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

S Chand Class 12 ICSE Maths Solutions Chapter 10 Mean Value Theorems Ex 10(b)

Question 1.
(i) f(x) = x(x – 2) in [1, 2]
(ii) f(x) = x² – 2x + 4 in [1, 5].
Answer:
(i) Given \( f(x) = x(x – 2) \) in \( [1, 2] \).
Since \( f(x) \) is a polynomial in \( x \), it is continuous on \( [1, 2] \) and differentiable on \( (1, 2) \). This means its derivative exists for all \( x \in (1, 2) \).
So, by Lagrange's Mean Value Theorem (LMVT), there must be at least one real number \( c \in (1, 2) \) such that \( \frac{f(2)-f(1)}{2-1} = f'(c) \).
First, find \( f(1) \) and \( f(2) \):
\( f(1) = 1(1-2) = -1 \)
\( f(2) = 2(2-2) = 0 \)
Next, find the derivative \( f'(x) \):
\( f(x) = x^2 - 2x \)
\( f'(x) = 2x - 2 \)
Now, substitute these into the LMVT equation:
\( \frac{0 - (-1)}{2-1} = 2c - 2 \)
\( \implies \frac{1}{1} = 2c - 2 \)
\( \implies 1 = 2c - 2 \)
\( \implies 2c = 3 \)
\( \implies c = \frac{3}{2} \)
Since \( c = \frac{3}{2} \) (or 1.5) is in the interval \( (1, 2) \), Lagrange's Mean Value Theorem is verified for this function.
(ii) Given \( f(x) = x^2 – 2x + 4 \) in \( [1, 5] \).
This function is a polynomial, so it is continuous everywhere and differentiable everywhere. Thus, it is continuous on \( [1, 5] \) and differentiable on \( (1, 5) \). Both conditions for LMVT are satisfied.
Therefore, there exists at least one real number \( c \in (1, 5) \) such that \( f'(c) = \frac{f(5)-f(1)}{5-1} \).
First, find \( f(1) \) and \( f(5) \):
\( f(1) = 1^2 - 2(1) + 4 = 1 - 2 + 4 = 3 \)
\( f(5) = 5^2 - 2(5) + 4 = 25 - 10 + 4 = 19 \)
Next, find the derivative \( f'(x) \):
\( f'(x) = 2x - 2 \)
Now, substitute these into the LMVT equation:
\( 2c - 2 = \frac{19 - 3}{5 - 1} \)
\( \implies 2c - 2 = \frac{16}{4} \)
\( \implies 2c - 2 = 4 \)
\( \implies 2c = 6 \)
\( \implies c = 3 \)
Since \( c = 3 \) is in the interval \( (1, 5) \), Lagrange's Mean Value Theorem is verified for this function.
In simple words: For both parts, we checked if the function is smooth and continuous in the given range. Then, we used Lagrange's Mean Value Theorem, which says there's a point 'c' where the slope of the tangent line equals the slope of the line connecting the endpoints. We found this 'c' and made sure it falls within the given interval.

🎯 Exam Tip: Always state the continuity and differentiability conditions explicitly. Calculate \( f(a) \), \( f(b) \), and \( f'(x) \) correctly before substituting them into the LMVT formula. Verify that the value of \( c \) obtained lies within the open interval \( (a, b) \).

Question 2. f(x) = x² + x − 1 in the interval [0, 4].
Answer:
Given \( f(x) = x^2 + x - 1 \) in the interval \( [0, 4] \).
Since \( f(x) \) is a polynomial function, it is continuous on \( [0, 4] \) and differentiable on \( (0, 4) \). Thus, both conditions of LMVT are satisfied.
So, there exists at least one real number \( c \in (0, 4) \) such that \( f'(c) = \frac{f(4)-f(0)}{4-0} \).
First, find \( f(0) \) and \( f(4) \):
\( f(0) = 0^2 + 0 - 1 = -1 \)
\( f(4) = 4^2 + 4 - 1 = 16 + 4 - 1 = 19 \)
Next, find the derivative \( f'(x) \):
\( f'(x) = 2x + 1 \)
Now, substitute these into the LMVT equation:
\( 2c + 1 = \frac{19 - (-1)}{4 - 0} \)
\( \implies 2c + 1 = \frac{20}{4} \)
\( \implies 2c + 1 = 5 \)
\( \implies 2c = 4 \)
\( \implies c = 2 \)
Since \( c = 2 \) is in the interval \( (0, 4) \), Lagrange's Mean Value Theorem is verified for this function.
In simple words: This function is a simple polynomial, so it's smooth and continuous. We found a point 'c' in the interval where the slope of the tangent matches the overall slope of the line connecting the start and end points, as the theorem predicts.

🎯 Exam Tip: When dealing with polynomial functions, remember they are always continuous and differentiable over any real interval. This simplifies checking the initial conditions for LMVT. Double-check your arithmetic for function evaluations and the derivative.

Question 3. f(x) = 2x² - 10x + 29 in [2, 7].
Answer:
Given \( f(x) = 2x^2 - 10x + 29 \) in \( [2, 7] \).
Since \( f(x) \) is a polynomial in \( x \), it is continuous on \( [2, 7] \) and differentiable on \( (2, 7) \). Thus, both conditions of LMVT are satisfied.
So, there exists at least one real number \( c \in (2, 7) \) such that \( f'(c) = \frac{f(7)-f(2)}{7-2} \).
First, find \( f(2) \) and \( f(7) \):
\( f(2) = 2(2)^2 - 10(2) + 29 = 2(4) - 20 + 29 = 8 - 20 + 29 = 17 \)
\( f(7) = 2(7)^2 - 10(7) + 29 = 2(49) - 70 + 29 = 98 - 70 + 29 = 57 \)
Next, find the derivative \( f'(x) \):
\( f'(x) = 4x - 10 \)
Now, substitute these into the LMVT equation:
\( 4c - 10 = \frac{57 - 17}{7 - 2} \)
\( \implies 4c - 10 = \frac{40}{5} \)
\( \implies 4c - 10 = 8 \)
\( \implies 4c = 18 \)
\( \implies c = \frac{18}{4} = \frac{9}{2} \)
Since \( c = \frac{9}{2} \) (or 4.5) is in the interval \( (2, 7) \), Lagrange's Mean Value Theorem is verified for this function.
In simple words: This is another polynomial function, which means it behaves nicely everywhere. We applied the LMVT to find a specific point 'c' within the given range where the instantaneous rate of change (derivative) matches the average rate of change between the two endpoints.

🎯 Exam Tip: Pay attention to the calculations for \( f(a) \) and \( f(b) \), especially with negative numbers or larger values. A small error in these steps can lead to an incorrect value for \( c \).

Question 4. f(x) = x - \( \frac { 1 }{ x } \); x ∈ [1, 3].
Answer:
Given \( f(x) = x - \frac{1}{x} \) for \( x \in [1, 3] \).
This function can be rewritten as \( f(x) = \frac{x^2-1}{x} \). This is a rational function. It is continuous on \( [1, 3] \) because \( x \neq 0 \) in this interval.
Next, find the derivative \( f'(x) \):
\( f'(x) = 1 - (-1)x^{-2} = 1 + \frac{1}{x^2} \).
This derivative exists for all \( x \in (1, 3) \) (since \( x \neq 0 \)). Thus, \( f(x) \) is derivable on \( (1, 3) \). Both conditions of LMVT are satisfied.
So, there exists at least one real number \( c \in (1, 3) \) such that \( f'(c) = \frac{f(3)-f(1)}{3-1} \).
First, find \( f(1) \) and \( f(3) \):
\( f(1) = 1 - \frac{1}{1} = 0 \)
\( f(3) = 3 - \frac{1}{3} = \frac{9-1}{3} = \frac{8}{3} \)
Now, substitute these into the LMVT equation:
\( 1 + \frac{1}{c^2} = \frac{\frac{8}{3} - 0}{3 - 1} \)
\( \implies 1 + \frac{1}{c^2} = \frac{\frac{8}{3}}{2} \)
\( \implies 1 + \frac{1}{c^2} = \frac{8}{6} = \frac{4}{3} \)
\( \implies \frac{1}{c^2} = \frac{4}{3} - 1 \)
\( \implies \frac{1}{c^2} = \frac{4-3}{3} \)
\( \implies \frac{1}{c^2} = \frac{1}{3} \)
\( \implies c^2 = 3 \)
\( \implies c = \pm\sqrt{3} \)
Since the interval is \( [1, 3] \), we must choose the positive value for \( c \).
\( c = \sqrt{3} \)
We know that \( 1 < \sqrt{3} < 3 \). So \( c = \sqrt{3} \) is in the interval \( (1, 3) \). Lagrange's Mean Value Theorem is verified.
In simple words: This function involves a fraction, but it's well-behaved in the given positive range. We found a spot 'c' where the curve's steepness (tangent slope) matches the average steepness between the starting and ending points. We chose the positive square root because our interval only includes positive numbers.

🎯 Exam Tip: For rational functions, ensure that the denominator does not become zero within the interval. When solving for \( c^2 \), remember to consider both positive and negative roots, but select only the one that falls within the specified open interval.

Question 5. f(x) = (x – 1)(x – 2)(x – 3) in [1, 4].
Answer:
Given \( f(x) = (x – 1)(x – 2)(x – 3) \) in \( [1, 4] \).
First, expand the function to identify it as a polynomial:
\( f(x) = (x^2 - 3x + 2)(x - 3) \)
\( f(x) = x^3 - 3x^2 + 2x - 3x^2 + 9x - 6 \)
\( f(x) = x^3 - 6x^2 + 11x - 6 \)
Since \( f(x) \) is a polynomial in \( x \), it is continuous on \( [1, 4] \) and differentiable on \( (1, 4) \). Thus, both conditions of LMVT are satisfied.
So, there exists at least one real number \( c \in (1, 4) \) such that \( f'(c) = \frac{f(4)-f(1)}{4-1} \).
First, find \( f(1) \) and \( f(4) \):
\( f(1) = (1-1)(1-2)(1-3) = 0 \cdot (-1) \cdot (-2) = 0 \)
\( f(4) = (4-1)(4-2)(4-3) = 3 \cdot 2 \cdot 1 = 6 \)
Next, find the derivative \( f'(x) \):
\( f'(x) = 3x^2 - 12x + 11 \)
Now, substitute these into the LMVT equation:
\( 3c^2 - 12c + 11 = \frac{6 - 0}{4 - 1} \)
\( \implies 3c^2 - 12c + 11 = \frac{6}{3} \)
\( \implies 3c^2 - 12c + 11 = 2 \)
\( \implies 3c^2 - 12c + 9 = 0 \)
Divide the entire equation by 3:
\( \implies c^2 - 4c + 3 = 0 \)
Factor the quadratic equation:
\( \implies (c - 1)(c - 3) = 0 \)
\( \implies c = 1 \) or \( c = 3 \)
According to LMVT, \( c \) must be in the *open* interval \( (1, 4) \). Therefore, \( c=1 \) is not included, but \( c=3 \) is in \( (1, 4) \).
Since \( c = 3 \) is in the interval \( (1, 4) \), Lagrange's Mean Value Theorem is verified for this function.
In simple words: This function is a polynomial, so it's continuous and smooth. We used the Mean Value Theorem to find a point 'c' where the slope of the curve is exactly the same as the average slope from the start of the interval to the end. One of the 'c' values matched an endpoint, so we only chose the other one which was inside the interval.

