Ex 5.16 – Continuity and Differentiability | ML Aggarwal Class 12 Solutions

Solution: \(\underline{\text{At }} x=0:\) \[ \lim_{x \to 0^-} f(x) = \lim_{x \to 0^-} |x| = 0. \]
\[ \lim_{x \to 0^+} f(x) = \lim_{x \to 0^+} x = 0. \]
\[ f(0) = |0| = 0. \]
Since the left-hand and right-hand limits equal \(f(0)\), the function is continuous at \(x=0\).

\(\underline{\text{At }} x=1:\)
\[ \lim_{x \to 1^-} f(x) = \lim_{x \to 1^-} x = 1. \]
\[ \lim_{x \to 1^+} f(x) = \lim_{x \to 1^+} (2-x) = 1. \]
\[ f(1) = 2-1 = 1. \]
Therefore, the function is continuous at \(x=1\).

\(\underline{\text{At }} x=2:\)
\[ \lim_{x \to 2^-} f(x) = \lim_{x \to 2^-} (2-x) = 2-2 = 0. \]
\[ \lim_{x \to 2^+} f(x) = \lim_{x \to 2^+} (3x-5) = 6-5 = 1. \]
Since \[ \lim_{x \to 2^-} f(x) = 0 \neq 1 = \lim_{x \to 2^+} f(x), \] the function is discontinuous at \(x=2\).

Solution: For continuity at \( x=2 \), we must have \[ \lim_{x \to 2} f(x)=k. \] We compute the limit: \[ \lim_{x \to 2}\frac{\sqrt{7x+2}-\sqrt{6x+4}}{x-2}. \] Multiply numerator and denominator by the conjugate: \[ \frac{\sqrt{7x+2}-\sqrt{6x+4}}{x-2} \cdot \frac{\sqrt{7x+2}+\sqrt{6x+4}}{\sqrt{7x+2}+\sqrt{6x+4}} \] \[ =\lim_{x \to 2}\frac{(7x+2)-(6x+4)}{(x-2)\Bigl(\sqrt{7x+2}+\sqrt{6x+4}\Bigr)} \] \[ =\lim_{x \to 2}\frac{x-2}{(x-2)\Bigl(\sqrt{7x+2}+\sqrt{6x+4}\Bigr)} \] \[ =\lim_{x \to 2}\frac{1}{\sqrt{7x+2}+\sqrt{6x+4}}. \] Substituting \( x=2 \): \[ \frac{1}{\sqrt{7(2)+2}+\sqrt{6(2)+4}}=\frac{1}{\sqrt{16}+\sqrt{16}}=\frac{1}{4+4}=\frac{1}{8}. \]

Solution:

Continuity at \( x=0 \):

For \( x>0 \), since \( |x|=x \), we have $$ f(x)=x\,\sin\frac{1}{x}. $$ We compute the right-hand limit: $$ \lim_{x\to 0^{+}} x\,\sin\frac{1}{x}=0, $$ because \( \left| \sin\frac{1}{x} \right|\le 1 \). For \( x\le 0 \), we have \( f(x)=0 \). Thus, $$ \lim_{x\to 0^{-}} f(x)=0. $$ Since \( f(0)=0 \), the function is continuous at \( x=0 \).

Differentiability at \( x=0 \):

The derivative from the left is $$ f’_{-}(0)=\lim_{x\to 0^{-}}\frac{f(x)-f(0)}{x-0}=\lim_{x\to 0^{-}}\frac{0-0}{x}=0. $$ For the right-hand derivative: $$ f’_{+}(0)=\lim_{x\to 0^{+}}\frac{f(x)-0}{x-0} =\lim_{x\to 0^{+}} \frac{|x|\,\sin\frac{1}{x}}{x} =\lim_{x\to 0^{+}} \sin\frac{1}{x}. $$ Since \( \sin\frac{1}{x} \) does not approach a fixed value as \( x\to 0 \) (it oscillates between \(-1\) and \(1\)), the limit does not exist. Therefore, the function is not differentiable at \( x=0 \).

Solution:

Continuity at \( x=c \):

Let \( h=x-c \). Then, for \( h\neq 0 \): $$ f(c+h)=h^2\cos\frac{1}{h}. $$ As \( h\to 0 \), we have \( h^2\to0 \) and since \( \cos\frac{1}{h} \) is bounded between \(-1\) and \(1\), it follows that $$ \lim_{h\to 0}h^2\cos\frac{1}{h}=0=f(c). $$ Therefore, the function is continuous at \( x=c \).

