents a local extremum where the function's rate of change is zero.In simple words: For this function, we checked that it is smooth and has the same value at its start and end points. We then found a point in between where its slope is completely flat, just like Rolle's Theorem says.

🎯 Exam Tip: Always specify the open interval for differentiability and the closed interval for continuity when applying Rolle's Theorem.

Question 4. \( y = 16 - x^2 \), \( x \in [-1, 1] \).
Answer: We are given the function \( y = f(x) = 16 - x^2 \) on the interval \([-1, 1]\). Since \( f(x) \) is a polynomial function, it is continuous on the closed interval \([-1, 1]\) and differentiable on the open interval \( (-1, 1) \). Polynomials are always smooth curves. Let's find the derivative of \( f(x) \): \( f'(x) = -2x \) This derivative exists for all real numbers, including all \( x \in (-1, 1) \). Now, we evaluate the function at the endpoints of the interval: \( f(-1) = 16 - (-1)^2 = 16 - 1 = 15 \) \( f(1) = 16 - (1)^2 = 16 - 1 = 15 \) So, we find that \( f(-1) = f(1) = 15 \). All three conditions of Rolle's Theorem are satisfied. This means there is at least one real number \( c \) in the open interval \( (-1, 1) \) such that \( f'(c) = 0 \). Set the derivative equal to zero: \( -2c = 0 \)
\( \implies \) \( c = 0 \) Since \( c = 0 \) is within the interval \( (-1, 1) \), Rolle's Theorem is verified. This point \( c=0 \) is the vertex of the parabola, where its slope is zero.In simple words: This problem showed us how Rolle's Theorem works for a simple curve. Because the curve is smooth and starts and ends at the same height, we found a point where its slope is flat, which matches the theorem.

🎯 Exam Tip: When the derivative is simple, like \( -2x \), it's easy to find \( c \) by setting it to zero. Always double-check that your \( c \) value falls within the open interval \((a, b)\).

Question 5. \( f(x) = x(x - 3)^2 \) on \([0, 3]\).
Answer: We are given the function \( f(x) = x(x - 3)^2 \) on the interval \([0, 3]\). First, expand the function: \( f(x) = x(x^2 - 6x + 9) = x^3 - 6x^2 + 9x \). Since \( f(x) \) is a polynomial, it is continuous on the closed interval \([0, 3]\) and differentiable on the open interval \( (0, 3) \). Polynomials always behave smoothly without any breaks or sharp corners. Now, evaluate the function at the endpoints: \( f(0) = 0(0 - 3)^2 = 0 \) \( f(3) = 3(3 - 3)^2 = 3(0)^2 = 0 \) So, we have \( f(0) = f(3) = 0 \). All three conditions of Rolle's Theorem are satisfied. This means there exists at least one real number \( c \) in the open interval \( (0, 3) \) such that \( f'(c) = 0 \). Let's find the derivative of \( f(x) \). We can use the product rule or differentiate the expanded form: Using the product rule on \( f(x) = x(x-3)^2 \): \( f'(x) = 1 \cdot (x-3)^2 + x \cdot 2(x-3) \cdot 1 \) \( f'(x) = (x-3)^2 + 2x(x-3) \) Factor out \( (x-3) \): \( f'(x) = (x-3)[(x-3) + 2x] \) \( f'(x) = (x-3)(3x-3) \) \( f'(x) = 3(x-3)(x-1) \) Now, set \( f'(c) = 0 \): \( 3(c-3)(c-1) = 0 \) This gives us two possible values for \( c \): \( c - 3 = 0 \implies c = 3 \) \( c - 1 = 0 \implies c = 1 \) Rolle's Theorem requires \( c \) to be in the *open* interval \( (0, 3) \). For \( c = 3 \), it is not in \( (0, 3) \) because it is an endpoint. For \( c = 1 \), it is in \( (0, 3) \). So, \( c = 1 \) is the value that verifies Rolle's Theorem. This function has a local maximum or minimum at \( c=1 \).In simple words: This problem shows that even for more complex polynomial functions, Rolle's Theorem still applies. We found two possible points where the slope could be zero, but only one was inside the given range, which confirms the theorem.

🎯 Exam Tip: When solving for \( c \), always check if the calculated values are strictly within the *open* interval \((a, b)\), not just the closed interval \([a, b]\). Endpoints are not valid for \( c \).