🎯 Exam Tip: Remember to express polynomial functions in expanded form if given in factored form before differentiating, to avoid errors. When you get multiple values for \( c \), always check which one(s) fall within the *open* interval \( (a, b) \) specified by the theorem.

Question 6. f(x) = logx in [1, e].
Answer:
Given \( f(x) = \log x \) in \( [1, e] \).
The logarithmic function \( \log x \) is continuous and differentiable in its domain, which is \( x > 0 \). Since the interval \( [1, e] \) is within this domain (as \( 1 > 0 \) and \( e \approx 2.718 > 0 \)), \( f(x) \) is continuous on \( [1, e] \) and differentiable on \( (1, e) \). Both conditions for LMVT are satisfied.
So, there exists at least one real number \( c \in (1, e) \) such that \( f'(c) = \frac{f(e)-f(1)}{e-1} \).
First, find \( f(1) \) and \( f(e) \):
\( f(1) = \log 1 = 0 \)
\( f(e) = \log e = 1 \) (assuming natural logarithm, base \( e \))
Next, find the derivative \( f'(x) \):
\( f'(x) = \frac{1}{x} \)
Now, substitute these into the LMVT equation:
\( \frac{1}{c} = \frac{1 - 0}{e - 1} \)
\( \implies \frac{1}{c} = \frac{1}{e - 1} \)
\( \implies c = e - 1 \)
To confirm if \( c = e - 1 \) is in the interval \( (1, e) \):
We know \( e \approx 2.718 \).
So, \( c \approx 2.718 - 1 = 1.718 \).
Since \( 1 < 1.718 < 2.718 \), \( c = e - 1 \) is indeed in the interval \( (1, e) \). Lagrange's Mean Value Theorem is verified.
In simple words: The logarithm function works well for positive numbers. Here, we found a point 'c' in the interval \( (1, e) \) where the slope of the function's tangent matches the average slope between \( x=1 \) and \( x=e \). This shows the theorem holds true for logarithms too.

🎯 Exam Tip: Remember that \( \log x \) is often assumed to be \( \ln x \) (natural logarithm) unless a different base is specified. Crucially, recall that \( \log 1 = 0 \) and \( \log_e e = 1 \). Always verify that the derived value of \( c \) lies strictly within the open interval \( (a,b) \).

Question 7. f(x) = eˣ in [0, 1].
Answer:
Given \( f(x) = e^x \) in \( [0, 1] \).
The exponential function \( e^x \) is continuous and differentiable everywhere for all real numbers. Thus, it is continuous on \( [0, 1] \) and differentiable on \( (0, 1) \). Both conditions for LMVT are satisfied.
So, there exists at least one real number \( c \in (0, 1) \) such that \( f'(c) = \frac{f(1)-f(0)}{1-0} \).
First, find \( f(0) \) and \( f(1) \):
\( f(0) = e^0 = 1 \)
\( f(1) = e^1 = e \)
Next, find the derivative \( f'(x) \):
\( f'(x) = e^x \)
Now, substitute these into the LMVT equation:
\( e^c = \frac{e - 1}{1 - 0} \)
\( \implies e^c = e - 1 \)
To find \( c \), take the natural logarithm of both sides:
\( \ln(e^c) = \ln(e - 1) \)
\( \implies c = \ln(e - 1) \)
To confirm if \( c = \ln(e - 1) \) is in the interval \( (0, 1) \):
We know \( e \approx 2.718 \), so \( e - 1 \approx 1.718 \).
Since \( 1 < 1.718 < e \approx 2.718 \), and \( \ln x \) is an increasing function, we can compare its values:
\( \ln 1 < \ln(e - 1) < \ln e \)
\( \implies 0 < \ln(e - 1) < 1 \)
So, \( c = \ln(e - 1) \) is in the interval \( (0, 1) \). Lagrange's Mean Value Theorem is verified.
In simple words: The exponential function is very smooth and has no breaks. We found a point 'c' between 0 and 1 where the slope of the curve at that point matches the overall average slope between \( x=0 \) and \( x=1 \).

🎯 Exam Tip: When dealing with exponential functions, remember that their derivative is itself. Also, be careful with logarithmic properties when solving for \( c \). It's helpful to approximate values for \( e \) and \( \ln \) to quickly check if \( c \) is within the correct interval.

Question 8. f(x) = [x] in [-1, 1],
Answer:
Given \( f(x) = [x] \) (the greatest integer function) in \( [-1, 1] \).
The greatest integer function is discontinuous at every integer value. In the interval \( [-1, 1] \), it is discontinuous at \( x = -1, 0, 1 \).
For Lagrange's Mean Value Theorem to apply, the function must be continuous on the closed interval \( [a, b] \) and differentiable on the open interval \( (a, b) \).
Since \( f(x) = [x] \) is not continuous at \( x=0 \) (which is inside \( (-1, 1) \)), and not differentiable at integer points, it fails the continuity condition on \( [-1, 1] \) and differentiability condition on \( (-1, 1) \).
Therefore, Lagrange's Mean Value Theorem is not applicable to this function in the given interval.
In simple words: This function is like steps on a staircase, jumping at each whole number. Because it's not smooth and has these jumps, we cannot use Lagrange's Mean Value Theorem. The theorem only works for functions that are continuous and smooth.

🎯 Exam Tip: Always remember that the greatest integer function (floor function) is discontinuous at integer points. This immediately tells you that LMVT and Rolle's Theorem will not apply for intervals containing such points of discontinuity or non-differentiability.

Question 9.
(i) Find c so that f'(c) = \( \frac{f(6)-f(4)}{6-4} \), where f(x) = \( \sqrt{x+2} \) and c \( \epsilon \) (4, 6).
(ii) Given f(x) = x³/2, find the value of c \( \in \) (0,1) such that f'(c) = \( \frac{f(1)-f(0)}{1-0} \).
Answer:
(i) Given \( f(x) = \sqrt{x+2} \).
First, find the derivative \( f'(x) \):
\( f(x) = (x+2)^{1/2} \)
\( f'(x) = \frac{1}{2}(x+2)^{-1/2} \cdot 1 = \frac{1}{2\sqrt{x+2}} \)
Next, find \( f(4) \) and \( f(6) \):
\( f(4) = \sqrt{4+2} = \sqrt{6} \)
\( f(6) = \sqrt{6+2} = \sqrt{8} \)
Now, substitute these into the given equation \( f'(c) = \frac{f(6)-f(4)}{6-4} \):
\( \frac{1}{2\sqrt{c+2}} = \frac{\sqrt{8} - \sqrt{6}}{6 - 4} \)
\( \implies \frac{1}{2\sqrt{c+2}} = \frac{\sqrt{8} - \sqrt{6}}{2} \)
Multiply both sides by 2:
\( \frac{1}{\sqrt{c+2}} = \sqrt{8} - \sqrt{6} \)
Rationalize the denominator by multiplying by \( \frac{\sqrt{8} + \sqrt{6}}{\sqrt{8} + \sqrt{6}} \):
\( \frac{1}{\sqrt{c+2}} = \frac{(\sqrt{8} - \sqrt{6})(\sqrt{8} + \sqrt{6})}{\sqrt{8} + \sqrt{6}} \)
\( \implies \frac{1}{\sqrt{c+2}} = \frac{8 - 6}{\sqrt{8} + \sqrt{6}} = \frac{2}{\sqrt{8} + \sqrt{6}} \)
So, \( \sqrt{c+2} = \frac{\sqrt{8} + \sqrt{6}}{2} \)
Square both sides:
\( c+2 = \left( \frac{\sqrt{8} + \sqrt{6}}{2} \right)^2 \)
\( c+2 = \frac{(\sqrt{8})^2 + (\sqrt{6})^2 + 2\sqrt{8}\sqrt{6}}{4} \)
\( c+2 = \frac{8 + 6 + 2\sqrt{48}}{4} \)
\( c+2 = \frac{14 + 2\sqrt{16 \cdot 3}}{4} \)
\( c+2 = \frac{14 + 2 \cdot 4\sqrt{3}}{4} \)
\( c+2 = \frac{14 + 8\sqrt{3}}{4} \)
\( c+2 = \frac{2(7 + 4\sqrt{3})}{4} \)
\( c+2 = \frac{7 + 4\sqrt{3}}{2} \)
Now solve for \( c \):
\( c = \frac{7 + 4\sqrt{3}}{2} - 2 \)
\( c = \frac{7 + 4\sqrt{3} - 4}{2} \)
\( c = \frac{3 + 4\sqrt{3}}{2} \)
To check if \( c \) is in \( (4, 6) \), we can approximate \( \sqrt{3} \approx 1.732 \):
\( c \approx \frac{3 + 4(1.732)}{2} = \frac{3 + 6.928}{2} = \frac{9.928}{2} = 4.964 \)
Since \( 4 < 4.964 < 6 \), the value of \( c = \frac{3 + 4\sqrt{3}}{2} \) is valid.
(ii) Given \( f(x) = x^{3/2} \).
First, find the derivative \( f'(x) \):
\( f'(x) = \frac{3}{2}x^{3/2 - 1} = \frac{3}{2}x^{1/2} = \frac{3}{2}\sqrt{x} \)
Next, find \( f(0) \) and \( f(1) \):
\( f(0) = 0^{3/2} = 0 \)
\( f(1) = 1^{3/2} = 1 \)
Now, substitute these into the given equation \( f'(c) = \frac{f(1)-f(0)}{1-0} \):
\( \frac{3}{2}\sqrt{c} = \frac{1 - 0}{1 - 0} \)
\( \implies \frac{3}{2}\sqrt{c} = 1 \)
\( \implies \sqrt{c} = \frac{2}{3} \)
Square both sides:
\( c = \left(\frac{2}{3}\right)^2 \)
\( \implies c = \frac{4}{9} \)
To check if \( c = \frac{4}{9} \) is in \( (0, 1) \):
Since \( 0 < \frac{4}{9} < 1 \), the value of \( c = \frac{4}{9} \) is valid.
In simple words: For both parts, we were asked to find a specific 'c' value where the instantaneous slope matches the average slope between two points. We calculated the derivative, found the function values at the given points, and then solved for 'c' to make sure it fit in the specified range.