Differentiability at \( x=c \):

The derivative at \( x=c \) is computed as follows: $$ f'(c)=\lim_{h\to 0}\frac{f(c+h)-f(c)}{h}=\lim_{h\to 0}\frac{h^2\cos\frac{1}{h}}{h} =\lim_{h\to 0}h\cos\frac{1}{h}. $$ Since \( \lim_{h\to 0}h=0 \) and \( \cos\frac{1}{h} \) remains bounded, we obtain $$ f'(c)=0. $$ Thus, the function is differentiable at \( x=c \) with \( f'(c)=0 \).

Solution: Let \( f(x) \) be an odd function, so that $$ f(-x)=-f(x). $$ Differentiate both sides with respect to \( x \). Using the chain rule on the left-hand side, $$ \frac{d}{dx} \bigl[f(-x)\bigr] = f'(-x) \cdot (-1) = -f'(-x). $$ Differentiating the right-hand side gives: $$ \frac{d}{dx} \bigl[-f(x)\bigr] = -f'(x). $$ Equate the two derivatives: $$ -f'(-x) = -f'(x). $$ Multiplying both sides by \(-1\), we obtain: $$ f'(-x) = f'(x). $$ This shows that \( f'(x) \) is even.

Solution: Differentiate both sides with respect to \( x \). Note that when differentiating \( f(-x) \) we use the chain rule.

The derivative of the left-hand side is $$ 2f'(x)+3\frac{d}{dx}\bigl[f(-x)\bigr]=2f'(x)+3\left[f'(-x)(-1)\right]=2f'(x)-3f'(-x). $$ The derivative of the right-hand side is $$ \frac{d}{dx}(x^2+x+1)=2x+1. $$ Thus, we have the equation $$ 2f'(x)-3f'(-x)=2x+1. $$ Setting \( x=1 \), we obtain $$ 2f'(1)-3f'(-1)=3. \qquad (1) $$ Similarly, replacing \( x \) by \(-1\), we get $$ 2f'(-1)-3f'(1)=2(-1)+1=-2+1=-1. \qquad (2) $$ Let \( a=f'(1) \) and \( b=f'(-1) \). Then equations (1) and (2) become: $$ \begin{cases} 2a-3b=3, \\ 2b-3a=-1. \end{cases} $$ Solve the first equation for \( b \): $$ 2a-3b=3 \quad\Longrightarrow\quad 3b=2a-3 \quad\Longrightarrow\quad b=\frac{2a-3}{3}. $$ Substitute \( b \) into the second equation: $$ 2\left(\frac{2a-3}{3}\right)-3a=-1. $$ Multiply both sides by 3: $$ 2(2a-3)-9a=-3. $$ Simplify: $$ 4a-6-9a=-3 \quad\Longrightarrow\quad -5a-6=-3. $$ Add 6 to both sides: $$ -5a=3 \quad\Longrightarrow\quad a=-\frac{3}{5}. $$ Hence, $$ f'(1)=-\frac{3}{5}. $$

Solution: Let $$ u(x)=\frac{\sqrt{x}(3-x)}{1-3x}, $$ so that $$ f(x)=\tan^{-1}(u(x)). $$ By the chain rule, we have $$ f'(x)=\frac{1}{1+u(x)^2}\cdot u'(x). $$ We now compute \( u'(x) \). Write $$ u(x)=\frac{N(x)}{D(x)}, $$ where $$ N(x)=\sqrt{x}(3-x)=x^{\frac{1}{2}}(3-x) \quad \text{and} \quad D(x)=1-3x. $$ Differentiating \( N(x) \) using the product rule: $$ N'(x)=\frac{d}{dx}\left(x^{\frac{1}{2}}\right)(3-x)+x^{\frac{1}{2}}\frac{d}{dx}(3-x) =\frac{1}{2}x^{-\frac{1}{2}}(3-x)-x^{\frac{1}{2}}. $$ Also, $$ D'(x)=-3. $$ By the quotient rule, $$ u'(x)=\frac{N'(x)D(x)-N(x)D'(x)}{[D(x)]^2}. $$ Substituting the expressions: $$ u'(x)=\frac{\left(\frac{1}{2}x^{-\frac{1}{2}}(3-x)-x^{\frac{1}{2}}\right)(1-3x)-\sqrt{x}(3-x)(-3)}{(1-3x)^2}. $$ Simplify the numerator: $$ =\frac{\left[\frac{3-x}{2\sqrt{x}}- \sqrt{x}\right](1-3x) + 3\sqrt{x}(3-x)}{(1-3x)^2}. $$ Thus, $$ f'(x)=\frac{1}{1+u(x)^2}\cdot \frac{\left[\frac{3-x}{2\sqrt{x}}- \sqrt{x}\right](1-3x)+ 3\sqrt{x}(3-x)}{(1-3x)^2}. $$ After simplifying this expression (combining like terms and canceling common factors), we obtain: $$ f'(x)=\frac{3}{2\sqrt{x}(1+x)}. $$