Question 6. \( f(x) = x(x - 3)^2 \) on \([2, 4]\).
Answer: We are given the function \( f(x) = x(x - 3)^2 \) on the interval \([2, 4]\). First, expand the function: \( f(x) = x(x^2 - 6x + 9) = x^3 - 6x^2 + 9x \). Since \( f(x) \) is a polynomial, it is continuous on the closed interval \([2, 4]\) and differentiable on the open interval \( (2, 4) \). All polynomial functions are smooth and have well-defined derivatives everywhere. Now, let's evaluate the function at the endpoints: \( f(2) = 2(2 - 3)^2 = 2(-1)^2 = 2(1) = 2 \) \( f(4) = 4(4 - 3)^2 = 4(1)^2 = 4(1) = 4 \) Here we notice that \( f(2) = 2 \) and \( f(4) = 4 \). Since \( f(2) \neq f(4) \), the third condition of Rolle's Theorem is *not* satisfied. Therefore, Rolle's Theorem is not applicable to this function on the given interval \([2, 4]\). This doesn't mean there is no point where \( f'(x)=0 \), but Rolle's Theorem does not guarantee it. *Note: The original solution seems to have skipped this conclusion and proceeded to find roots of \( f'(x)=0 \), implying Rolle's theorem *is* applicable. However, based on calculation, \( f(2) \neq f(4) \). I will correct the answer to reflect that Rolle's Theorem is not applicable.* Let's check the derivative and its roots as the provided solution does, but still conclude non-applicability of Rolle's Theorem. We found \( f'(x) = 3(x-3)(x-1) \) from the previous problem. If we set \( f'(c) = 0 \): \( 3(c-3)(c-1) = 0 \) This gives \( c = 3 \) or \( c = 1 \). The interval for this problem is \([2, 4]\). For \( c = 3 \), it is in the open interval \( (2, 4) \). For \( c = 1 \), it is not in the open interval \( (2, 4) \). So, while we found a point \( c=3 \) where the derivative is zero within the interval, Rolle's Theorem cannot be used to *guarantee* its existence because \( f(2) \neq f(4) \). This shows that finding a point \( c \) does not automatically mean the theorem applies if the conditions are not met.In simple words: This problem did not meet all the rules for Rolle's Theorem because the function had different values at the start and end of the interval. So, even if we could find a point with a flat slope, the theorem itself could not be used to prove it.

🎯 Exam Tip: Always double-check *all three* conditions for Rolle's Theorem before proceeding to find \( c \). If any condition fails, state that the theorem is not applicable.

Question 7. \( f(x) = x^3 - 7x^2 + 16x - 12 \) on \([2, 3]\).
Answer: We are given the function \( f(x) = x^3 - 7x^2 + 16x - 12 \) on the interval \([2, 3]\). Since \( f(x) \) is a polynomial, it is continuous on the closed interval \([2, 3]\) and differentiable on the open interval \( (2, 3) \). Polynomials are always smooth and well-behaved functions. Next, let's evaluate the function at the endpoints: \( f(2) = (2)^3 - 7(2)^2 + 16(2) - 12 = 8 - 7(4) + 32 - 12 = 8 - 28 + 32 - 12 = 0 \) \( f(3) = (3)^3 - 7(3)^2 + 16(3) - 12 = 27 - 7(9) + 48 - 12 = 27 - 63 + 48 - 12 = 0 \) So, we have \( f(2) = f(3) = 0 \). All three conditions of Rolle's Theorem are satisfied. This guarantees that there exists at least one real number \( c \) in the open interval \( (2, 3) \) such that \( f'(c) = 0 \). Let's find the derivative of \( f(x) \): \( f'(x) = 3x^2 - 14x + 16 \) Now, set \( f'(c) = 0 \): \( 3c^2 - 14c + 16 = 0 \) We can solve this quadratic equation for \( c \) using the quadratic formula \( c = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \): \( c = \frac{-(-14) \pm \sqrt{(-14)^2 - 4(3)(16)}}{2(3)} \) \( c = \frac{14 \pm \sqrt{196 - 192}}{6} \) \( c = \frac{14 \pm \sqrt{4}}{6} \) \( c = \frac{14 \pm 2}{6} \) This gives two possible values for \( c \): \( c_1 = \frac{14 + 2}{6} = \frac{16}{6} = \frac{8}{3} \) \( c_2 = \frac{14 - 2}{6} = \frac{12}{6} = 2 \) Rolle's Theorem requires \( c \) to be in the *open* interval \( (2, 3) \). For \( c_1 = \frac{8}{3} \approx 2.67 \), it is in \( (2, 3) \). For \( c_2 = 2 \), it is not in \( (2, 3) \) because it is an endpoint. So, \( c = \frac{8}{3} \) is the value that verifies Rolle's Theorem. This is a point where the function's slope is horizontal.In simple words: This problem used Rolle's Theorem for a cube function. After checking that the function was smooth and had the same value at its ends, we found a spot inside where its slope was zero, confirming the theorem.

🎯 Exam Tip: For quadratic derivatives, use the quadratic formula carefully. Remember to exclude any values of \( c \) that fall on the boundary of the *open* interval.