🎯 Exam Tip: When dealing with square roots or fractional exponents, be careful with differentiation and algebraic manipulation. Rationalizing the denominator or squaring both sides correctly is crucial. Always verify that the obtained value of \( c \) lies within the given interval.

Question 10.
(i) Find a point on the graph of y = x³, where tangent is parallel to the chord joining (1, 1) and (3, 27).
(ii) Use L.M.V. to determine the point on the curve y = x³ – 3x, where the tangent to the curve is parallel to the chord joining (1, -2) and (2, 2).
Answer:
(i) Given \( y = f(x) = x^3 \). The interval is implicitly \( [1, 3] \).
Since \( f(x) \) is a polynomial, it is continuous on \( [1, 3] \) and differentiable on \( (1, 3) \).
The slope of the chord joining \( (1, 1) \) and \( (3, 27) \) is given by:
\( \text{Slope} = \frac{f(3) - f(1)}{3 - 1} = \frac{27 - 1}{2} = \frac{26}{2} = 13 \)
Next, find the derivative \( f'(x) \):
\( f'(x) = 3x^2 \)
For the tangent to be parallel to the chord, their slopes must be equal:
\( f'(c) = \text{Slope of chord} \)
\( 3c^2 = 13 \)
\( \implies c^2 = \frac{13}{3} \)
\( \implies c = \pm\sqrt{\frac{13}{3}} \)
Since the interval is \( (1, 3) \), we choose the positive value:
\( c = \sqrt{\frac{13}{3}} \)
(Note: \( \sqrt{\frac{13}{3}} \approx \sqrt{4.33} \approx 2.08 \), which is between 1 and 3).
Now, find the y-coordinate of the point using \( f(c) = c^3 \):
\( y = \left(\sqrt{\frac{13}{3}}\right)^3 = \left(\frac{13}{3}\right)^{1/2 \cdot 3} = \frac{13}{3}\sqrt{\frac{13}{3}} \)
So, the required point is \( \left(\sqrt{\frac{13}{3}}, \frac{13}{3}\sqrt{\frac{13}{3}}\right) \).
(ii) Given \( y = f(x) = x^3 - 3x \). The interval is implicitly \( [1, 2] \).
Since \( f(x) \) is a polynomial, it is continuous on \( [1, 2] \) and differentiable on \( (1, 2) \).
The slope of the chord joining \( (1, -2) \) and \( (2, 2) \) is given by:
\( \text{Slope} = \frac{f(2) - f(1)}{2 - 1} = \frac{2 - (-2)}{1} = \frac{4}{1} = 4 \)
Next, find the derivative \( f'(x) \):
\( f'(x) = 3x^2 - 3 \)
For the tangent to be parallel to the chord, their slopes must be equal:
\( f'(c) = \text{Slope of chord} \)
\( 3c^2 - 3 = 4 \)
\( \implies 3c^2 = 7 \)
\( \implies c^2 = \frac{7}{3} \)
\( \implies c = \pm\sqrt{\frac{7}{3}} \)
Since the interval is \( (1, 2) \), we choose the positive value:
\( c = \sqrt{\frac{7}{3}} \)
(Note: \( \sqrt{\frac{7}{3}} \approx \sqrt{2.33} \approx 1.52 \), which is between 1 and 2).
Now, find the y-coordinate of the point using \( f(c) = c^3 - 3c \):
\( y = \left(\sqrt{\frac{7}{3}}\right)^3 - 3\left(\sqrt{\frac{7}{3}}\right) \)
\( y = \frac{7}{3}\sqrt{\frac{7}{3}} - 3\sqrt{\frac{7}{3}} \)
\( y = \sqrt{\frac{7}{3}} \left( \frac{7}{3} - 3 \right) \)
\( y = \sqrt{\frac{7}{3}} \left( \frac{7 - 9}{3} \right) \)
\( y = \sqrt{\frac{7}{3}} \left( -\frac{2}{3} \right) \)
\( y = -\frac{2}{3}\sqrt{\frac{7}{3}} \)
So, the required point is \( \left(\sqrt{\frac{7}{3}}, -\frac{2}{3}\sqrt{\frac{7}{3}}\right) \).
In simple words: For both parts, we used the idea that if a tangent line is parallel to a chord, their slopes are the same. We calculated the slope of the chord (the line connecting two points on the curve), then found the derivative of the function (which gives the slope of the tangent). Setting these equal, we solved for 'c', the x-coordinate, and then found the y-coordinate to get the specific point on the curve.

🎯 Exam Tip: The core idea here is that the slope of the tangent at point \( c \) is equal to the slope of the chord connecting the endpoints. Ensure you correctly calculate both slopes and solve the resulting equation for \( c \). Remember to then find the corresponding y-coordinate to give the complete point.

Question 11. Explain why Lagrange's Mean Value theorem is not applicable to the following functions:
(i) f(x) = \( \left\{\begin{array}{c} |x| \text { if } x \neq 0 \\ 0 \text { if } x=0 \end{array}, x \in[-1,1]\right. \)
(ii) f(x) = |x| in [- 1, 1]
Answer:
Lagrange's Mean Value Theorem (LMVT) requires two conditions to be met for a function \( f(x) \) on an interval \( [a, b] \):
1. \( f(x) \) must be continuous on the closed interval \( [a, b] \).
2. \( f(x) \) must be differentiable on the open interval \( (a, b) \).

(i) Given \( f(x) = \left\{\begin{array}{c} |x| \text { if } x \neq 0 \\ 0 \text { if } x=0 \end{array} \) in \( [-1,1] \).
This function is equivalent to \( f(x) = |x| \) for all \( x \).
To check differentiability at \( x=0 \), we look at the left-hand derivative (LHD) and right-hand derivative (RHD).
LHD at \( x=0 \): \( Lf'(0) = \lim_{x \to 0^-} \frac{f(x)-f(0)}{x-0} = \lim_{x \to 0^-} \frac{|x|-0}{x-0} = \lim_{x \to 0^-} \frac{-x}{x} = -1 \)
RHD at \( x=0 \): \( Rf'(0) = \lim_{x \to 0^+} \frac{f(x)-f(0)}{x-0} = \lim_{x \to 0^+} \frac{|x|-0}{x-0} = \lim_{x \to 0^+} \frac{x}{x} = 1 \)
Since \( Lf'(0) \neq Rf'(0) \), the function \( f(x) = |x| \) is not differentiable at \( x=0 \).
The point \( x=0 \) lies within the open interval \( (-1, 1) \). Because the function is not differentiable at \( x=0 \), it fails the second condition for LMVT.
Therefore, LMVT is not applicable.

(ii) Given \( f(x) = |x| \) in \( [-1, 1] \).
This is the same function as in part (i).
As shown above, \( f(x) = |x| \) is not differentiable at \( x=0 \), which is an interior point of the interval \( (-1, 1) \).
Therefore, the condition of differentiability on the open interval \( (a,b) \) is not satisfied.
Thus, Lagrange's Mean Value Theorem is not applicable.
In simple words: Lagrange's Mean Value Theorem needs a function to be smooth everywhere in the middle of its interval. The absolute value function, \( |x| \), has a sharp corner at \( x=0 \), which means it's not smooth or "differentiable" there. Since \( x=0 \) is inside the interval \( [-1, 1] \), the theorem cannot be used for this function.

🎯 Exam Tip: The absolute value function \( |x| \) is a classic example of a continuous function that is not differentiable at \( x=0 \). This is a common trap in MVT/Rolle's theorem questions. Always check for points where the derivative might not exist, especially at "corners" or "cusps."

Question 12. f(x) = x(1 – logx); x > 0, show that (a-b)logc = b(1-logb)-a(1-loga); 0 < a < b.
Answer:
Given \( f(x) = x(1 - \log x) \) for \( x > 0 \) in \( [a, b] \), where \( 0 < a < b \).
For LMVT, the function must be continuous on \( [a, b] \) and differentiable on \( (a, b) \).
Since \( x \) is a polynomial and \( \log x \) is continuous and differentiable for \( x > 0 \), their product \( x(1-\log x) \) is also continuous and differentiable for \( x > 0 \). Given \( 0 < a < b \), the interval \( [a, b] \) is within the domain where the function is continuous and differentiable.
So, by LMVT, there exists at least one real number \( c \in (a, b) \) such that \( f'(c) = \frac{f(b)-f(a)}{b-a} \).
First, find \( f(a) \) and \( f(b) \):
\( f(a) = a(1 - \log a) \)
\( f(b) = b(1 - \log b) \)
Next, find the derivative \( f'(x) \) using the product rule:
\( f'(x) = \frac{d}{dx}(x) \cdot (1 - \log x) + x \cdot \frac{d}{dx}(1 - \log x) \)
\( f'(x) = 1 \cdot (1 - \log x) + x \cdot \left(-\frac{1}{x}\right) \)
\( f'(x) = 1 - \log x - 1 \)
\( f'(x) = -\log x \)
Now, substitute these into the LMVT equation:
\( -\log c = \frac{b(1 - \log b) - a(1 - \log a)}{b - a} \)
Multiply both sides by \( -(b-a) \):
\( -(b-a)\log c = b(1 - \log b) - a(1 - \log a) \)
\( (a-b)\log c = b(1 - \log b) - a(1 - \log a) \)
This is the desired result, showing the relationship as per LMVT.
In simple words: This problem asks us to prove a formula based on Lagrange's Mean Value Theorem for a function involving a logarithm. We confirmed the function is smooth, then found its derivative. We used the LMVT formula to connect the average change of the function over the interval \( [a, b] \) to its instantaneous rate of change at point 'c', leading directly to the required equation.

🎯 Exam Tip: When proving an identity using LMVT, first ensure the function satisfies the theorem's conditions. Carefully apply the product rule for differentiation if necessary. The main goal is to equate \( f'(c) \) with the average rate of change \( \frac{f(b)-f(a)}{b-a} \) and then algebraically rearrange to match the target identity.