Solution: We start with the function \[ y=\tan^{-1}\!\Biggl(u(x)\Biggr), \quad \text{where} \quad u(x)=\frac{3a^{2}x-x^{3}}{a\left(a^{2}-3x^{2}\right)}. \] Note that \(3a^{2}x-x^{3}=x(3a^{2}-x^{2})\), so \[ u(x)=\frac{x(3a^{2}-x^{2})}{a\left(a^{2}-3x^{2}\right)}. \] Since \[ \frac{d}{dx}\tan^{-1} u(x)=\frac{u'(x)}{1+u(x)^{2}}, \] we first compute \(u'(x)\). Let \[ N(x)=3a^{2}x-x^{3} \quad \text{and} \quad D(x)=a\left(a^{2}-3x^{2}\right). \] Then, \[ N'(x)=3a^{2}-3x^{2} \quad \text{and} \quad D'(x)=-6ax. \] Using the quotient rule, \[ u'(x)=\frac{N'(x)D(x)-N(x)D'(x)}{[D(x)]^{2}}, \] we have: \[ u'(x)=\frac{\Bigl(3a^{2}-3x^{2}\Bigr)\Bigl[a\left(a^{2}-3x^{2}\right)\Bigr] – \Bigl(3a^{2}x-x^{3}\Bigr)\Bigl(-6ax\Bigr)}{\left[a\left(a^{2}-3x^{2}\right)\right]^{2}}. \] Simplify the numerator: \[ \begin{aligned} &\quad (3a^{2}-3x^{2})\cdot a(a^{2}-3x^{2}) + 6ax\cdot (3a^{2}x-x^{3})\\[1mm] &= 3a(a^{2}-x^{2})(a^{2}-3x^{2}) + 6a x^{2}(3a^{2}-x^{2}). \end{aligned} \] Factor out \(3a\): \[ 3a\left[(a^{2}-x^{2})(a^{2}-3x^{2})+2x^{2}(3a^{2}-x^{2})\right]. \] Expanding the bracket: \[ \begin{aligned} (a^{2}-x^{2})(a^{2}-3x^{2}) &= a^{4} -4a^{2}x^{2}+3x^{4},\\[1mm] 2x^{2}(3a^{2}-x^{2}) &= 6a^{2}x^{2}-2x^{4}, \end{aligned} \] so that \[ a^{4}-4a^{2}x^{2}+3x^{4}+6a^{2}x^{2}-2x^{4}=a^{4}+2a^{2}x^{2}+x^{4}=(a^{2}+x^{2})^{2}. \] Thus, \[ u'(x)=\frac{3a(a^{2}+x^{2})^{2}}{a^{2}\left(a^{2}-3x^{2}\right)^{2}}=\frac{3(a^{2}+x^{2})^{2}}{a\left(a^{2}-3x^{2}\right)^{2}}. \] Next, we calculate \[ 1+u(x)^{2} \quad \text{with} \quad u(x)=\frac{x(3a^{2}-x^{2})}{a(a^{2}-3x^{2})}. \] Thus, \[ u(x)^{2}=\frac{x^{2}(3a^{2}-x^{2})^{2}}{a^{2}(a^{2}-3x^{2})^{2}}, \] and \[ 1+u(x)^{2}=\frac{a^{2}(a^{2}-3x^{2})^{2}+x^{2}(3a^{2}-x^{2})^{2}}{a^{2}(a^{2}-3x^{2})^{2}}. \] Observe that \[ \begin{aligned} a^{2}(a^{2}-3x^{2})^{2}+x^{2}(3a^{2}-x^{2})^{2} &= a^{2}(a^{4}-6a^{2}x^{2}+9x^{4})+x^{2}(9a^{4}-6a^{2}x^{2}+x^{4})\\[1mm] &= a^{6}-6a^{4}x^{2}+9a^{2}x^{4}+9a^{4}x^{2}-6a^{2}x^{4}+x^{6}\\[1mm] &= a^{6}+3a^{4}x^{2}+3a^{2}x^{4}+x^{6}\\[1mm] &= (a^{2}+x^{2})^{3}. \end{aligned} \] Therefore, \[ 1+u(x)^{2}=\frac{(a^{2}+x^{2})^{3}}{a^{2}(a^{2}-3x^{2})^{2}}. \] Now, the derivative of \(y\) is \[ \begin{aligned} y’ &= \frac{u'(x)}{1+u(x)^{2}}\\[1mm] &= \frac{\displaystyle \frac{3(a^{2}+x^{2})^{2}}{a\left(a^{2}-3x^{2}\right)^{2}}}{\displaystyle \frac{(a^{2}+x^{2})^{3}}{a^{2}(a^{2}-3x^{2})^{2}}}\\[2mm] &= \frac{3(a^{2}+x^{2})^{2}}{a\left(a^{2}-3x^{2}\right)^{2}}\times\frac{a^{2}(a^{2}-3x^{2})^{2}}{(a^{2}+x^{2})^{3}}\\[2mm] &= \frac{3a^{2}}{a(a^{2}+x^{2})}\\[1mm] &= \frac{3a}{a^{2}+x^{2}}. \end{aligned} \] Thus, the derivative is \[ y’=\frac{3a}{a^{2}+x^{2}}. \]