Question 8. \( f(x) = 4 \sin x \), \( x \in [0, \pi] \).
Answer: We are given the function \( f(x) = 4 \sin x \) on the interval \([0, \pi]\). The sine function is continuous and differentiable everywhere. Therefore, \( f(x) = 4 \sin x \) is continuous on the closed interval \([0, \pi]\) and differentiable on the open interval \( (0, \pi) \). Next, we evaluate the function at the endpoints: \( f(0) = 4 \sin(0) = 4(0) = 0 \) \( f(\pi) = 4 \sin(\pi) = 4(0) = 0 \) So, we have \( f(0) = f(\pi) = 0 \). All three conditions of Rolle's Theorem are satisfied. This guarantees that there exists at least one real number \( c \) in the open interval \( (0, \pi) \) such that \( f'(c) = 0 \). Let's find the derivative of \( f(x) \): \( f'(x) = \frac{d}{dx} (4 \sin x) = 4 \cos x \) Now, set \( f'(c) = 0 \): \( 4 \cos c = 0 \)
\( \implies \) \( \cos c = 0 \) In the interval \( (0, \pi) \), the value of \( c \) for which \( \cos c = 0 \) is: \( c = \frac{\pi}{2} \) Since \( c = \frac{\pi}{2} \) is indeed in the interval \( (0, \pi) \), Rolle's Theorem is verified. This point is where the sine wave reaches its peak or trough, making its slope flat.In simple words: We used Rolle's Theorem for a sine wave. The sine function is smooth and starts and ends at the same height here. We found a spot exactly in the middle where its slope was zero, which matches the theorem.

🎯 Exam Tip: For trigonometric functions, know the values where sine and cosine are zero or one. Remember to consider only the values of \( c \) that lie within the specified open interval.

Question 9. \( f(x) = \sin 2x \), \( x \in [0, \frac{\pi}{2}] \).
Answer: We are given the function \( f(x) = \sin 2x \) on the interval \( [0, \frac{\pi}{2}] \). The sine function is continuous and differentiable everywhere. Therefore, \( f(x) = \sin 2x \) is continuous on the closed interval \( [0, \frac{\pi}{2}] \) and differentiable on the open interval \( (0, \frac{\pi}{2}) \). Next, we evaluate the function at the endpoints: \( f(0) = \sin(2 \cdot 0) = \sin(0) = 0 \) \( f(\frac{\pi}{2}) = \sin(2 \cdot \frac{\pi}{2}) = \sin(\pi) = 0 \) So, we have \( f(0) = f(\frac{\pi}{2}) = 0 \). All three conditions of Rolle's Theorem are satisfied. This ensures that there exists at least one real number \( c \) in the open interval \( (0, \frac{\pi}{2}) \) such that \( f'(c) = 0 \). Let's find the derivative of \( f(x) \): \( f'(x) = \frac{d}{dx} (\sin 2x) = \cos 2x \cdot 2 = 2 \cos 2x \) Now, set \( f'(c) = 0 \): \( 2 \cos 2c = 0 \)
\( \implies \) \( \cos 2c = 0 \) For \( \cos \theta = 0 \), \( \theta \) can be \( \frac{\pi}{2}, \frac{3\pi}{2}, \dots \) So, \( 2c = \frac{\pi}{2} \) (as we are looking for \( c \) in \( (0, \frac{\pi}{2}) \), which means \( 2c \) would be in \( (0, \pi) \)).
\( \implies \) \( c = \frac{\pi}{4} \) Since \( c = \frac{\pi}{4} \) is indeed in the interval \( (0, \frac{\pi}{2}) \), Rolle's Theorem is verified. This point marks where the function's rate of change is momentarily zero.In simple words: We applied Rolle's Theorem to a compressed sine wave. Since it was smooth and had the same value at both ends, we found a point inside where its slope was flat, just as the theorem stated.

🎯 Exam Tip: Be careful with chain rule when differentiating trigonometric functions like \( \sin 2x \). Also, remember to adjust the range for \( 2c \) when finding values for \( c \).

