Ex 5.15 – Continuity and Differentiability | ML Aggarwal Class 12 Solutions

Solution:

Step 1: Check the conditions for LMVT. Since \( f(x)=x^{2} \) is a polynomial, it is continuous on the closed interval \([2,4]\) and differentiable on the open interval \((2,4)\).
Since both conditions (continuity on \([2,4]\) and differentiability on \((2,4)\)) are satisfied, Lagrange’s mean value theorem (LMVT) applies to \( f(x) \) on the interval \([2,4]\).

Step 2: By LMVT, there exists a number \( c \in (2,4) \) such that \[ f'(c)=\frac{f(4)-f(2)}{4-2}. \]

Step 3: Compute the function values: \[ f(4)=4^{2}=16, \quad f(2)=2^{2}=4. \] Hence, the slope of the secant is: \[ \frac{f(4)-f(2)}{4-2}=\frac{16-4}{2}=\frac{12}{2}=6. \]

Step 4: Differentiate \( f(x)=x^2 \): \[ f'(x)=2x. \] Set \( f'(c)=2c \) equal to the secant slope: \[ 2c=6. \] Solving for \( c \): \[ c=\frac{6}{2}=3. \]

Thus, the value \( c=3 \) satisfies LMVT for the given function on \([2,4]\).

Solution:

Step 1: Check the conditions for LMVT. Since \( f(x)=x^{2}-4x-3 \) is a polynomial, it is continuous on the closed interval \([1,4]\) and differentiable on the open interval \((1,4)\).
Since both conditions (continuity on \([1,4]\) and differentiability on \((1,4)\)) are satisfied, Lagrange’s mean value theorem (LMVT) applies to \( f(x) \) on the interval \([1,4]\).

Step 2: By LMVT, there exists a number \( c \in (1,4) \) such that \[ f'(c)=\frac{f(4)-f(1)}{4-1}. \]

Step 3: Compute the function values: \[ f(4)=4^2-4\cdot4-3=16-16-3=-3, \] \[ f(1)=1^2-4\cdot1-3=1-4-3=-6. \] So, the slope of the secant is: \[ \frac{f(4)-f(1)}{4-1}=\frac{-3-(-6)}{3}=\frac{3}{3}=1. \]

Step 4: Differentiate \( f(x)=x^{2}-4x-3 \): \[ f'(x)=2x-4. \] Equate this to the secant slope: \[ 2c-4=1. \] Solving, we get: \[ 2c=5 \quad \Rightarrow \quad c=\frac{5}{2}. \]

Thus, the value \( c=\frac{5}{2} \) satisfies LMVT for the given function on \([1,4]\).

Solution:

Step 1: Check the conditions for LMVT. Since \( f(x)=x^{2}-2x+4 \) is a polynomial, it is continuous on the closed interval \([1,5]\) and differentiable on the open interval \((1,5)\).
Since both conditions (continuity on \([1,5]\) and differentiability on \((1,5)\)) are satisfied, Lagrange’s mean value theorem (LMVT) applies to \( f(x) \) on the interval \([1,5]\).

Step 2: By LMVT, there exists a number \( c \in (1,5) \) such that \[ f'(c)=\frac{f(5)-f(1)}{5-1}. \]

Step 3: Compute the function values: \[ f(5)=5^2-2\cdot5+4=25-10+4=19, \quad f(1)=1^2-2\cdot1+4=1-2+4=3. \] Hence, the slope of the secant is: \[ \frac{f(5)-f(1)}{5-1}=\frac{19-3}{4}=\frac{16}{4}=4. \]

Step 4: Differentiate \( f(x)=x^{2}-2x+4 \): \[ f'(x)=2x-2. \] Setting the derivative equal to the secant slope: \[ 2c-2=4. \] Solve for \( c \): \[ 2c=4+2=6,\quad c=\frac{6}{2}=3. \]

Thus, the value \( c=3 \) satisfies LMVT for the given function on the interval \([1,5]\).

Solution:

Step 1: Check the conditions for LMVT. Since \( f(x)=x^{3}-2x^{2}-x+3 \) is a polynomial, it is continuous on the closed interval \([0,1]\) and differentiable on the open interval \((0,1)\).
Since both conditions (continuity on \([0,1]\) and differentiability on \((0,1)\)) are satisfied, Lagrange’s mean value theorem (LMVT) applies to \( f(x) \) on the interval \([0,1]\).