Example Question 1. Find 'c' of the Lagrange's Mean Value Theorem, if f(x) = x² – 3x – 1 \( \in \left[\frac{-11}{7}, \frac{13}{7}\right] \)
Answer:
Given \( f(x) = x^2 – 3x – 1 \) in the interval \( \left[\frac{-11}{7}, \frac{13}{7}\right] \).
Since \( f(x) \) is a polynomial in \( x \), it is continuous on \( \left[\frac{-11}{7}, \frac{13}{7}\right] \) and differentiable on \( \left(\frac{-11}{7}, \frac{13}{7}\right) \). Both conditions of LMVT are satisfied.
So, there exists at least one real number \( c \in \left(\frac{-11}{7}, \frac{13}{7}\right) \) such that \( f'(c) = \frac{f(\frac{13}{7})-f(\frac{-11}{7})}{\frac{13}{7}-(-\frac{11}{7})} \).
First, find \( f\left(\frac{-11}{7}\right) \) and \( f\left(\frac{13}{7}\right) \):
\( f\left(\frac{-11}{7}\right) = \left(\frac{-11}{7}\right)^2 - 3\left(\frac{-11}{7}\right) - 1 = \frac{121}{49} + \frac{33}{7} - 1 = \frac{121 + 33 \cdot 7 - 49}{49} = \frac{121 + 231 - 49}{49} = \frac{303}{49} \)
\( f\left(\frac{13}{7}\right) = \left(\frac{13}{7}\right)^2 - 3\left(\frac{13}{7}\right) - 1 = \frac{169}{49} - \frac{39}{7} - 1 = \frac{169 - 39 \cdot 7 - 49}{49} = \frac{169 - 273 - 49}{49} = \frac{-153}{49} \)
Next, find the derivative \( f'(x) \):
\( f'(x) = 2x - 3 \)
Now, substitute these into the LMVT equation:
\( 2c - 3 = \frac{\frac{-153}{49} - \frac{303}{49}}{\frac{13}{7} + \frac{11}{7}} \)
\( \implies 2c - 3 = \frac{\frac{-153 - 303}{49}}{\frac{24}{7}} \)
\( \implies 2c - 3 = \frac{-456}{49} \cdot \frac{7}{24} \)
\( \implies 2c - 3 = \frac{-456 \cdot 7}{49 \cdot 24} \)
\( \implies 2c - 3 = \frac{-19 \cdot 24 \cdot 7}{7 \cdot 7 \cdot 24} \)
\( \implies 2c - 3 = \frac{-19}{7} \)
\( \implies 2c = 3 - \frac{19}{7} \)
\( \implies 2c = \frac{21 - 19}{7} \)
\( \implies 2c = \frac{2}{7} \)
\( \implies c = \frac{1}{7} \)
To check if \( c = \frac{1}{7} \) is in \( \left(\frac{-11}{7}, \frac{13}{7}\right) \):
Since \( -\frac{11}{7} < \frac{1}{7} < \frac{13}{7} \), the value of \( c = \frac{1}{7} \) is valid. Lagrange's Mean Value Theorem is verified.
In simple words: This is a polynomial, so it's smooth and continuous. We followed the steps of the LMVT: calculated the function values at the endpoints, found the derivative, and then solved for 'c' by equating the derivative at 'c' to the average slope of the function over the given interval. The calculated 'c' value falls within the required range.

🎯 Exam Tip: When dealing with fractional endpoints or function values, be extra careful with arithmetic operations, especially when adding or subtracting fractions. Simplify intermediate steps to prevent calculation errors. Always confirm that the final value of \( c \) lies within the specified open interval.

Question 2. Verify Rolle's Theorem for the function f(x) = log\( \frac{x^2+a b}{x(a+b)} \), x \( \in \) [a, b]
Answer:
Given \( f(x) = \log\left(\frac{x^2+ab}{x(a+b)}\right) \), \( x \in [a, b] \). We assume \( a,b > 0 \).
First, simplify the function using logarithm properties:
\( f(x) = \log(x^2+ab) - \log(x) - \log(a+b) \)
To verify Rolle's Theorem, we need to check three conditions:
1. **Continuity:** Since \( x^2+ab \), \( x \), and \( a+b \) are all positive and continuous for \( x \in [a, b] \) (given \( a,b > 0 \)), and \( \log \) is continuous for positive arguments, \( f(x) \) is continuous on \( [a, b] \).
2. **Differentiability:** Find the derivative \( f'(x) \):
\( f'(x) = \frac{2x}{x^2+ab} - \frac{1}{x} - 0 \)
\( f'(x) = \frac{2x^2 - (x^2+ab)}{x(x^2+ab)} = \frac{x^2-ab}{x(x^2+ab)} \)
Since \( x \neq 0 \) and \( x^2+ab \neq 0 \) for \( x \in (a, b) \) (given \( a,b > 0 \)), \( f'(x) \) exists for all \( x \in (a,b) \). So, \( f(x) \) is differentiable on \( (a,b) \).
3. **\( f(a) = f(b) \):**
\( f(a) = \log\left(\frac{a^2+ab}{a(a+b)}\right) = \log\left(\frac{a(a+b)}{a(a+b)}\right) = \log 1 = 0 \)
\( f(b) = \log\left(\frac{b^2+ab}{b(a+b)}\right) = \log\left(\frac{b(b+a)}{b(a+b)}\right) = \log 1 = 0 \)
Thus, \( f(a) = f(b) = 0 \).
All three conditions of Rolle's Theorem are satisfied. Therefore, by Rolle's Theorem, there exists at least one real number \( c \in (a, b) \) such that \( f'(c) = 0 \).
Set \( f'(c) = 0 \):
\( \frac{c^2-ab}{c(c^2+ab)} = 0 \)
This implies \( c^2 - ab = 0 \)
\( \implies c^2 = ab \)
\( \implies c = \pm\sqrt{ab} \)
Since \( a,b > 0 \), their geometric mean \( \sqrt{ab} \) is also positive. Also, the geometric mean of two positive numbers lies between them (i.e., \( a < \sqrt{ab} < b \)).
So, \( c = \sqrt{ab} \) is in the interval \( (a,b) \). (The negative root \( -\sqrt{ab} \) is not in the interval \( (a,b) \) since \( a,b>0 \)).
Thus, Rolle's Theorem is verified, and the value of \( c \) is \( \sqrt{ab} \).
In simple words: Rolle's Theorem requires three things: the function must be smooth, continuous, and have the same value at its start and end points. This logarithmic function meets all these. We found its derivative, set it to zero, and solved for 'c'. The 'c' we found, \( \sqrt{ab} \), is exactly what the theorem predicted, meaning there's a horizontal tangent somewhere in the middle.

🎯 Exam Tip: When dealing with logarithmic functions for Rolle's Theorem, first simplify the logarithm if possible. Always verify the three conditions: continuity, differentiability, and \( f(a) = f(b) \). For the geometric mean \( \sqrt{ab} \), remember its property that it lies between \( a \) and \( b \) when \( a,b \) are positive.

Question 3. Examine the validity and conclusion of Lagrange's Mean Value Theorem for the function f(x) = x(x – 1)(x – 2) for every x \( \in \left[0, \frac{1}{2}\right] \).
Answer:
Given \( f(x) = x(x – 1)(x – 2) \) in the interval \( \left[0, \frac{1}{2}\right] \).
First, expand the function:
\( f(x) = x(x^2 - 3x + 2) = x^3 - 3x^2 + 2x \)
For LMVT, the function must satisfy two conditions:
1. **Continuity:** Since \( f(x) \) is a polynomial function, it is continuous on \( \left[0, \frac{1}{2}\right] \).
2. **Differentiability:** Find the derivative \( f'(x) \):
\( f'(x) = 3x^2 - 6x + 2 \)
Since \( f'(x) \) exists for all \( x \in \left(0, \frac{1}{2}\right) \), \( f(x) \) is differentiable on \( \left(0, \frac{1}{2}\right) \).
Both conditions of Lagrange's Mean Value Theorem are satisfied. Therefore, there exists at least one real number \( c \in \left(0, \frac{1}{2}\right) \) such that \( f'(c) = \frac{f(\frac{1}{2})-f(0)}{\frac{1}{2}-0} \).
First, find \( f(0) \) and \( f\left(\frac{1}{2}\right) \):
\( f(0) = 0(0-1)(0-2) = 0 \)
\( f\left(\frac{1}{2}\right) = \frac{1}{2}\left(\frac{1}{2}-1\right)\left(\frac{1}{2}-2\right) = \frac{1}{2}\left(-\frac{1}{2}\right)\left(-\frac{3}{2}\right) = \frac{3}{8} \)
Now, substitute these into the LMVT equation:
\( 3c^2 - 6c + 2 = \frac{\frac{3}{8} - 0}{\frac{1}{2} - 0} \)
\( \implies 3c^2 - 6c + 2 = \frac{\frac{3}{8}}{\frac{1}{2}} \)
\( \implies 3c^2 - 6c + 2 = \frac{3}{8} \cdot 2 \)
\( \implies 3c^2 - 6c + 2 = \frac{3}{4} \)
Multiply the entire equation by 4 to clear the fraction:
\( \implies 12c^2 - 24c + 8 = 3 \)
\( \implies 12c^2 - 24c + 5 = 0 \)
Use the quadratic formula to solve for \( c \): \( c = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \)
Here, \( a=12, b=-24, c=5 \).
\( c = \frac{-(-24) \pm \sqrt{(-24)^2 - 4(12)(5)}}{2(12)} \)
\( c = \frac{24 \pm \sqrt{576 - 240}}{24} \)
\( c = \frac{24 \pm \sqrt{336}}{24} \)
\( c = \frac{24 \pm \sqrt{16 \cdot 21}}{24} \)
\( c = \frac{24 \pm 4\sqrt{21}}{24} \)
\( c = \frac{6 \pm \sqrt{21}}{6} \)
Now, we check which of these values are in the interval \( \left(0, \frac{1}{2}\right) \).
Approximate \( \sqrt{21} \approx 4.58 \).
\( c_1 = \frac{6 + 4.58}{6} = \frac{10.58}{6} \approx 1.76 \)
\( c_2 = \frac{6 - 4.58}{6} = \frac{1.42}{6} \approx 0.236 \)
The interval is \( \left(0, \frac{1}{2}\right) = (0, 0.5) \).
\( c_1 \approx 1.76 \) is not in \( (0, 0.5) \).
\( c_2 \approx 0.236 \) is in \( (0, 0.5) \).
Therefore, \( c = \frac{6 - \sqrt{21}}{6} \) is the valid value for \( c \). The conclusion of Lagrange's Mean Value Theorem holds.
In simple words: This function is a polynomial, so it's smooth and continuous, making Lagrange's Mean Value Theorem applicable. We calculated the average slope across the interval and set it equal to the function's derivative. Solving this equation gave us two possible 'c' values, but only one of them fell within the specific open interval, confirming the theorem.