Question 10. \( f(x) = \sin^2 x \), \( 0 \le x \le \pi \).
Answer: We are given the function \( f(x) = \sin^2 x \) on the interval \( [0, \pi] \). The sine function is continuous and differentiable everywhere. Since \( f(x) = (\sin x)^2 \), it is a composition of continuous and differentiable functions, making \( f(x) \) continuous on the closed interval \( [0, \pi] \) and differentiable on the open interval \( (0, \pi) \). Next, we evaluate the function at the endpoints: \( f(0) = \sin^2(0) = (0)^2 = 0 \) \( f(\pi) = \sin^2(\pi) = (0)^2 = 0 \) So, we have \( f(0) = f(\pi) = 0 \). All three conditions of Rolle's Theorem are satisfied. This guarantees that there exists at least one real number \( c \) in the open interval \( (0, \pi) \) such that \( f'(c) = 0 \). Let's find the derivative of \( f(x) \): \( f'(x) = \frac{d}{dx} (\sin^2 x) = 2 \sin x \cos x \) We know the identity \( \sin 2x = 2 \sin x \cos x \). So, \( f'(x) = \sin 2x \). Now, set \( f'(c) = 0 \): \( \sin 2c = 0 \) For \( \sin \theta = 0 \), \( \theta \) can be \( 0, \pi, 2\pi, \dots \) We need \( c \) in the interval \( (0, \pi) \), which means \( 2c \) would be in \( (0, 2\pi) \). Possible values for \( 2c \) in \( (0, 2\pi) \) are \( \pi \). So, \( 2c = \pi \)
\( \implies \) \( c = \frac{\pi}{2} \) Since \( c = \frac{\pi}{2} \) is in the interval \( (0, \pi) \), Rolle's Theorem is verified. This value of \( c \) represents a point where the squared sine function momentarily flattens out.In simple words: We used Rolle's Theorem for \( \sin^2 x \). It was smooth and had the same value at both ends. We found a point in the middle where its slope was zero, which proved the theorem works here.

🎯 Exam Tip: Recognizing trigonometric identities like \( 2 \sin x \cos x = \sin 2x \) can simplify differentiation and solving for \( c \). Always ensure your \( c \) value falls strictly within the open interval.

Question 11. \( f(x) = e^x \cos x \), \( x \in [-\frac{\pi}{2}, \frac{\pi}{2}] \).
Answer: We are given the function \( f(x) = e^x \cos x \) on the interval \( [-\frac{\pi}{2}, \frac{\pi}{2}] \). The exponential function \( e^x \) and the cosine function \( \cos x \) are both continuous and differentiable everywhere. The product of continuous and differentiable functions is also continuous and differentiable. Therefore, \( f(x) = e^x \cos x \) is continuous on the closed interval \( [-\frac{\pi}{2}, \frac{\pi}{2}] \) and differentiable on the open interval \( (-\frac{\pi}{2}, \frac{\pi}{2}) \). Next, we evaluate the function at the endpoints: \( f(-\frac{\pi}{2}) = e^{-\frac{\pi}{2}} \cos(-\frac{\pi}{2}) = e^{-\frac{\pi}{2}} \cdot 0 = 0 \) \( f(\frac{\pi}{2}) = e^{\frac{\pi}{2}} \cos(\frac{\pi}{2}) = e^{\frac{\pi}{2}} \cdot 0 = 0 \) So, we have \( f(-\frac{\pi}{2}) = f(\frac{\pi}{2}) = 0 \). All three conditions of Rolle's Theorem are satisfied. This guarantees that there exists at least one real number \( c \) in the open interval \( (-\frac{\pi}{2}, \frac{\pi}{2}) \) such that \( f'(c) = 0 \). Let's find the derivative of \( f(x) \) using the product rule: \( f'(x) = \frac{d}{dx} (e^x \cos x) = e^x (\cos x) + e^x (-\sin x) \) \( f'(x) = e^x (\cos x - \sin x) \) Now, set \( f'(c) = 0 \): \( e^c (\cos c - \sin c) = 0 \) Since \( e^c \) is never zero for any real \( c \), we must have: \( \cos c - \sin c = 0 \)
\( \implies \) \( \cos c = \sin c \)
\( \implies \) \( \tan c = 1 \) (dividing by \( \cos c \), assuming \( \cos c \neq 0 \). If \( \cos c = 0 \), then \( \sin c \) must also be \( 0 \), which is impossible for the same angle.) In the interval \( (-\frac{\pi}{2}, \frac{\pi}{2}) \), the value of \( c \) for which \( \tan c = 1 \) is: \( c = \frac{\pi}{4} \) Since \( c = \frac{\pi}{4} \) is indeed in the interval \( (-\frac{\pi}{2}, \frac{\pi}{2}) \), Rolle's Theorem is verified. This point is where the product function has a horizontal tangent.In simple words: This problem involved a mix of exponential and trig functions. We showed it was smooth and had the same value at its ends. Then, we found a point where its slope was zero, proving Rolle's Theorem.

🎯 Exam Tip: When \( e^x \) is a factor, remember that \( e^x \) is always positive, so it cannot be zero. Focus on the other factor to find the critical points.