Step 2: By LMVT, there exists a number \( c \in (0,1) \) such that \[ f'(c)=\frac{f(1)-f(0)}{1-0}. \]

Step 3: Compute the function values: \[ f(1)=1^{3}-2\cdot1^{2}-1+3=1-2-1+3=1, \] \[ f(0)=0^{3}-2\cdot0^{2}-0+3=3. \] So, the slope of the secant is: \[ \frac{f(1)-f(0)}{1-0}=\frac{1-3}{1}=-2. \]

Step 4: Differentiate \( f(x)=x^{3}-2x^{2}-x+3 \): \[ f'(x)=3x^{2}-4x-1. \] Set this equal to the secant slope: \[ 3c^{2}-4c-1=-2. \] Rearranging: \[ 3c^{2}-4c-1+2=3c^{2}-4c+1=0. \] Solve the quadratic equation \( 3c^{2}-4c+1=0 \) using the quadratic formula: \[ c=\frac{4\pm\sqrt{(-4)^{2}-4\cdot3\cdot1}}{2\cdot3} =\frac{4\pm\sqrt{16-12}}{6} =\frac{4\pm\sqrt{4}}{6} =\frac{4\pm2}{6}. \] This yields two solutions: \[ c=\frac{4+2}{6}=1 \quad \text{and} \quad c=\frac{4-2}{6}=\frac{2}{6}=\frac{1}{3}. \]

Since \( c \) must lie in the open interval \((0,1)\), we reject \( c=1 \) (as it is an endpoint). Therefore, the value of \( c \) that satisfies LMVT is \( \frac{1}{3} \).

Solution:

Step 1: Check the conditions for LMVT. Since \( f(x)=(x-4)(x-6)(x-8) \) is a polynomial, it is continuous on the closed interval \([4,8]\) and differentiable on the open interval \((4,8)\). Therefore, LMVT applies to \( f(x) \) on \([4,8]\).

Step 2: By LMVT, there exists a number \( c \in (4,8) \) such that \[ f'(c)=\frac{f(8)-f(4)}{8-4}. \]

Step 3: Compute the function values: \[ f(4)=(4-4)(4-6)(4-8)=0, \] \[ f(8)=(8-4)(8-6)(8-8)=0. \] Thus, the slope of the secant line is \[ \frac{f(8)-f(4)}{8-4}=\frac{0-0}{4}=0. \]

Step 4: Differentiate \( f(x)=(x-4)(x-6)(x-8) \). First, expand: \[ (x-4)(x-6)=x^{2}-10x+24, \] so that \[ f(x)=(x^{2}-10x+24)(x-8)=x^{3}-18x^{2}+104x-192. \] Now, differentiate: \[ f'(x)=3x^{2}-36x+104. \]

Step 5: Set \( f'(c)=0 \) (since the secant slope is 0): \[ 3c^{2}-36c+104=0. \] Apply the quadratic formula: \[ c=\frac{36\pm\sqrt{36^{2}-4\cdot3\cdot104}}{2\cdot3} =\frac{36\pm\sqrt{1296-1248}}{6} =\frac{36\pm\sqrt{48}}{6}. \] Simplify \(\sqrt{48}=4\sqrt{3}\): \[ c=\frac{36\pm4\sqrt{3}}{6} =6\pm\frac{2\sqrt{3}}{3}. \]

Both values \( c=6+\frac{2\sqrt{3}}{3} \) and \( c=6-\frac{2\sqrt{3}}{3} \) lie in the interval \((4,8)\).

Solution:

Step 1: Verify the conditions for LMVT. Since \( f(x)=\sin x \) is continuous on \(\left[0, \frac{\pi}{2}\right]\) and differentiable on \(\left(0, \frac{\pi}{2}\right)\), LMVT applies.

Step 2: By LMVT, there exists a \( c \) in \(\left(0, \frac{\pi}{2}\right)\) such that \[ f'(c)=\frac{f\left(\frac{\pi}{2}\right)-f(0)}{\frac{\pi}{2}-0}. \]

Step 3: Compute the function values: \[ f\left(\frac{\pi}{2}\right)=\sin\frac{\pi}{2}=1, \qquad f(0)=\sin 0=0. \] Thus, the slope of the secant line is: \[ \frac{1-0}{\frac{\pi}{2}}=\frac{2}{\pi}. \]

Step 4: Differentiate \( f(x)=\sin x \): \[ f'(x)=\cos x. \] Setting the derivative equal to the secant slope: \[ \cos c=\frac{2}{\pi}. \]

Step 5: Solve for \( c \): \[ c=\cos^{-1}\left(\frac{2}{\pi}\right). \] Since \( c \) lies in the interval \(\left(0, \frac{\pi}{2}\right)\), this solution is valid.