🎯 Exam Tip: When evaluating LMVT for polynomial functions, ensure the expansion is correct. Solving for \( c \) can sometimes lead to a quadratic equation; use the quadratic formula carefully. It's vital to check which solutions for \( c \) fall strictly within the *open* interval \( (a,b) \).

Question 4. Show that the function f(x) = x² – 6x + 1 on [1, 3] satisfies Lagrange's mean value theorem. Also find the coordinates of a point at which the tangent to the curve represented by the above function is parallel to the chord joining A (1, – 4) and B (3, – 8).
Answer: Given function is \( f(x) = x^2 - 6x + 1 \). This function is a polynomial, so it is continuous on the interval [1, 3] and differentiable on (1, 3).
First, find the function values at the interval endpoints:
\( f(1) = 1^2 - 6(1) + 1 = 1 - 6 + 1 = -4 \)
\( f(3) = 3^2 - 6(3) + 1 = 9 - 18 + 1 = -8 \)
Next, find the derivative of the function:
\( f'(x) = 2x - 6 \)
According to Lagrange's Mean Value Theorem, there exists a point \( c \in (1, 3) \) such that:
\( f'(c) = \frac{f(b) - f(a)}{b - a} \)
\( 2c - 6 = \frac{f(3) - f(1)}{3 - 1} \)
\( 2c - 6 = \frac{-8 - (-4)}{2} \)
\( 2c - 6 = \frac{-8 + 4}{2} \)
\( 2c - 6 = \frac{-4}{2} \)
\( 2c - 6 = -2 \)
\( 2c = -2 + 6 \)
\( 2c = 4 \)
\( c = \frac{4}{2} \)
\( c = 2 \)
Since \( c = 2 \) is within the interval (1, 3), the theorem is satisfied. To find the coordinates of the point where the tangent is parallel to the chord, we use \( x = c = 2 \).
Substitute \( x = 2 \) back into the original function to find \( y \):
\( f(2) = 2^2 - 6(2) + 1 = 4 - 12 + 1 = -7 \)
So, the required point on the curve is (2, -7). This means that at the point (2, -7), the tangent line to the curve has the same slope as the line connecting (1, -4) and (3, -8).
In simple words: The function meets the rules for Lagrange's Mean Value Theorem. We found a point where the slope of the curve is the same as the slope of the straight line connecting the two ends of the curve. This point is (2, -7).

🎯 Exam Tip: Always show that both continuity and differentiability conditions are met for the given interval before applying the Mean Value Theorem. Clearly state the value of 'c' and confirm it lies within the open interval.

Question 5. Examine the validity and conclusion of Rolle's Theorem for the f(x) = eˣ sin x in x ∈[0, π].
Answer: Given the function \( f(x) = e^x \sin x \) on the interval \( x \in [0, \pi] \).
First, we check the conditions for Rolle's Theorem:
1. **Continuity:** The exponential function \( e^x \) is continuous everywhere, and the sine function \( \sin x \) is continuous everywhere. The product of two continuous functions is also continuous. Therefore, \( f(x) = e^x \sin x \) is continuous on the closed interval \( [0, \pi] \).
2. **Differentiability:** The derivative of \( e^x \) is \( e^x \), and the derivative of \( \sin x \) is \( \cos x \). Both are differentiable everywhere. Using the product rule, the derivative of \( f(x) \) is:
\( f'(x) = e^x \cos x + \sin x \cdot e^x = e^x (\cos x + \sin x) \)
This derivative exists for all \( x \in (0, \pi) \), so \( f(x) \) is differentiable on the open interval \( (0, \pi) \).
3. **Function values at endpoints:** Evaluate the function at the endpoints of the interval:
\( f(0) = e^0 \sin(0) = 1 \cdot 0 = 0 \)
\( f(\pi) = e^\pi \sin(\pi) = e^\pi \cdot 0 = 0 \)
Since \( f(0) = f(\pi) = 0 \), all three conditions of Rolle's Theorem are satisfied.

**Conclusion of Rolle's Theorem:** Since all conditions are met, there must exist at least one real number \( c \in (0, \pi) \) such that \( f'(c) = 0 \).
We set the derivative to zero and solve for \( c \):
\( f'(c) = e^c (\cos c + \sin c) = 0 \)
Since \( e^c \) is always positive and never zero, we must have:
\( \cos c + \sin c = 0 \)
\( \sin c = -\cos c \)
\( \frac{\sin c}{\cos c} = -1 \)
\( \tan c = -1 \)
For \( c \in (0, \pi) \), the value of \( c \) where \( \tan c = -1 \) is \( c = \frac{3\pi}{4} \).
Since \( c = \frac{3\pi}{4} \) lies within the interval \( (0, \pi) \), Rolle's theorem is verified. This means there is a point on the curve where the tangent line is perfectly flat.
In simple words: The function \( e^x \sin x \) is smooth and has the same value at \( x=0 \) and \( x=\pi \). Because of this, we found a point \( x = \frac{3\pi}{4} \) between 0 and \( \pi \) where the curve's slope is exactly zero.

🎯 Exam Tip: When applying Rolle's Theorem, clearly state and justify each of the three conditions: continuity, differentiability, and \( f(a) = f(b) \). Then, show the calculation for \( f'(c) = 0 \) and confirm that the resulting 'c' value falls within the open interval.

Question 6. Verify Rolle's theorem for the function f(x) = e^(2x)(sin 2x − cos 2x) in \left[\frac{\pi}{8}, \frac{5 \pi}{8}\right].
Answer: Given the function \( f(x) = e^{2x} (\sin 2x - \cos 2x) \) on the interval \( \left[\frac{\pi}{8}, \frac{5 \pi}{8}\right] \).
Let's check the conditions for Rolle's Theorem:
1. **Continuity:** The exponential function \( e^{2x} \), \( \sin 2x \), and \( \cos 2x \) are all continuous everywhere. The difference and product of continuous functions are continuous. Therefore, \( f(x) \) is continuous on the closed interval \( \left[\frac{\pi}{8}, \frac{5 \pi}{8}\right] \).
2. **Differentiability:** We need to find the derivative \( f'(x) \) using the product rule: \( (uv)' = u'v + uv' \).
Let \( u = e^{2x} \Rightarrow u' = 2e^{2x} \)
Let \( v = \sin 2x - \cos 2x \Rightarrow v' = 2\cos 2x - (-2\sin 2x) = 2\cos 2x + 2\sin 2x \)
So, \( f'(x) = (2e^{2x})(\sin 2x - \cos 2x) + (e^{2x})(2\cos 2x + 2\sin 2x) \)
\( f'(x) = 2e^{2x}\sin 2x - 2e^{2x}\cos 2x + 2e^{2x}\cos 2x + 2e^{2x}\sin 2x \)
\( f'(x) = 4e^{2x}\sin 2x \)
This derivative exists for all \( x \) in the open interval \( \left(\frac{\pi}{8}, \frac{5 \pi}{8}\right) \). So, \( f(x) \) is differentiable.
3. **Function values at endpoints:** Evaluate \( f(x) \) at \( x = \frac{\pi}{8} \) and \( x = \frac{5\pi}{8} \).
For \( x = \frac{\pi}{8} \):
\( f\left(\frac{\pi}{8}\right) = e^{2(\frac{\pi}{8})} \left(\sin\left(2\cdot\frac{\pi}{8}\right) - \cos\left(2\cdot\frac{\pi}{8}\right)\right) \)
\( f\left(\frac{\pi}{8}\right) = e^{\frac{\pi}{4}} \left(\sin\left(\frac{\pi}{4}\right) - \cos\left(\frac{\pi}{4}\right)\right) \)
\( f\left(\frac{\pi}{8}\right) = e^{\frac{\pi}{4}} \left(\frac{1}{\sqrt{2}} - \frac{1}{\sqrt{2}}\right) = e^{\frac{\pi}{4}} \cdot 0 = 0 \)
For \( x = \frac{5\pi}{8} \):
\( f\left(\frac{5\pi}{8}\right) = e^{2(\frac{5\pi}{8})} \left(\sin\left(2\cdot\frac{5\pi}{8}\right) - \cos\left(2\cdot\frac{5\pi}{8}\right)\right) \)
\( f\left(\frac{5\pi}{8}\right) = e^{\frac{5\pi}{4}} \left(\sin\left(\frac{5\pi}{4}\right) - \cos\left(\frac{5\pi}{4}\right)\right) \)
\( f\left(\frac{5\pi}{8}\right) = e^{\frac{5\pi}{4}} \left(-\frac{1}{\sqrt{2}} - \left(-\frac{1}{\sqrt{2}}\right)\right) = e^{\frac{5\pi}{4}} \left(-\frac{1}{\sqrt{2}} + \frac{1}{\sqrt{2}}\right) = e^{\frac{5\pi}{4}} \cdot 0 = 0 \)
Since \( f\left(\frac{\pi}{8}\right) = f\left(\frac{5\pi}{8}\right) = 0 \), all three conditions of Rolle's Theorem are satisfied.

**Conclusion of Rolle's Theorem:** There must exist at least one real number \( c \in \left(\frac{\pi}{8}, \frac{5 \pi}{8}\right) \) such that \( f'(c) = 0 \).
Set \( f'(c) = 0 \):
\( 4e^{2c}\sin 2c = 0 \)
Since \( 4e^{2c} \) is never zero, we must have \( \sin 2c = 0 \).
This means \( 2c = n\pi \) for some integer \( n \).
\( c = \frac{n\pi}{2} \)
We need \( c \) to be in the interval \( \left(\frac{\pi}{8}, \frac{5 \pi}{8}\right) \).
If \( n=1 \), \( c = \frac{\pi}{2} \). Let's check if \( \frac{\pi}{8} < \frac{\pi}{2} < \frac{5\pi}{8} \).
\( 0.125\pi < 0.5\pi < 0.625\pi \). This is true.
So, \( c = \frac{\pi}{2} \) is a value in the given interval that satisfies the theorem. The derivative is zero at this point. This shows that the theorem holds true for this function and interval.
In simple words: The function for this problem is smooth and has the same value at both ends of the given range. So, Rolle's Theorem says there must be a point in between where the curve's slope is flat, which we found at \( x = \frac{\pi}{2} \).