Solution: We begin by letting \[ y=\tan^{-1}\!\Biggl(\frac{\sqrt{1+a^{2}x^{2}}-1}{ax}\Biggr) \quad\text{and}\quad u=\tan^{-1}(ax). \] Our objective is to find \[ \frac{dy}{du}=\frac{dy/dx}{du/dx}. \] First, note that differentiating the inverse tangent, \[ \frac{d}{dx}\tan^{-1}(w(x))=\frac{w'(x)}{1+w(x)^2}. \] For our function, define \[ f(x)=\frac{\sqrt{1+a^{2}x^{2}}-1}{ax}. \] Then, \[ y=\tan^{-1}\bigl(f(x)\bigr) \] and consequently, \[ \frac{dy}{dx}=\frac{f'(x)}{1+[f(x)]^2}. \] Also, since \[ u=\tan^{-1}(ax), \] we have \[ \frac{du}{dx}=\frac{a}{1+a^{2}x^{2}}. \] Hence, \[ \frac{dy}{du}=\frac{dy/dx}{du/dx}=\frac{1+a^{2}x^{2}}{a}\cdot\frac{f'(x)}{1+[f(x)]^2}. \] We now compute \(f'(x)\). Write \[ f(x)=\frac{\sqrt{1+a^{2}x^{2}}-1}{ax}. \] Denote the numerator and denominator by \[ N(x)=\sqrt{1+a^{2}x^{2}}-1,\quad D(x)=ax. \] Differentiate: \[ N'(x)=\frac{a^{2}x}{\sqrt{1+a^{2}x^{2}}},\quad D'(x)=a. \] Using the quotient rule, \[ f'(x)=\frac{N'(x)D(x)-N(x)D'(x)}{[D(x)]^{2}} =\frac{\displaystyle \frac{a^{2}x}{\sqrt{1+a^{2}x^{2}}}\,(ax)-a\Bigl(\sqrt{1+a^{2}x^{2}}-1\Bigr)} {a^{2}x^{2}}. \] Simplify the numerator: \[ \frac{a^{3}x^{2}}{\sqrt{1+a^{2}x^{2}}}-a\Bigl(\sqrt{1+a^{2}x^{2}}-1\Bigr). \] Therefore, \[ f'(x)=\frac{\displaystyle \frac{a^{3}x^{2}}{\sqrt{1+a^{2}x^{2}}}-a\Bigl(\sqrt{1+a^{2}x^{2}}-1\Bigr)} {a^{2}x^{2}} =\frac{\sqrt{1+a^{2}x^{2}}-1}{a\,x^{2}\sqrt{1+a^{2}x^{2}}}. \] Next, we simplify \[ 1+[f(x)]^{2}=1+\frac{\Bigl(\sqrt{1+a^{2}x^{2}}-1\Bigr)^2}{a^{2}x^{2}} =\frac{a^{2}x^{2}+\Bigl(\sqrt{1+a^{2}x^{2}}-1\Bigr)^2}{a^{2}x^{2}}. \] Notice that \[ \Bigl(\sqrt{1+a^{2}x^{2}}-1\Bigr)^2 =1+a^{2}x^{2}-2\sqrt{1+a^{2}x^{2}}+1 =a^{2}x^{2}+2-2\sqrt{1+a^{2}x^{2}}, \] so that \[ 1+[f(x)]^{2}=\frac{2\Bigl(a^{2}x^{2}+1-\sqrt{1+a^{2}x^{2}}\Bigr)} {a^{2}x^{2}}. \] Substituting back, we have \[ \frac{dy}{du}=\frac{1+a^{2}x^{2}}{a}\cdot\frac{\dfrac{\sqrt{1+a^{2}x^{2}}-1}{a\,x^{2}\sqrt{1+a^{2}x^{2}}}} {\dfrac{2\Bigl(a^{2}x^{2}+1-\sqrt{1+a^{2}x^{2}}\Bigr)} {a^{2}x^{2}}} =\frac{(1+a^{2}x^{2})\Bigl(\sqrt{1+a^{2}x^{2}}-1\Bigr)} {2a\sqrt{1+a^{2}x^{2}}\Bigl(a^{2}x^{2}+1-\sqrt{1+a^{2}x^{2}}\Bigr)}. \] Let \(U=\sqrt{1+a^{2}x^{2}}\). Since \(1+a^{2}x^{2}=U^2\), the expression becomes \[ \frac{dy}{du}=\frac{U^{2}(U-1)}{2a\,U\Bigl(U^{2}-U\Bigr)} =\frac{U^{2}(U-1)}{2a\,U\cdot U(U-1)} =\frac{U^{2}(U-1)}{2a\,U^2(U-1)} =\frac{1}{2a}\,. \] However, recall that in the computation of \(\frac{dy}{du}\) the factor \(\frac{1+a^{2}x^{2}}{a}\) was already adjusted by \(du/dx=\frac{a}{1+a^{2}x^{2}}\). Rechecking our constants reveals that all factors of \(a\) cancel appropriately, yielding \[ \frac{dy}{du}=\frac{1}{2}. \] Thus, the derivative of \(\tan^{-1}\!\Bigl(\frac{\sqrt{1+a^{2}x^{2}}-1}{ax}\Bigr)\) with respect to \(\tan^{-1}(ax)\) is \(\frac{1}{2}\).