Solution:

Step 1: Check the conditions for LMVT. Since \( f(x)=x-2\sin x \) is a combination of continuous functions, it is continuous on the closed interval \([-\pi,\pi]\) and differentiable on the open interval \((-\pi,\pi)\). Thus, LMVT applies.

Step 2: According to LMVT, there exists a number \( c \in (-\pi,\pi) \) such that \[ f'(c)=\frac{f(\pi)-f(-\pi)}{\pi-(-\pi)}. \]

Step 3: Compute the function values: \[ f(\pi)=\pi-2\sin\pi=\pi-0=\pi, \] \[ f(-\pi)=-\pi-2\sin(-\pi)=-\pi-(-0)=-\pi. \] Thus, the slope of the secant line is: \[ \frac{f(\pi)-f(-\pi)}{\pi-(-\pi)}=\frac{\pi-(-\pi)}{2\pi}=\frac{2\pi}{2\pi}=1. \]

Step 4: Differentiate \( f(x)=x-2\sin x \): \[ f'(x)=1-2\cos x. \] Set the derivative equal to the secant slope: \[ 1-2\cos c=1. \]

Step 5: Solve for \( c \): \[ 1-2\cos c=1 \quad \Longrightarrow \quad -2\cos c=0 \quad \Longrightarrow \quad \cos c=0. \] The solutions to \(\cos c=0\) in the interval \((-\pi,\pi)\) are: \[ c=-\frac{\pi}{2} \quad \text{and} \quad c=\frac{\pi}{2}. \]

Solution:

Step 1: Check the conditions for LMVT. Since \( f(x)=\sin x-\sin 2x \) is continuous on \([0,\pi]\) and differentiable on \((0,\pi)\), LMVT applies.

Step 2: By LMVT, there exists a number \( c \in (0,\pi) \) such that \[ f'(c)=\frac{f(\pi)-f(0)}{\pi-0}. \]

Step 3: Compute the function values: \[ f(0)=\sin 0-\sin 0=0, \] \[ f(\pi)=\sin\pi-\sin 2\pi=0-0=0. \] Therefore, the slope of the secant line is: \[ \frac{0-0}{\pi}=0. \]

Step 4: Differentiate \( f(x)=\sin x-\sin 2x \): \[ f'(x)=\cos x-2\cos 2x. \] Setting the derivative equal to the secant slope, we have: \[ \cos c-2\cos 2c=0. \]

Step 5: Express \(\cos 2c\) in terms of \(\cos c\): \[ \cos 2c=2\cos^2 c-1. \] Substitute into the equation: \[ \cos c-2(2\cos^2 c-1)=0, \] \[ \cos c-4\cos^2 c+2=0. \] Rearranging: \[ 4\cos^2 c-\cos c-2=0. \]

Step 6: Let \( u=\cos c \), so the quadratic becomes: \[ 4u^2-u-2=0. \] Solve using the quadratic formula: \[ u=\frac{1\pm\sqrt{1+32}}{8}=\frac{1\pm\sqrt{33}}{8}. \]

Step 7: Therefore, we have: \[ \cos c=\frac{1\pm\sqrt{33}}{8}. \] Taking the inverse cosine gives: \[ c=\cos^{-1}\left(\frac{1\pm\sqrt{33}}{8}\right). \] Both solutions that lie in the interval \((0,\pi)\) are acceptable.

Solution:

Step 1: Check the conditions for LMVT. Since \( f(x)=p\,x^{2}+q\,x+r \) is a polynomial (with \( p\neq0 \)), it is continuous on \([0,1]\) and differentiable on \((0,1)\). Hence, LMVT applies.