🎯 Exam Tip: For trigonometric functions, carefully evaluate the function at the endpoints to ensure \( f(a) = f(b) \). When solving \( f'(c) = 0 \), remember to consider all possible values of 'c' within the specific open interval, especially for periodic functions.

Question 7. Verify Lagrange's Mean Value Theorem for the function f(x) = 3x² – 5x + 1 defined in the interval [2, 5].
Answer: Given the function \( f(x) = 3x^2 - 5x + 1 \) on the interval \( [2, 5] \).
Let's check the conditions for Lagrange's Mean Value Theorem:
1. **Continuity:** The function \( f(x) = 3x^2 - 5x + 1 \) is a polynomial. Polynomial functions are continuous everywhere. Therefore, \( f(x) \) is continuous on the closed interval \( [2, 5] \).
2. **Differentiability:** The derivative of \( f(x) \) is \( f'(x) = 6x - 5 \). This derivative exists for all real numbers. Therefore, \( f(x) \) is differentiable on the open interval \( (2, 5) \).
Since both conditions are satisfied, Lagrange's Mean Value Theorem applies.

**Conclusion of Lagrange's Mean Value Theorem:** There must exist at least one real number \( c \in (2, 5) \) such that \( f'(c) = \frac{f(b) - f(a)}{b - a} \).
First, calculate \( f(a) \) and \( f(b) \):
\( f(2) = 3(2)^2 - 5(2) + 1 = 3(4) - 10 + 1 = 12 - 10 + 1 = 3 \)
\( f(5) = 3(5)^2 - 5(5) + 1 = 3(25) - 25 + 1 = 75 - 25 + 1 = 51 \)
Now, apply the theorem formula:
\( f'(c) = \frac{f(5) - f(2)}{5 - 2} \)
\( 6c - 5 = \frac{51 - 3}{3} \)
\( 6c - 5 = \frac{48}{3} \)
\( 6c - 5 = 16 \)
\( 6c = 16 + 5 \)
\( 6c = 21 \)
\( c = \frac{21}{6} \)
\( c = \frac{7}{2} = 3.5 \)
Since \( c = 3.5 \) lies within the interval \( (2, 5) \), Lagrange's Mean Value Theorem is verified. This means there's a point on the curve where the slope of the tangent matches the average slope of the function over the entire interval.
In simple words: The function is smooth and continuous. So, we can find a point in the middle of the interval where the curve's slope is the same as the slope of the straight line connecting the points at the ends of the interval. We found this point at \( x=3.5 \).

🎯 Exam Tip: For polynomial functions, continuity and differentiability are usually easy to confirm. The main steps are correctly calculating \( f(a) \), \( f(b) \), and \( f'(x) \), then solving for 'c' and verifying its interval. Remember that Mean Value Theorem applies to a wider range of functions than Rolle's Theorem, not requiring \( f(a) = f(b) \).

Question 8. Use Lagrange's Mean Value Theorem to determine a point P on the curve y = \sqrt{x-2} defined in the interval [2, 3] where the tangent is parallel to the chord joining the end points on the curve.
Answer: Given the function \( f(x) = \sqrt{x-2} \) on the interval \( [2, 3] \).
Let's check the conditions for Lagrange's Mean Value Theorem:
1. **Continuity:** The square root function \( \sqrt{x-2} \) is defined and continuous when \( x-2 \ge 0 \), i.e., \( x \ge 2 \). Thus, it is continuous on the closed interval \( [2, 3] \).
2. **Differentiability:** The derivative of \( f(x) \) is:
\( f'(x) = \frac{d}{dx}(x-2)^{1/2} = \frac{1}{2}(x-2)^{-1/2} \cdot 1 = \frac{1}{2\sqrt{x-2}} \)
This derivative exists for \( x-2 > 0 \), i.e., \( x > 2 \). Thus, \( f(x) \) is differentiable on the open interval \( (2, 3) \).
Since both conditions are satisfied, Lagrange's Mean Value Theorem applies.

**Conclusion of Lagrange's Mean Value Theorem:** There exists at least one real number \( c \in (2, 3) \) such that \( f'(c) = \frac{f(b) - f(a)}{b - a} \).
First, calculate \( f(a) \) and \( f(b) \):
\( f(2) = \sqrt{2-2} = \sqrt{0} = 0 \)
\( f(3) = \sqrt{3-2} = \sqrt{1} = 1 \)
Now, apply the theorem formula:
\( f'(c) = \frac{f(3) - f(2)}{3 - 2} \)
\( \frac{1}{2\sqrt{c-2}} = \frac{1 - 0}{1} \)
\( \frac{1}{2\sqrt{c-2}} = 1 \)
\( 1 = 2\sqrt{c-2} \)
\( \sqrt{c-2} = \frac{1}{2} \)
Square both sides:
\( c-2 = \left(\frac{1}{2}\right)^2 \)
\( c-2 = \frac{1}{4} \)
\( c = 2 + \frac{1}{4} \)
\( c = \frac{8}{4} + \frac{1}{4} \)
\( c = \frac{9}{4} \)
Since \( c = \frac{9}{4} = 2.25 \) lies within the interval \( (2, 3) \), the theorem is verified. This specific value of 'c' tells us the x-coordinate where the tangent is parallel to the chord.
To find the y-coordinate of point P, substitute \( c = \frac{9}{4} \) back into the original function:
\( f\left(\frac{9}{4}\right) = \sqrt{\frac{9}{4} - 2} = \sqrt{\frac{9}{4} - \frac{8}{4}} = \sqrt{\frac{1}{4}} = \frac{1}{2} \)
Therefore, the point P on the curve where the tangent is parallel to the chord joining the endpoints is \( \left(\frac{9}{4}, \frac{1}{2}\right) \).
In simple words: The function \( y = \sqrt{x-2} \) is continuous and smooth. We found a point on this curve, \( \left(\frac{9}{4}, \frac{1}{2}\right) \), where its tangent line runs parallel to the straight line connecting the start and end of the curve over the given interval.

🎯 Exam Tip: When dealing with square root functions, pay close attention to the domain for continuity and differentiability. Always confirm that the value of 'c' you find is strictly within the open interval for differentiability.

Question 10. Using Rolle's Theorem, find a point on the curve sin x + cos x − 1 = 0, x ∈ \left[0, \frac{\pi}{2}\right] where the tangent is parallel to the x-axis.
Answer: Let the function be \( f(x) = \sin x + \cos x - 1 \) on the interval \( \left[0, \frac{\pi}{2}\right] \). A tangent parallel to the x-axis means the derivative is zero, which is the conclusion of Rolle's Theorem.
Let's check the conditions for Rolle's Theorem:
1. **Continuity:** The functions \( \sin x \) and \( \cos x \) are continuous everywhere. The sum of continuous functions is continuous. Therefore, \( f(x) \) is continuous on the closed interval \( \left[0, \frac{\pi}{2}\right] \).
2. **Differentiability:** The derivative of \( f(x) \) is:
\( f'(x) = \cos x - \sin x \)
This derivative exists for all real numbers. Therefore, \( f(x) \) is differentiable on the open interval \( \left(0, \frac{\pi}{2}\right) \).
3. **Function values at endpoints:** Evaluate \( f(x) \) at the endpoints:
\( f(0) = \sin(0) + \cos(0) - 1 = 0 + 1 - 1 = 0 \)
\( f\left(\frac{\pi}{2}\right) = \sin\left(\frac{\pi}{2}\right) + \cos\left(\frac{\pi}{2}\right) - 1 = 1 + 0 - 1 = 0 \)
Since \( f(0) = f\left(\frac{\pi}{2}\right) = 0 \), all three conditions of Rolle's Theorem are satisfied.

**Conclusion of Rolle's Theorem:** There exists at least one real number \( c \in \left(0, \frac{\pi}{2}\right) \) such that \( f'(c) = 0 \).
Set the derivative to zero and solve for \( c \):
\( \cos c - \sin c = 0 \)
\( \cos c = \sin c \)
Since \( \cos c \neq 0 \) in this interval, we can divide by \( \cos c \):
\( \frac{\sin c}{\cos c} = 1 \)
\( \tan c = 1 \)
For \( c \in \left(0, \frac{\pi}{2}\right) \), the value of \( c \) where \( \tan c = 1 \) is \( c = \frac{\pi}{4} \).
Since \( c = \frac{\pi}{4} \) lies within the interval \( \left(0, \frac{\pi}{2}\right) \), the theorem is verified. This point represents where the tangent line is horizontal.
To find the y-coordinate of this point, substitute \( c = \frac{\pi}{4} \) back into the original function \( f(x) \):
\( f\left(\frac{\pi}{4}\right) = \sin\left(\frac{\pi}{4}\right) + \cos\left(\frac{\pi}{4}\right) - 1 = \frac{1}{\sqrt{2}} + \frac{1}{\sqrt{2}} - 1 = \frac{2}{\sqrt{2}} - 1 = \sqrt{2} - 1 \)
Therefore, the point on the curve where the tangent is parallel to the x-axis is \( \left(\frac{\pi}{4}, \sqrt{2} - 1\right) \). This point marks a local maximum or minimum of the function.
In simple words: For the given function, all conditions of Rolle's Theorem are met. This means there's a point within the interval where the curve's slope is zero, making the tangent line flat. That point is \( \left(\frac{\pi}{4}, \sqrt{2} - 1\right) \).

🎯 Exam Tip: For problems involving Rolle's Theorem and trigonometric functions, it's crucial to correctly identify the interval for 'c' and solve the trigonometric equation \( f'(c) = 0 \) within that specific range. Don't forget to find the y-coordinate to provide the complete point.

Question 11. Verify Lagrange's mean value theorem for the function f (x) = sin x − sin 2x in the interval [0, π].
Answer: Given the function \( f(x) = \sin x - \sin 2x \) on the interval \( [0, \pi] \).
Let's check the conditions for Lagrange's Mean Value Theorem:
1. **Continuity:** The functions \( \sin x \) and \( \sin 2x \) are continuous everywhere. The difference of continuous functions is continuous. Therefore, \( f(x) \) is continuous on the closed interval \( [0, \pi] \).
2. **Differentiability:** The derivative of \( f(x) \) is:
\( f'(x) = \cos x - 2\cos 2x \)
This derivative exists for all real numbers. Therefore, \( f(x) \) is differentiable on the open interval \( (0, \pi) \).
Since both conditions are satisfied, Lagrange's Mean Value Theorem applies.