Question 12. \( f(x) = \frac{\sin x}{e^x} \), \( x \in [0, \pi] \).
Answer: We are given the function \( f(x) = \frac{\sin x}{e^x} \) on the interval \( [0, \pi] \). The sine function (\( \sin x \)) is continuous and differentiable everywhere. The exponential function (\( e^x \)) is also continuous and differentiable everywhere and is never zero. The quotient of two continuous and differentiable functions is also continuous and differentiable where the denominator is not zero. Therefore, \( f(x) = \frac{\sin x}{e^x} \) is continuous on the closed interval \( [0, \pi] \) and differentiable on the open interval \( (0, \pi) \). Next, we evaluate the function at the endpoints: \( f(0) = \frac{\sin 0}{e^0} = \frac{0}{1} = 0 \) \( f(\pi) = \frac{\sin \pi}{e^\pi} = \frac{0}{e^\pi} = 0 \) So, we have \( f(0) = f(\pi) = 0 \). All three conditions of Rolle's Theorem are satisfied. This means there must exist at least one real number \( c \) in the open interval \( (0, \pi) \) such that \( f'(c) = 0 \). Let's find the derivative of \( f(x) \) using the quotient rule: \( f'(x) = \frac{e^x \frac{d}{dx}(\sin x) - \sin x \frac{d}{dx}(e^x)}{(e^x)^2} \) \( f'(x) = \frac{e^x \cos x - \sin x e^x}{e^{2x}} \) \( f'(x) = \frac{e^x (\cos x - \sin x)}{e^{2x}} \) \( f'(x) = \frac{\cos x - \sin x}{e^x} \) Now, set \( f'(c) = 0 \): \( \frac{\cos c - \sin c}{e^c} = 0 \) Since \( e^c \) is never zero, the numerator must be zero: \( \cos c - \sin c = 0 \)
\( \implies \) \( \cos c = \sin c \)
\( \implies \) \( \tan c = 1 \) (assuming \( \cos c \neq 0 \)) In the interval \( (0, \pi) \), the value of \( c \) for which \( \tan c = 1 \) is: \( c = \frac{\pi}{4} \) Since \( c = \frac{\pi}{4} \) is indeed in the interval \( (0, \pi) \), Rolle's Theorem is verified. This point indicates a local extremum for the function.In simple words: This problem used Rolle's Theorem on a fraction that combined sine and exponential functions. We proved it was smooth and had the same value at its ends, and then found a spot in between where its slope was zero, just as expected.

🎯 Exam Tip: When using the quotient rule, simplify the derivative as much as possible before setting it to zero. Remember that the denominator \( e^x \) is never zero, so you only need to focus on the numerator.

Question 13. \( f(x) = x(x + 3)e^{-x/2} \) defined in the interval \([-3, 0]\).
Answer: We are given the function \( f(x) = x(x + 3)e^{-x/2} \) on the interval \([-3, 0]\). This function is a product of a polynomial \( x(x+3) = x^2+3x \) and an exponential function \( e^{-x/2} \). Both polynomial and exponential functions are continuous and differentiable everywhere. The product of continuous and differentiable functions is also continuous and differentiable. Therefore, \( f(x) \) is continuous on the closed interval \([-3, 0]\) and differentiable on the open interval \( (-3, 0) \). Next, we evaluate the function at the endpoints: \( f(-3) = (-3)(-3 + 3)e^{-(-3)/2} = (-3)(0)e^{3/2} = 0 \) \( f(0) = (0)(0 + 3)e^{-0/2} = (0)(3)e^0 = 0 \) So, we have \( f(-3) = f(0) = 0 \). All three conditions of Rolle's Theorem are satisfied. This guarantees that there exists at least one real number \( c \) in the open interval \( (-3, 0) \) such that \( f'(c) = 0 \). Let's find the derivative of \( f(x) = (x^2+3x)e^{-x/2} \) using the product rule: \( f'(x) = (2x + 3)e^{-x/2} + (x^2 + 3x)e^{-x/2} (-\frac{1}{2}) \) Factor out \( e^{-x/2} \): \( f'(x) = e^{-x/2} [(2x + 3) - \frac{1}{2}(x^2 + 3x)] \) \( f'(x) = e^{-x/2} [2x + 3 - \frac{1}{2}x^2 - \frac{3}{2}x] \) \( f'(x) = e^{-x/2} [-\frac{1}{2}x^2 + (2 - \frac{3}{2})x + 3] \) \( f'(x) = e^{-x/2} [-\frac{1}{2}x^2 + \frac{1}{2}x + 3] \) To make it easier to solve, multiply the bracketed term by \( -2 \): \( f'(x) = -\frac{1}{2}e^{-x/2} [x^2 - x - 6] \) Now, set \( f'(c) = 0 \): \( -\frac{1}{2}e^{-c/2} [c^2 - c - 6] = 0 \) Since \( -\frac{1}{2}e^{-c/2} \) is never zero (because \( e^c \) is always positive), we must have: \( c^2 - c - 6 = 0 \) Factor the quadratic: \( (c - 3)(c + 2) = 0 \) This gives two possible values for \( c \): \( c - 3 = 0 \implies c = 3 \) \( c + 2 = 0 \implies c = -2 \) Rolle's Theorem requires \( c \) to be in the *open* interval \( (-3, 0) \). For \( c = 3 \), it is not in \( (-3, 0) \). For \( c = -2 \), it is in \( (-3, 0) \). So, \( c = -2 \) is the value that verifies Rolle's Theorem. This point represents a turning point in the function's graph.In simple words: This problem used Rolle's Theorem on a function that combined a simple curve with an exponential part. We checked it was smooth and had the same value at its ends, and then found a spot inside the given range where its slope was zero, confirming the theorem.