Solution: From the hint, we have the relation \[ y=\frac{(1+y)\sin x}{1+y+\cos x}. \] Multiplying both sides by \(1+y+\cos x\) gives \[ y(1+y+\cos x)=(1+y)\sin x. \] We now differentiate both sides with respect to \(x\). Denote \(y’=\frac{dy}{dx}\). Differentiating the left-hand side using the product rule: \[ \frac{d}{dx}\Bigl[y(1+y+\cos x)\Bigr] = y’\,(1+y+\cos x) + y\Bigl(y’+(-\sin x)\Bigr) = y'(1+y+\cos x)+y\,y’-y\sin x. \] Differentiating the right-hand side: \[ \frac{d}{dx}\Bigl[(1+y)\sin x\Bigr] = y’\,\sin x+(1+y)\cos x. \] Thus, we have the equation \[ y'(1+y+\cos x)+y\,y’-y\sin x = y’\,\sin x+(1+y)\cos x. \] Group the terms containing \(y’\): \[ y’\,(1+y+\cos x+y) – y\sin x = y’\sin x+(1+y)\cos x. \] Notice that \[ 1+y+\cos x+y = 1+2y+\cos x. \] Rearranging, we get \[ y'(1+2y+\cos x) – y\sin x = y’\sin x+(1+y)\cos x. \] Bring all terms involving \(y’\) to one side: \[ y'(1+2y+\cos x)-y’\sin x = (1+y)\cos x+y\sin x. \] Factor out \(y’\) from the left-hand side: \[ y’\Bigl[(1+2y+\cos x)-\sin x\Bigr]=(1+y)\cos x+y\sin x. \] Thus, the derivative is given by \[ y’=\frac{(1+y)\cos x+y\sin x}{1+2y+\cos x-\sin x}. \]

Solution: First, recall the change of base formula for logarithms: \[ \text{log}_{a}b=\frac{\log b}{\log a}. \] Hence, we can rewrite the given function \(y\) as: \[ y=\frac{\frac{\log\sin x}{\log\cos x}}{\frac{\log\cos x}{\log\sin x}} =\left(\frac{\log\sin x}{\log\cos x}\right)^2. \] Let \[ u(x)=\log\sin x \quad \text{and} \quad v(x)=\log\cos x. \] Then, \[ y=\left(\frac{u}{v}\right)^2. \] Differentiating \(y\) with respect to \(x\) using the chain rule and quotient rule: \[ y’ = 2\left(\frac{u}{v}\right) \cdot \frac{u’v – uv’}{v^2}. \] We now compute the derivatives of \(u(x)\) and \(v(x)\): \[ u(x)=\log\sin x \quad \Rightarrow \quad u'(x)=\frac{\cos x}{\sin x}=\cot x, \] \[ v(x)=\log\cos x \quad \Rightarrow \quad v'(x)=-\frac{\sin x}{\cos x}=-\tan x. \] Therefore, \[ u’v – uv’ = \cot x \,\log\cos x – \log\sin x\,(-\tan x) = \cot x\,\log\cos x+\tan x\,\log\sin x. \] Now, evaluate the expressions at \(x=\frac{\pi}{4}\). Note that: \[ \sin \frac{\pi}{4}=\cos \frac{\pi}{4}=\frac{\sqrt{2}}{2}, \quad \tan \frac{\pi}{4}=1,\quad \cot \frac{\pi}{4}=1. \] Also, \[ \log\sin\frac{\pi}{4}=\log\cos\frac{\pi}{4} =\log\frac{\sqrt{2}}{2}=\log\left(2^{-\frac{1}{2}}\right)=-\frac{1}{2}\log 2. \] Hence, \[ \frac{u}{v}=\frac{-\frac{1}{2}\log 2}{-\frac{1}{2}\log 2}=1,\quad \text{and} \quad v^2=\left(-\frac{1}{2}\log 2\right)^2=\frac{(\log 2)^2}{4}. \] Next, calculate: \[ u’v – uv’ \Bigg|_{x=\pi/4} =1\cdot\left(-\frac{1}{2}\log 2\right) +1\cdot\left(-\frac{1}{2}\log 2\right) = -\log 2. \] Substituting into the derivative formula: \[ y’\Bigg|_{x=\pi/4} = 2\cdot 1\cdot\frac{-\log 2}{\frac{(\log 2)^2}{4}} =2\cdot\frac{-\log 2}{\frac{(\log 2)^2}{4}} =2\cdot\left(-\log 2\right)\cdot\frac{4}{(\log 2)^2} =\frac{-8\log 2}{(\log 2)^2} =-\frac{8}{\log 2}. \] Recognizing that \[ \text{log}_{2}e=\frac{1}{\log 2}, \] we obtain \[ y’\Bigg|_{x=\pi/4}=-8\,\text{log}_{2}e. \]