**Conclusion of Lagrange's Mean Value Theorem:** There exists at least one real number \( c \in (0, \pi) \) such that \( f'(c) = \frac{f(b) - f(a)}{b - a} \).
First, calculate \( f(a) \) and \( f(b) \):
\( f(0) = \sin(0) - \sin(2 \cdot 0) = 0 - 0 = 0 \)
\( f(\pi) = \sin(\pi) - \sin(2\pi) = 0 - 0 = 0 \)
Now, apply the theorem formula:
\( f'(c) = \frac{f(\pi) - f(0)}{\pi - 0} \)
\( \cos c - 2\cos 2c = \frac{0 - 0}{\pi} \)
\( \cos c - 2\cos 2c = 0 \)
Use the double angle identity \( \cos 2c = 2\cos^2 c - 1 \):
\( \cos c - 2(2\cos^2 c - 1) = 0 \)
\( \cos c - 4\cos^2 c + 2 = 0 \)
Rearrange into a quadratic equation in terms of \( \cos c \):
\( 4\cos^2 c - \cos c - 2 = 0 \)
Let \( y = \cos c \). Then \( 4y^2 - y - 2 = 0 \). Use the quadratic formula \( y = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \):
\( y = \frac{-(-1) \pm \sqrt{(-1)^2 - 4(4)(-2)}}{2(4)} \)
\( y = \frac{1 \pm \sqrt{1 + 32}}{8} \)
\( y = \frac{1 \pm \sqrt{33}}{8} \)
So, \( \cos c = \frac{1 + \sqrt{33}}{8} \) or \( \cos c = \frac{1 - \sqrt{33}}{8} \).
Approximate values:
\( \sqrt{33} \approx 5.74 \)
\( \cos c \approx \frac{1 + 5.74}{8} = \frac{6.74}{8} \approx 0.8425 \)
\( \cos c \approx \frac{1 - 5.74}{8} = \frac{-4.74}{8} \approx -0.5925 \)
Since \( c \in (0, \pi) \), \( \cos c \) can range from -1 to 1. Both \( 0.8425 \) and \( -0.5925 \) are within this range. Thus, there exist two values of \( c \) in \( (0, \pi) \) that satisfy \( \cos c = 0.8425 \) and \( \cos c = -0.5925 \). The Lagrange Mean Value Theorem is verified because at least one such 'c' exists.
In simple words: The function is smooth and continuous. Since it satisfies the conditions of Lagrange's Mean Value Theorem, we found that there are two points in the interval \( (0, \pi) \) where the curve's slope is zero. This happens because the starting and ending values of the function are the same, which means the average slope over the interval is zero.

🎯 Exam Tip: When solving for 'c' in trigonometric equations, ensure you consider all possible solutions within the specified open interval. The use of double angle identities is common in such problems, so be prepared to apply them.

Question 12. Verify Rolle's theorem for f(x) = log[(x² + ab) /(a + b)x] in [a b]; 0 ∉ [a, b].
Answer: Given the function \( f(x) = \log\left(\frac{x^2+ab}{x(a+b)}\right) \) on the interval \( [a, b] \), where \( 0 \notin [a, b] \).
We can rewrite \( f(x) \) using logarithm properties:
\( f(x) = \log(x^2 + ab) - \log(x) - \log(a+b) \)
Let's check the conditions for Rolle's Theorem:
1. **Continuity:** Logarithmic functions are continuous in their domain. For \( f(x) \) to be defined and continuous on \( [a, b] \):
- \( x^2+ab \) must be positive. Since \( 0 \notin [a, b] \), \( x \) is either always positive or always negative. \( x^2 \) is always positive. If \( a \) and \( b \) have the same sign (which they must, as 0 is not between them), then \( ab \) is positive. So \( x^2+ab \) is always positive.
- \( x \) must be positive for \( \log(x) \). This implies that \( [a, b] \) must be a subset of \( (0, \infty) \).
Assuming \( a, b > 0 \), then \( f(x) \) is continuous on \( [a, b] \).
2. **Differentiability:** Find the derivative \( f'(x) \):
\( f'(x) = \frac{2x}{x^2+ab} - \frac{1}{x} - 0 \) (since \( \log(a+b) \) is a constant)
\( f'(x) = \frac{2x^2 - (x^2+ab)}{x(x^2+ab)} = \frac{x^2-ab}{x(x^2+ab)} \)
This derivative exists for all \( x \in (a, b) \) provided \( x \neq 0 \) and \( x^2+ab \neq 0 \), which are true given \( 0 \notin [a, b] \) and \( x^2+ab > 0 \). So, \( f(x) \) is differentiable on \( (a, b) \).
3. **Function values at endpoints:** Evaluate \( f(x) \) at \( x=a \) and \( x=b \).
\( f(a) = \log\left(\frac{a^2+ab}{a(a+b)}\right) = \log\left(\frac{a(a+b)}{a(a+b)}\right) = \log(1) = 0 \)
\( f(b) = \log\left(\frac{b^2+ab}{b(a+b)}\right) = \log\left(\frac{b(b+a)}{b(a+b)}\right) = \log(1) = 0 \)
Since \( f(a) = f(b) = 0 \), all three conditions of Rolle's Theorem are satisfied.

**Conclusion of Rolle's Theorem:** There exists at least one real number \( c \in (a, b) \) such that \( f'(c) = 0 \).
Set the derivative to zero and solve for \( c \):
\( \frac{c^2-ab}{c(c^2+ab)} = 0 \)
For this fraction to be zero, the numerator must be zero:
\( c^2 - ab = 0 \)
\( c^2 = ab \)
\( c = \pm \sqrt{ab} \)
Since \( 0 \notin [a, b] \), \( a \) and \( b \) must have the same sign. The interval \( [a, b] \) also implies \( a, b > 0 \) for \( \log x \) to be defined. If \( a, b > 0 \), then \( \sqrt{ab} \) is a positive real number. The geometric mean \( \sqrt{ab} \) always lies between \( a \) and \( b \) (for \( a, b > 0 \)).
Thus, \( c = \sqrt{ab} \) is in the interval \( (a, b) \). The negative value \( -\sqrt{ab} \) would not be in \( (a, b) \) because \( a, b \) are positive. The theorem is verified. This point \( c = \sqrt{ab} \) is where the function's rate of change becomes zero.
In simple words: The function \( f(x) \) is continuous and smooth, and its value is zero at both ends of the interval \( [a, b] \). Because of this, Rolle's Theorem tells us there's a point inside this interval, specifically \( c = \sqrt{ab} \), where the curve becomes flat (its slope is zero).

🎯 Exam Tip: When dealing with logarithmic functions, always simplify the expression using log properties before differentiating. Crucially, ensure that the domain of the logarithm is strictly positive throughout the interval, often by verifying that 0 is not included in or between the endpoints.

Question 13. Verify Lagranges' Mean Value theorem for the function f(x) = \sqrt{x^2-x} in the interval [1, 4].
Answer: Given the function \( f(x) = \sqrt{x^2-x} \) on the interval \( [1, 4] \).
We can rewrite \( f(x) = \sqrt{x(x-1)} \).
Let's check the conditions for Lagrange's Mean Value Theorem:
1. **Continuity:** For \( f(x) \) to be defined and continuous, \( x^2-x \ge 0 \). This means \( x(x-1) \ge 0 \), which holds for \( x \le 0 \) or \( x \ge 1 \). Since our interval is \( [1, 4] \), \( x^2-x \) is non-negative throughout the interval, and the square root function is continuous for non-negative values. Thus, \( f(x) \) is continuous on the closed interval \( [1, 4] \).
2. **Differentiability:** Find the derivative \( f'(x) \):
\( f'(x) = \frac{d}{dx}(x^2-x)^{1/2} = \frac{1}{2}(x^2-x)^{-1/2} \cdot (2x-1) = \frac{2x-1}{2\sqrt{x^2-x}} \)
This derivative exists for \( x^2-x > 0 \), i.e., \( x(x-1) > 0 \). This holds for \( x < 0 \) or \( x > 1 \). In our interval \( [1, 4] \), the function is differentiable on the open interval \( (1, 4) \). (Note: it is not differentiable at \( x=1 \) where the denominator would be zero, but differentiability on the *open* interval is what is required).
Since both conditions are satisfied, Lagrange's Mean Value Theorem applies.

**Conclusion of Lagrange's Mean Value Theorem:** There exists at least one real number \( c \in (1, 4) \) such that \( f'(c) = \frac{f(b) - f(a)}{b - a} \).
First, calculate \( f(a) \) and \( f(b) \):
\( f(1) = \sqrt{1^2-1} = \sqrt{1-1} = \sqrt{0} = 0 \)
\( f(4) = \sqrt{4^2-4} = \sqrt{16-4} = \sqrt{12} \)
Now, apply the theorem formula:
\( f'(c) = \frac{f(4) - f(1)}{4 - 1} \)
\( \frac{2c-1}{2\sqrt{c^2-c}} = \frac{\sqrt{12} - 0}{3} \)
\( \frac{2c-1}{2\sqrt{c^2-c}} = \frac{2\sqrt{3}}{3} \)
Square both sides to solve for \( c \):
\( \left(\frac{2c-1}{2\sqrt{c^2-c}}\right)^2 = \left(\frac{2\sqrt{3}}{3}\right)^2 \)
\( \frac{(2c-1)^2}{4(c^2-c)} = \frac{4 \cdot 3}{9} \)
\( \frac{(2c-1)^2}{4(c^2-c)} = \frac{12}{9} = \frac{4}{3} \)
\( 3(2c-1)^2 = 16(c^2-c) \)
\( 3(4c^2 - 4c + 1) = 16c^2 - 16c \)
\( 12c^2 - 12c + 3 = 16c^2 - 16c \)
Rearrange into a quadratic equation:
\( 0 = 16c^2 - 12c^2 - 16c + 12c - 3 \)
\( 0 = 4c^2 - 4c - 3 \)
Factor the quadratic:
\( 4c^2 - 6c + 2c - 3 = 0 \)
\( 2c(2c-3) + 1(2c-3) = 0 \)
\( (2c+1)(2c-3) = 0 \)
This gives two possible values for \( c \):
\( 2c+1 = 0 \Rightarrow c = -\frac{1}{2} \)
\( 2c-3 = 0 \Rightarrow c = \frac{3}{2} \)
We need \( c \in (1, 4) \).
The value \( c = -\frac{1}{2} \) is not in the interval \( (1, 4) \).
The value \( c = \frac{3}{2} = 1.5 \) is in the interval \( (1, 4) \).
Therefore, \( c = \frac{3}{2} \) verifies Lagrange's Mean Value Theorem. This means at this specific point, the instantaneous rate of change of the function equals its average rate of change over the interval.
In simple words: The function \( f(x) = \sqrt{x^2-x} \) is continuous and smooth enough for the theorem to apply. We found a point \( x = 1.5 \) within the interval where the slope of the curve is the same as the average slope of the line connecting the start and end points of the curve.