🎯 Exam Tip: When differentiating a product of functions, don't forget the chain rule for terms like \( e^{-x/2} \). Always factor out the common exponential term to simplify solving for \( c \).

Question 14. \( f(x) = \sqrt{4-x^2} \) on \([-2, 2]\).
Answer: We are given the function \( f(x) = \sqrt{4-x^2} \) on the interval \([-2, 2]\). This function represents the upper semicircle of a circle with radius 2 centered at the origin. For \( f(x) \) to be defined, we need \( 4 - x^2 \ge 0 \), which means \( x^2 \le 4 \), or \( -2 \le x \le 2 \). So the domain is \([-2, 2]\). The function \( f(x) = \sqrt{4-x^2} \) is continuous on its closed domain \([-2, 2]\). The square root function is continuous wherever its argument is non-negative. Let's find the derivative of \( f(x) \): \( f'(x) = \frac{d}{dx} (4-x^2)^{1/2} = \frac{1}{2}(4-x^2)^{-1/2}(-2x) \) \( f'(x) = \frac{-x}{\sqrt{4-x^2}} \) For \( f(x) \) to be differentiable, \( f'(x) \) must exist. The derivative \( f'(x) \) is defined when \( 4-x^2 > 0 \), which means \( -2 < x < 2 \). This implies that \( f(x) \) is differentiable on the open interval \( (-2, 2) \). It's crucial that the derivative exists for all points in the open interval. Next, we evaluate the function at the endpoints: \( f(-2) = \sqrt{4 - (-2)^2} = \sqrt{4 - 4} = \sqrt{0} = 0 \) \( f(2) = \sqrt{4 - (2)^2} = \sqrt{4 - 4} = \sqrt{0} = 0 \) So, we have \( f(-2) = f(2) = 0 \). All three conditions of Rolle's Theorem are satisfied. This guarantees that there exists at least one real number \( c \) in the open interval \( (-2, 2) \) such that \( f'(c) = 0 \). Now, set \( f'(c) = 0 \): \( \frac{-c}{\sqrt{4-c^2}} = 0 \) For this fraction to be zero, the numerator must be zero: \( -c = 0 \)
\( \implies \) \( c = 0 \) Since \( c = 0 \) is indeed in the interval \( (-2, 2) \), Rolle's Theorem is verified. This point is the highest point of the semicircle, where the tangent is horizontal.In simple words: This problem used Rolle's Theorem on a semicircle. We checked it was continuous, differentiable (inside the interval), and had the same value at its ends. Then, we found the highest point where its slope was flat, confirming the theorem.

🎯 Exam Tip: For functions involving square roots, pay close attention to the domain and where the derivative exists. The derivative of \( \sqrt{u} \) is \( \frac{u'}{2\sqrt{u}} \), so it won't exist where \( u=0 \).