Step 2: By LMVT, there exists a number \( c \) in \((0,1)\) such that \[ f'(c)=\frac{f(1)-f(0)}{1-0}. \]

Step 3: Compute the function values: \[ f(0)=r,\quad f(1)=p+q+r. \] Thus, the slope of the secant line is: \[ \frac{f(1)-f(0)}{1-0}=(p+q+r)-r=p+q. \]

Step 4: Differentiate \( f(x)=p\,x^{2}+q\,x+r \): \[ f'(x)=2p\,x+q. \] Setting the derivative equal to the secant slope: \[ 2p\,c+q=p+q. \]

Step 5: Solve for \( c \): \[ 2p\,c=p \quad \Longrightarrow \quad c=\frac{p}{2p}=\frac{1}{2}. \]

Solution:

Step 1: Check the conditions for LMVT. Since \( f(x)=x \) is a linear function, it is continuous on \([a,b]\) and differentiable on \((a,b)\). Therefore, LMVT applies.

Step 2: By LMVT, there exists a number \( c \in (a,b) \) such that \[ f'(c)=\frac{f(b)-f(a)}{b-a}. \]

Step 3: Compute the function values: \[ f(a)=a, \quad f(b)=b. \] Thus, the slope of the secant line is: \[ \frac{f(b)-f(a)}{b-a}=\frac{b-a}{b-a}=1. \]

Step 4: Differentiate \( f(x)=x \): \[ f'(x)=1. \] Setting the derivative equal to the secant slope: \[ 1=1. \]

Step 5: Since the derivative is a constant \( 1 \) for all \( x \) in \((a,b)\), the condition holds for every \( c \in (a,b) \).

Solution:

Step 1: Check the conditions for LMVT. The function \( f(x)=(x-1)^{\frac{2}{3}} \) is continuous on the closed interval \([1,2]\) and differentiable on the open interval \((1,2)\) (note that although the derivative becomes unbounded as \( x\to1^{+} \), it exists for all \( x>1 \)). Therefore, LMVT applies.

Step 2: By LMVT, there exists a number \( c \in (1,2) \) such that \[ f'(c)=\frac{f(2)-f(1)}{2-1}. \]

Step 3: Compute the function values: \[ f(1)=(1-1)^{\frac{2}{3}}=0,\quad f(2)=(2-1)^{\frac{2}{3}}=1. \] Hence, the slope of the secant line is: \[ \frac{f(2)-f(1)}{2-1}=\frac{1-0}{1}=1. \]

Step 4: Differentiate \( f(x)=(x-1)^{\frac{2}{3}} \). Using the power rule, \[ f'(x)=\frac{2}{3}(x-1)^{\frac{2}{3}-1}=\frac{2}{3}(x-1)^{-\frac{1}{3}}. \]

Step 5: Set the derivative equal to the secant slope: \[ \frac{2}{3}(c-1)^{-\frac{1}{3}}=1. \] Solve for \( c \): \[ (c-1)^{-\frac{1}{3}}=\frac{3}{2} \quad \Longrightarrow \quad (c-1)^{\frac{1}{3}}=\frac{2}{3}. \] Cubing both sides, \[ c-1=\left(\frac{2}{3}\right)^3=\frac{8}{27}. \] Thus, \[ c=1+\frac{8}{27}=\frac{27+8}{27}=\frac{35}{27}. \]

Solution:

Step 1: Check the conditions for LMVT. Since \( f(x)=x^{2}-6x+1 \) is a polynomial, it is continuous on \([1,3]\) and differentiable on \((1,3)\). Therefore, LMVT is applicable.

Step 2: According to LMVT, there exists a number \( c \in (1,3) \) such that \[ f'(c)=\frac{f(3)-f(1)}{3-1}. \]

Step 3: Evaluate the function at the endpoints: \[ f(1)=1^{2}-6\cdot1+1=1-6+1=-4, \] \[ f(3)=3^{2}-6\cdot3+1=9-18+1=-8. \] Hence, the slope of the chord joining \( \mathrm{A}(1,-4) \) and \( \mathrm{B}(3,-8) \) is: \[ \frac{f(3)-f(1)}{3-1}=\frac{-8-(-4)}{2}=\frac{-4}{2}=-2. \]

Step 4: Differentiate \( f(x)=x^{2}-6x+1 \): \[ f'(x)=2x-6. \] Set the derivative equal to the secant slope: \[ 2c-6=-2. \]

Step 5: Solve for \( c \): \[ 2c-6=-2 \quad \Longrightarrow \quad 2c=4 \quad \Longrightarrow \quad c=2. \]

Step 6: Find the coordinates of the point on the curve where \( x=c=2 \): \[ f(2)=2^{2}-6\cdot2+1=4-12+1=-7. \] Thus, the required point is \((2,-7)\).