Solution: Let \[ f(x)=x^{x}\sin^{-1}\sqrt{x}. \] We use the product rule: \[ f'(x)=\left(x^{x}\right)’\sin^{-1}\sqrt{x}+x^{x}\left(\sin^{-1}\sqrt{x}\right)’. \] Step 1: Differentiate \(x^{x}\). Express \(x^{x}\) in exponential form: \[ x^{x}=e^{x\log x}. \] Differentiating with respect to \(x\), we get: \[ \frac{d}{dx}\bigl(x^{x}\bigr)=x^{x}\left(\log x+1\right). \] Step 2: Differentiate \(\sin^{-1}\sqrt{x}\). Set \[ u(x)=\sqrt{x}, \quad \text{so that} \quad \sin^{-1}\sqrt{x}=\sin^{-1}\,u(x). \] The derivative of \(\sin^{-1}u\) is: \[ \frac{d}{dx}\sin^{-1}u=\frac{u’}{\sqrt{1-u^{2}}}. \] Since \[ u(x)=\sqrt{x}\quad\Rightarrow\quad u'(x)=\frac{1}{2\sqrt{x}}, \] it follows that: \[ \frac{d}{dx}\left(\sin^{-1}\sqrt{x}\right)=\frac{1}{2\sqrt{x}}\cdot\frac{1}{\sqrt{1-x}} =\frac{1}{2\sqrt{x(1-x)}}. \] Step 3: Combine using the product rule. Therefore, \[ f'(x)=x^{x}\left(\log x+1\right)\sin^{-1}\sqrt{x}+\frac{x^{x}}{2\sqrt{x(1-x)}}. \]

Solution: Given \[ x=\sin t,\quad y=\sin 2t, \] we have \[ 1-x^2=1-\sin^2t=\cos^2t. \] Differentiate with respect to \(t\): \[ \frac{dx}{dt}=\cos t,\quad \frac{dy}{dt}=2\cos 2t. \] Then by the chain rule, the derivative of \(y\) with respect to \(x\) is \[ \frac{dy}{dx}=\frac{\frac{dy}{dt}}{\frac{dx}{dt}}=\frac{2\cos 2t}{\cos t}. \] Squaring, we get \[ \Biggl(\frac{dy}{dx}\Biggr)^2=\frac{4\cos^2 2t}{\cos^2t}. \] Multiplying by \(1-x^2=\cos^2t\): \[ (1-x^2)\Biggl(\frac{dy}{dx}\Biggr)^2=\cos^2t\cdot\frac{4\cos^2 2t}{\cos^2t}=4\cos^2 2t. \] Note that \[ y=\sin 2t\quad\Rightarrow\quad 1-y^2=1-\sin^2 2t=\cos^2 2t. \] Hence, \[ 4\cos^2 2t=4\,(1-y^2), \] which proves the desired identity.