Question 15. If Rolle's theorem holds for the function \( f(x) = x^3 + bx^2 + ax + 5 \) on \([1, 3]\), find the values of a and b.
Answer: We are given the function \( f(x) = x^3 + bx^2 + ax + 5 \) on the interval \([1, 3]\). Since Rolle's Theorem holds for \( f(x) \) on \([1, 3]\), the following conditions must be met: 1. \( f(x) \) is continuous on \([1, 3]\). (True, as \( f(x) \) is a polynomial) 2. \( f(x) \) is differentiable on \( (1, 3) \). (True, as \( f(x) \) is a polynomial) 3. \( f(1) = f(3) \). Let's use the third condition to form an equation: \( f(1) = (1)^3 + b(1)^2 + a(1) + 5 = 1 + b + a + 5 = a + b + 6 \) \( f(3) = (3)^3 + b(3)^2 + a(3) + 5 = 27 + 9b + 3a + 5 = 3a + 9b + 32 \) Since \( f(1) = f(3) \): \( a + b + 6 = 3a + 9b + 32 \) Rearrange the terms: \( 3a - a + 9b - b + 32 - 6 = 0 \) \( 2a + 8b + 26 = 0 \) Divide by 2: \( a + 4b + 13 = 0 \)
\( \implies \) \( a = -4b - 13 \) (Equation 1) Now, since Rolle's Theorem holds, there must exist at least one \( c \in (1, 3) \) such that \( f'(c) = 0 \). Let's find the derivative of \( f(x) \): \( f'(x) = 3x^2 + 2bx + a \) Set \( f'(c) = 0 \): \( 3c^2 + 2bc + a = 0 \) (Equation 2) The problem statement has a stray part `\frac{1}{\sqrt{3}}\right)` after `[1, 3]`. This could imply a specific value for \( c \). Let's assume \( c = 2 + \frac{1}{\sqrt{3}} \) as used in the original solution. The value \( c = 2 + \frac{1}{\sqrt{3}} \) is approximately \( 2 + 0.577 = 2.577 \), which is indeed within the interval \( (1, 3) \). Substitute \( c = 2 + \frac{1}{\sqrt{3}} \) into Equation 2: \( 3(2 + \frac{1}{\sqrt{3}})^2 + 2b(2 + \frac{1}{\sqrt{3}}) + a = 0 \) \( 3(4 + \frac{4}{\sqrt{3}} + \frac{1}{3}) + 2b(2 + \frac{1}{\sqrt{3}}) + a = 0 \) \( 3(4 + \frac{1}{3} + \frac{4\sqrt{3}}{3}) + 2b(2 + \frac{\sqrt{3}}{3}) + a = 0 \) \( 3(\frac{13}{3} + \frac{4\sqrt{3}}{3}) + 2b(\frac{6+\sqrt{3}}{3}) + a = 0 \) \( 13 + 4\sqrt{3} + \frac{2b(6+\sqrt{3})}{3} + a = 0 \) Multiply by 3 to clear the fraction: \( 3(13 + 4\sqrt{3}) + 2b(6+\sqrt{3}) + 3a = 0 \) \( 39 + 12\sqrt{3} + 12b + 2\sqrt{3}b + 3a = 0 \) Group terms with \( \sqrt{3} \) and without: \( (39 + 12b + 3a) + (12 + 2b)\sqrt{3} = 0 \) For this equation to hold, both the rational and irrational parts must be zero. So, \( 12 + 2b = 0 \)
\( \implies \) \( 2b = -12 \)
\( \implies \) \( b = -6 \) Now substitute \( b = -6 \) into Equation 1: \( a = -4(-6) - 13 \) \( a = 24 - 13 \) \( a = 11 \) So, the values are \( a = 11 \) and \( b = -6 \). Rolle's theorem gives us conditions that can be used to find unknown constants in a function.In simple words: Since Rolle's Theorem was confirmed for this function, we knew two things: the function values at the ends of the interval must be equal, and its slope must be zero at a certain point inside. Using these facts, along with a given point where the slope was zero, we solved for the unknown numbers 'a' and 'b'.

🎯 Exam Tip: When given that Rolle's Theorem holds, use both \( f(a) = f(b) \) and \( f'(c) = 0 \) to form a system of equations. If \( c \) is given or implied, substitute it carefully. Rationalize denominators for clearer calculations.

Question 16. Apply Rolle's theorem to find point (or points) on the given curve where the tangent is parallel to the x-axis.
(i) \( y = x^2 \) in \([-2, 2]\)
(ii) \( y = 16 - x^2 \), \( x \in [-1, 1]\)
Answer:
(i) Given \( y = f(x) = x^2 \) in \([-2, 2]\). This function is a polynomial, so it is continuous on \([-2, 2]\) and differentiable on \( (-2, 2) \). Let's find the function values at the endpoints: \( f(-2) = (-2)^2 = 4 \) \( f(2) = (2)^2 = 4 \) Since \( f(-2) = f(2) \), all three conditions of Rolle's Theorem are met. Now, find the derivative: \( f'(x) = 2x \). Set \( f'(c) = 0 \): \( 2c = 0 \)
\( \implies \) \( c = 0 \) Since \( c = 0 \) is in the interval \( (-2, 2) \), the point on the curve where the tangent is parallel to the x-axis is when \( x = 0 \). To find the y-coordinate, substitute \( x = 0 \) into \( y = x^2 \): \( y = (0)^2 = 0 \) So, the required point is \( (0, 0) \). This is the vertex of the parabola.
(ii) Given \( y = f(x) = 16 - x^2 \), \( x \in [-1, 1] \). This function is a polynomial, so it is continuous on \([-1, 1]\) and differentiable on \( (-1, 1) \). Let's find the function values at the endpoints: \( f(-1) = 16 - (-1)^2 = 16 - 1 = 15 \) \( f(1) = 16 - (1)^2 = 16 - 1 = 15 \) Since \( f(-1) = f(1) \), all three conditions of Rolle's Theorem are met. Now, find the derivative: \( f'(x) = -2x \). Set \( f'(c) = 0 \): \( -2c = 0 \)
\( \implies \) \( c = 0 \) Since \( c = 0 \) is in the interval \( (-1, 1) \), the point on the curve where the tangent is parallel to the x-axis is when \( x = 0 \). To find the y-coordinate, substitute \( x = 0 \) into \( y = 16 - x^2 \): \( y = 16 - (0)^2 = 16 \) So, the required point is \( (0, 16) \). This is the vertex of the downward-opening parabola.In simple words: This problem asked us to find points where the curve's slope is flat, which means the tangent line is parallel to the x-axis. We used Rolle's Theorem by checking the function's smoothness and equal endpoint values, then finding the point where its derivative was zero.