🎯 Exam Tip: Be careful when squaring both sides of an equation, as it can introduce extraneous solutions. Always check if the obtained values of 'c' actually lie within the required open interval and satisfy the original equation (if there were any domain restrictions due to squaring).

Question 14. Verify Rolle's theorem for the function f(x) = e^(sin x − cos x) on \left[\frac{\pi}{4}, \frac{5 \pi}{4}\right].
Answer: Given the function \( f(x) = e^{\sin x - \cos x} \) on the interval \( \left[\frac{\pi}{4}, \frac{5 \pi}{4}\right] \).
Let's check the conditions for Rolle's Theorem:
1. **Continuity:** The exponential function \( e^u \) is continuous everywhere, and the functions \( \sin x \) and \( \cos x \) are continuous everywhere. The difference of continuous functions is continuous. Therefore, the exponent \( \sin x - \cos x \) is continuous, and thus \( f(x) \) is continuous on the closed interval \( \left[\frac{\pi}{4}, \frac{5 \pi}{4}\right] \).
2. **Differentiability:** Find the derivative \( f'(x) \) using the chain rule: \( (e^u)' = e^u \cdot u' \).
The derivative of the exponent is \( \frac{d}{dx}(\sin x - \cos x) = \cos x - (-\sin x) = \cos x + \sin x \).
So, \( f'(x) = e^{\sin x - \cos x} (\cos x + \sin x) \)
This derivative exists for all real numbers. Therefore, \( f(x) \) is differentiable on the open interval \( \left(\frac{\pi}{4}, \frac{5 \pi}{4}\right) \).
3. **Function values at endpoints:** Evaluate \( f(x) \) at the endpoints:
For \( x = \frac{\pi}{4} \):
\( f\left(\frac{\pi}{4}\right) = e^{\sin(\frac{\pi}{4}) - \cos(\frac{\pi}{4})} = e^{\frac{1}{\sqrt{2}} - \frac{1}{\sqrt{2}}} = e^0 = 1 \)
For \( x = \frac{5\pi}{4} \):
\( f\left(\frac{5\pi}{4}\right) = e^{\sin(\frac{5\pi}{4}) - \cos(\frac{5\pi}{4})} = e^{-\frac{1}{\sqrt{2}} - (-\frac{1}{\sqrt{2}})} = e^{-\frac{1}{\sqrt{2}} + \frac{1}{\sqrt{2}}} = e^0 = 1 \)
Since \( f\left(\frac{\pi}{4}\right) = f\left(\frac{5\pi}{4}\right) = 1 \), all three conditions of Rolle's Theorem are satisfied.

**Conclusion of Rolle's Theorem:** There exists at least one real number \( c \in \left(\frac{\pi}{4}, \frac{5 \pi}{4}\right) \) such that \( f'(c) = 0 \).
Set the derivative to zero and solve for \( c \):
\( e^{\sin c - \cos c} (\cos c + \sin c) = 0 \)
Since \( e^{\sin c - \cos c} \) is always positive and never zero, we must have:
\( \cos c + \sin c = 0 \)
\( \sin c = -\cos c \)
\( \frac{\sin c}{\cos c} = -1 \)
\( \tan c = -1 \)
For \( c \in \left(\frac{\pi}{4}, \frac{5 \pi}{4}\right) \), the value of \( c \) where \( \tan c = -1 \) is \( c = \frac{3\pi}{4} \).
Let's check if \( \frac{\pi}{4} < \frac{3\pi}{4} < \frac{5\pi}{4} \). This is true. The interval is from 45 degrees to 225 degrees, and 135 degrees is in between. So, \( c = \frac{3\pi}{4} \) lies within the interval \( \left(\frac{\pi}{4}, \frac{5 \pi}{4}\right) \). Rolle's theorem is verified, showing a point where the tangent is horizontal.
In simple words: This function is smooth and has the same value at the start and end of the interval. Therefore, Rolle's Theorem guarantees a point in the middle, \( c = \frac{3\pi}{4} \), where the curve's slope is exactly zero, making the tangent line flat.

🎯 Exam Tip: When differentiating exponential functions with complex exponents, remember the chain rule: \( (e^{g(x)})' = e^{g(x)} \cdot g'(x) \). Double-check trigonometric values at critical angles for endpoint evaluations and solving for 'c'.

Question 15. Verify Lagranges' Mean Value Theorem for the function f(x) = 2 sin x + sin 2x in the interval [0, π].
Answer: Given the function \( f(x) = 2\sin x + \sin 2x \) on the interval \( [0, \pi] \).
Let's check the conditions for Lagrange's Mean Value Theorem:
1. **Continuity:** The functions \( \sin x \) and \( \sin 2x \) are continuous everywhere. The sum of continuous functions is continuous. Therefore, \( f(x) \) is continuous on the closed interval \( [0, \pi] \).
2. **Differentiability:** Find the derivative \( f'(x) \):
\( f'(x) = 2\cos x + 2\cos 2x \)
This derivative exists for all real numbers. Therefore, \( f(x) \) is differentiable on the open interval \( (0, \pi) \).
Since both conditions are satisfied, Lagrange's Mean Value Theorem applies.

**Conclusion of Lagrange's Mean Value Theorem:** There exists at least one real number \( c \in (0, \pi) \) such that \( f'(c) = \frac{f(b) - f(a)}{b - a} \).
First, calculate \( f(a) \) and \( f(b) \):
\( f(0) = 2\sin(0) + \sin(2 \cdot 0) = 2(0) + 0 = 0 \)
\( f(\pi) = 2\sin(\pi) + \sin(2\pi) = 2(0) + 0 = 0 \)
Now, apply the theorem formula:
\( f'(c) = \frac{f(\pi) - f(0)}{\pi - 0} \)
\( 2\cos c + 2\cos 2c = \frac{0 - 0}{\pi} \)
\( 2\cos c + 2\cos 2c = 0 \)
Divide by 2:
\( \cos c + \cos 2c = 0 \)
Use the double angle identity \( \cos 2c = 2\cos^2 c - 1 \):
\( \cos c + (2\cos^2 c - 1) = 0 \)
\( 2\cos^2 c + \cos c - 1 = 0 \)
Let \( y = \cos c \). Then \( 2y^2 + y - 1 = 0 \). Factor the quadratic:
\( 2y^2 + 2y - y - 1 = 0 \)
\( 2y(y+1) - 1(y+1) = 0 \)
\( (2y-1)(y+1) = 0 \)
This gives two possible values for \( y \):
\( 2y-1 = 0 \Rightarrow y = \frac{1}{2} \Rightarrow \cos c = \frac{1}{2} \)
\( y+1 = 0 \Rightarrow y = -1 \Rightarrow \cos c = -1 \)
Now find the values of \( c \) in the interval \( (0, \pi) \):
If \( \cos c = \frac{1}{2} \), then \( c = \frac{\pi}{3} \). This is in \( (0, \pi) \).
If \( \cos c = -1 \), then \( c = \pi \). This is an endpoint and not strictly in the open interval \( (0, \pi) \).
Therefore, \( c = \frac{\pi}{3} \) is a value in \( (0, \pi) \) that verifies Lagrange's Mean Value Theorem. This point indicates where the instantaneous rate of change matches the average rate of change over the interval. The theorem is verified because at least one such 'c' exists.
In simple words: This function is continuous and smooth. Since the function values at both ends of the interval are the same, the average slope is zero. We found that at \( c = \frac{\pi}{3} \), the curve's actual slope is also zero, which proves Lagrange's Theorem for this case.

🎯 Exam Tip: When applying Lagrange's Mean Value Theorem, always check if the found 'c' values fall strictly within the *open* interval \((a, b)\). Solutions that match the endpoints \(a\) or \(b\) are not valid for the theorem's conclusion regarding differentiability.

Question 16. Verify the conditions of Rolle's Theorem for the following function : f(x) = log (x² + 2) – log 3 on [- 1, 1].
Answer: Given the function \( f(x) = \log(x^2 + 2) - \log 3 \) on the interval \( [-1, 1] \).
Let's check the conditions for Rolle's Theorem:
1. **Continuity:** The function \( g(x) = x^2+2 \) is a polynomial, so it's continuous everywhere. Since \( x^2+2 \) is always greater than 0 (specifically, \( x^2+2 \ge 2 \) for all real \( x \)), \( \log(x^2+2) \) is defined and continuous everywhere. \( \log 3 \) is a constant, hence continuous. The difference of continuous functions is continuous. Therefore, \( f(x) \) is continuous on the closed interval \( [-1, 1] \).
2. **Differentiability:** Find the derivative \( f'(x) \):
\( f'(x) = \frac{d}{dx}(\log(x^2+2) - \log 3) \)
\( f'(x) = \frac{1}{x^2+2} \cdot (2x) - 0 \)
\( f'(x) = \frac{2x}{x^2+2} \)
This derivative exists for all real numbers since the denominator \( x^2+2 \) is never zero. Therefore, \( f(x) \) is differentiable on the open interval \( (-1, 1) \).
3. **Function values at endpoints:** Evaluate \( f(x) \) at the endpoints:
\( f(-1) = \log((-1)^2 + 2) - \log 3 = \log(1+2) - \log 3 = \log 3 - \log 3 = 0 \)
\( f(1) = \log(1^2 + 2) - \log 3 = \log(1+2) - \log 3 = \log 3 - \log 3 = 0 \)
Since \( f(-1) = f(1) = 0 \), all three conditions of Rolle's Theorem are satisfied.

**Conclusion of Rolle's Theorem:** There exists at least one real number \( c \in (-1, 1) \) such that \( f'(c) = 0 \).
Set the derivative to zero and solve for \( c \):
\( \frac{2c}{c^2+2} = 0 \)
For this fraction to be zero, the numerator must be zero:
\( 2c = 0 \)
\( c = 0 \)
Since \( c = 0 \) lies within the interval \( (-1, 1) \), Rolle's Theorem is verified. This point is where the tangent to the curve is perfectly horizontal.
In simple words: This function is smooth and continuous, and its value is zero at both ends of the interval. So, Rolle's Theorem confirms that there is a point in the middle, at \( x=0 \), where the curve's slope is flat or horizontal.

🎯 Exam Tip: For logarithmic functions, always ensure the argument of the logarithm is strictly positive to confirm continuity. Polynomial denominators (like \( x^2+2 \)) that are always positive ensure differentiability over the entire real line.