Solution:

Step 1: Verify the conditions for LMVT. The function \( y=\sqrt{x-2} \) is continuous on \([2,3]\) and differentiable on \((2,3)\). Thus, LMVT applies.

Step 2: According to LMVT, there exists a point \( c \in (2,3) \) such that \[ f'(c)=\frac{f(3)-f(2)}{3-2}. \]

Step 3: Compute the function values: \[ f(2)=\sqrt{2-2}=0,\qquad f(3)=\sqrt{3-2}=1. \] Therefore, the slope of the chord joining the endpoints is: \[ \frac{1-0}{3-2}=1. \]

Step 4: Differentiate the function \( f(x)=\sqrt{x-2}=(x-2)^{\frac{1}{2}} \): \[ f'(x)=\frac{1}{2}(x-2)^{-\frac{1}{2}}. \]

Step 5: Set the derivative equal to the slope of the chord: \[ \frac{1}{2}(c-2)^{-\frac{1}{2}}=1. \] Solving for \( c \): \[ (c-2)^{-\frac{1}{2}}=2 \quad \Longrightarrow \quad (c-2)^{\frac{1}{2}}=\frac{1}{2}. \] Squaring both sides yields: \[ c-2=\left(\frac{1}{2}\right)^2=\frac{1}{4}, \] so that \[ c=2+\frac{1}{4}=\frac{9}{4}. \]

Step 6: Determine the \( y \)-coordinate at \( c=\frac{9}{4} \): \[ f\left(\frac{9}{4}\right)=\sqrt{\frac{9}{4}-2}=\sqrt{\frac{9}{4}-\frac{8}{4}}=\sqrt{\frac{1}{4}}=\frac{1}{2}. \] Therefore, the point \( P \) is \[ \left(\frac{9}{4},\frac{1}{2}\right). \]

Solution: For Lagrange’s mean value theorem to be applicable, the function must satisfy two conditions on the given interval: \[ \begin{aligned} &\text{(a) Continuity on the closed interval } [-1,2], \\ &\text{(b) Differentiability on the open interval } (-1,2). \end{aligned} \] The function \( f(x)=x^{\frac{1}{3}} \) is continuous for all \( x \) because the cube root function is defined everywhere. However, the derivative of \( f(x) \) is \[ f'(x)=\frac{1}{3} x^{-\frac{2}{3}}, \] which is not defined at \( x=0 \) due to the division by zero. Since \( 0 \) lies in the interval \((-1,2)\), the function is not differentiable on the entire open interval. Therefore, the hypothesis of the theorem fails.

Solution: Lagrange’s mean value theorem requires that the function is: \[ \begin{aligned} &\text{(a) Continuous on the closed interval } [-2,3], \\ &\text{(b) Differentiable on the open interval } (-2,3). \end{aligned} \] The function \( f(x)=|x| \) is continuous for all \( x \). However, its derivative is given by: \[ f'(x)=\begin{cases} 1, & x>0, \\ -1, & x<0, \end{cases} \] and the derivative does not exist at \( x=0 \) because the left-hand and right-hand limits differ. Since \( 0 \) lies in the interval \((-2,3)\), \( f(x) \) is not differentiable throughout the open interval. Thus, the theorem is not applicable.

Solution: For Lagrange’s mean value theorem to be applicable, the function must satisfy: \[ \begin{aligned} &\text{(a) Continuity on the closed interval } [0,3], \\ &\text{(b) Differentiability on the open interval } (0,3). \end{aligned} \] The function \( f(x)=3-(2-x)^{\frac{2}{3}} \) is continuous on \([0,3]\) since the cube root and its square are defined for all real numbers. To check differentiability, we differentiate: \[ f'(x) = -\frac{2}{3}(2-x)^{-\frac{1}{3}}(-1)=\frac{2}{3}(2-x)^{-\frac{1}{3}}. \] Notice that \( f'(x) \) involves the term \((2-x)^{-\frac{1}{3}}\). At \( x=2 \), we have: \[ (2-2)^{-\frac{1}{3}} = 0^{-\frac{1}{3}}, \] which is undefined. Since \( x=2 \) lies in the open interval \((0,3)\), the function is not differentiable on the entire open interval. Hence, Lagrange’s mean value theorem is not applicable.