🎯 Exam Tip: "Tangent parallel to x-axis" is another way of saying \( f'(c) = 0 \). After finding \( c \), always substitute it back into the original function to get the full coordinate point \( (c, f(c)) \).

Question 17. Examine the applicability of Rolle's theorem on the following functions :
(i) \( f(x) = \sqrt{x} \) on \([-1, 1]\)
(ii) \( f(x) = 1 - (x - 1)^{2/3} \) on \([0, 2]\)
(iii) \( f(x) = x^{2/3} \) on \([-1, 1]\)
Answer:
(i) Given \( f(x) = \sqrt{x} \) on \([-1, 1]\). For \( f(x) = \sqrt{x} \) to be a real number, \( x \) must be greater than or equal to 0. The interval given is \([-1, 1]\). However, \( f(x) = \sqrt{x} \) is not defined for \( x \in [-1, 0) \). Since the function is not defined over the entire closed interval \([-1, 1]\), it cannot be continuous on this interval. Therefore, Rolle's Theorem is not applicable for \( f(x) = \sqrt{x} \) on \([-1, 1]\). The basic requirement of the function being defined and continuous over the entire closed interval fails.
(ii) Given \( f(x) = 1 - (x - 1)^{2/3} \) on \([0, 2]\). The function \( f(x) = 1 - (x - 1)^{2/3} \) is continuous on the closed interval \([0, 2]\). It involves a power function which is continuous. Let's find the derivative of \( f(x) \): \( f'(x) = \frac{d}{dx} [1 - (x - 1)^{2/3}] \) \( f'(x) = 0 - \frac{2}{3}(x - 1)^{2/3 - 1} \cdot 1 \) \( f'(x) = -\frac{2}{3}(x - 1)^{-1/3} \) \( f'(x) = -\frac{2}{3(x - 1)^{1/3}} \) For \( f'(x) \) to exist, the denominator cannot be zero. If \( x - 1 = 0 \), then \( x = 1 \). So, \( f'(x) \) does not exist at \( x = 1 \). Since \( x = 1 \) is in the open interval \( (0, 2) \), the function is not differentiable on \( (0, 2) \). This means there is a sharp point or vertical tangent at \( x=1 \). Therefore, Rolle's Theorem is not applicable for \( f(x) = 1 - (x - 1)^{2/3} \) on \([0, 2]\).
(iii) Given \( f(x) = x^{2/3} \) on \([-1, 1]\). The function \( f(x) = x^{2/3} = (x^2)^{1/3} \) or \( (x^{1/3})^2 \) is continuous on the closed interval \([-1, 1]\). Power functions are generally continuous. Let's find the derivative of \( f(x) \): \( f'(x) = \frac{d}{dx} (x^{2/3}) = \frac{2}{3}x^{2/3 - 1} \) \( f'(x) = \frac{2}{3}x^{-1/3} \) \( f'(x) = \frac{2}{3x^{1/3}} \) For \( f'(x) \) to exist, the denominator cannot be zero. If \( x^{1/3} = 0 \), then \( x = 0 \). So, \( f'(x) \) does not exist at \( x = 0 \). Since \( x = 0 \) is in the open interval \( (-1, 1) \), the function is not differentiable on \( (-1, 1) \). There is a cusp (sharp point) at \( x=0 \). Therefore, Rolle's Theorem is not applicable for \( f(x) = x^{2/3} \) on \([-1, 1]\).In simple words: This problem tested when Rolle's Theorem *doesn't* work. For the first function, it wasn't defined everywhere. For the second and third functions, they had sharp points or places where their slope wasn't defined, even though they were continuous. So, none of these functions could use Rolle's Theorem in the given ranges.

🎯 Exam Tip: When examining applicability, thoroughly check all three conditions. Watch out for square roots that make the function undefined, and look for fractional exponents (like 2/3 or 1/3) that can lead to non-differentiability at points where the base is zero.