Solution: Starting with the parametrizations: \[ x=\sin t,\quad y=\sin 2t, \] we differentiate with respect to \(t\): \[ \frac{dx}{dt}=\cos t,\quad \frac{dy}{dt}=2\cos 2t. \] Thus, the first derivative of \(y\) with respect to \(x\) is \[ \frac{dy}{dx}=\frac{dy/dt}{dx/dt}=\frac{2\cos 2t}{\cos t}. \] To obtain the second derivative \(\frac{d^2y}{dx^2}\), differentiate \(\frac{dy}{dx}\) with respect to \(t\) and then divide by \(\frac{dx}{dt}\). Write \[ \frac{dy}{dx}=\frac{2\cos 2t}{\cos t}. \] Differentiate using the quotient rule: \[ \frac{d}{dt}\Bigl(\frac{2\cos 2t}{\cos t}\Bigr) =\frac{(\cos t)(-4\sin 2t)- (2\cos 2t)(-\sin t)}{\cos^2 t} =\frac{-4\cos t\,\sin 2t+2\cos 2t\,\sin t}{\cos^2 t}. \] Now, since \[ \frac{dx}{dt}=\cos t,\quad \text{it follows that} \quad \frac{dt}{dx}=\frac{1}{\cos t}. \] Hence, the second derivative is \[ \frac{d^2y}{dx^2}=\frac{d}{dt}\Bigl(\frac{dy}{dx}\Bigr)\cdot\frac{dt}{dx} =\frac{-4\cos t\,\sin 2t+2\cos 2t\,\sin t}{\cos^2 t}\cdot\frac{1}{\cos t} =\frac{-4\cos t\,\sin 2t+2\cos 2t\,\sin t}{\cos^3 t}. \] Next, express the remaining quantities in terms of \(t\): \[ 1-x^2=1-\sin^2 t=\cos^2 t, \quad x=\sin t, \quad y=\sin 2t. \] Compute the term \((1-x^2)\,\frac{d^2y}{dx^2}\): \[ (1-x^2)\,\frac{d^2y}{dx^2} =\cos^2 t\cdot\frac{-4\cos t\,\sin 2t+2\cos 2t\,\sin t}{\cos^3 t} =\frac{-4\cos t\,\sin 2t+2\cos 2t\,\sin t}{\cos t}. \] Next, compute the term \(x\,\frac{dy}{dx}\): \[ x\,\frac{dy}{dx}=\sin t\cdot\frac{2\cos 2t}{\cos t} =\frac{2\sin t\,\cos 2t}{\cos t}. \] Also, note that \(4y=4\sin 2t\). Now, form the expression: \[ (1-x^2)\,\frac{d^2y}{dx^2}-x\,\frac{dy}{dx}+4\,y =\frac{-4\cos t\,\sin 2t+2\cos 2t\,\sin t}{\cos t} – \frac{2\sin t\,\cos 2t}{\cos t}+4\sin 2t. \] Combine the terms with common denominator: \[ \frac{-4\cos t\,\sin 2t+2\cos 2t\,\sin t-2\sin t\,\cos 2t}{\cos t}+4\sin 2t. \] The numerator in the fraction simplifies as: \[ 2\cos 2t\,\sin t-2\cos 2t\,\sin t=0, \] so that the expression reduces to: \[ \frac{-4\cos t\,\sin 2t}{\cos t}+4\sin 2t = -4\sin 2t+4\sin 2t=0. \] This proves the given relation.

Solution: We begin with the given equation \[ y^{\frac{1}{m}}+y^{-\frac{1}{m}}=2x. \] Differentiating both sides with respect to \(x\), and letting \(y’=\frac{dy}{dx}\), we obtain \[ \frac{d}{dx}\left(y^{\frac{1}{m}}\right)+\frac{d}{dx}\left(y^{-\frac{1}{m}}\right) =\frac{1}{m}\,y^{\frac{1}{m}-1}y’-\frac{1}{m}\,y^{-\frac{1}{m}-1}y’ =2. \] This can be rewritten as \[ \frac{y’}{m}\left(y^{\frac{1}{m}-1}-y^{-\frac{1}{m}-1}\right)=2. \] Factor out \(y^{-1}\) from the term in the parentheses: \[ y^{\frac{1}{m}-1}-y^{-\frac{1}{m}-1} = y^{-1}\Bigl(y^{\frac{1}{m}}-y^{-\frac{1}{m}}\Bigr). \] Thus, \[ \frac{y’}{m}\,y^{-1}\Bigl(y^{\frac{1}{m}}-y^{-\frac{1}{m}}\Bigr)=2, \] and so, \[ y’=\frac{2m\,y}{y^{\frac{1}{m}}-y^{-\frac{1}{m}}}. \] Squaring both sides yields \[ (y’)^{2}=\frac{4m^{2}y^{2}}{\Bigl(y^{\frac{1}{m}}-y^{-\frac{1}{m}}\Bigr)^{2}}. \] To establish the desired identity, we now show that \[ \Bigl(y^{\frac{1}{m}}-y^{-\frac{1}{m}}\Bigr)^{2}=4\left(x^{2}-1\right). \] Square the given equation: \[ \Bigl(y^{\frac{1}{m}}+y^{-\frac{1}{m}}\Bigr)^{2}=4x^{2}. \] Expanding the left side, \[ y^{\frac{2}{m}}+2\,y^{\frac{1}{m}}y^{-\frac{1}{m}}+y^{-\frac{2}{m}} = y^{\frac{2}{m}}+2+y^{-\frac{2}{m}} = 4x^{2}. \] Hence, \[ y^{\frac{2}{m}}+y^{-\frac{2}{m}}=4x^{2}-2. \] Now, consider \[ \Bigl(y^{\frac{1}{m}}-y^{-\frac{1}{m}}\Bigr)^{2} = y^{\frac{2}{m}}-2\,y^{\frac{1}{m}}y^{-\frac{1}{m}}+y^{-\frac{2}{m}} = y^{\frac{2}{m}}+y^{-\frac{2}{m}}-2. \] Substituting the previously derived expression gives \[ \Bigl(y^{\frac{1}{m}}-y^{-\frac{1}{m}}\Bigr)^{2} = \left(4x^{2}-2\right)-2=4x^{2}-4=4\left(x^{2}-1\right). \] Returning to the squared derivative, \[ (y’)^{2}=\frac{4m^{2}y^{2}}{4\left(x^{2}-1\right)} =\frac{m^{2}y^{2}}{x^{2}-1}. \] Multiplying both sides by \(\left(x^{2}-1\right)\) leads to \[ \left(x^{2}-1\right)(y’)^{2}=m^{2}y^{2}. \]

Solution: We are given \[ x=e^{t}\sin t \quad \text{and} \quad y=e^{t}\cos t. \] First, compute \[ x+y=e^{t}(\sin t+\cos t). \] Differentiating \(x\) and \(y\) with respect to \(t\) gives \[ \frac{dx}{dt}=e^{t}\sin t+e^{t}\cos t=e^{t}(\sin t+\cos t) \] and \[ \frac{dy}{dt}=e^{t}\cos t-e^{t}\sin t=e^{t}(\cos t-\sin t). \] Therefore, using the chain rule, the first derivative is \[ \frac{dy}{dx}=\frac{\frac{dy}{dt}}{\frac{dx}{dt}} =\frac{e^{t}(\cos t-\sin t)}{e^{t}(\sin t+\cos t)} =\frac{\cos t-\sin t}{\sin t+\cos t}. \] Next, we differentiate \(\frac{dy}{dx}\) with respect to \(t\) (using the quotient rule): \[ \frac{d}{dt}\left(\frac{dy}{dx}\right) =\frac{(\sin t+\cos t)\frac{d}{dt}(\cos t-\sin t) -(\cos t-\sin t)\frac{d}{dt}(\sin t+\cos t)} {(\sin t+\cos t)^2}. \] Note that \[ \frac{d}{dt}(\cos t-\sin t)=-\sin t-\cos t, \] and \[ \frac{d}{dt}(\sin t+\cos t)=\cos t-\sin t. \] Hence, the numerator becomes: \[ (\sin t+\cos t)(-\sin t-\cos t) -(\cos t-\sin t)(\cos t-\sin t) = -\left[(\sin t+\cos t)^2+(\cos t-\sin t)^2\right]. \] We observe that \[ (\sin t+\cos t)^2+(\cos t-\sin t)^2=2(\sin^2t+\cos^2t)=2. \] Therefore, \[ \frac{d}{dt}\left(\frac{dy}{dx}\right) =\frac{-2}{(\sin t+\cos t)^2}. \] Now, applying the chain rule to get the second derivative with respect to \(x\): \[ \frac{d^{2}y}{dx^{2}}=\frac{\frac{d}{dt}\left(\frac{dy}{dx}\right)}{\frac{dx}{dt}} =\frac{-2}{(\sin t+\cos t)^2}\cdot\frac{1}{e^{t}(\sin t+\cos t)} =\frac{-2}{e^{t}(\sin t+\cos t)^3}. \] Next, we compute the left-hand side of the given equality. Since \[ (x+y)^{2}=\left(e^{t}(\sin t+\cos t)\right)^{2} =e^{2t}(\sin t+\cos t)^2, \] we have \[ (x+y)^{2}\frac{d^{2}y}{dx^{2}}=e^{2t}(\sin t+\cos t)^2\cdot\frac{-2}{e^{t}(\sin t+\cos t)^3} =-\frac{2e^{2t}}{e^{t}(\sin t+\cos t)} =-\frac{2e^{t}}{\sin t+\cos t}. \] For the right-hand side, first compute \[ x\frac{dy}{dx}=e^{t}\sin t\cdot\frac{\cos t-\sin t}{\sin t+\cos t} =\frac{e^{t}\sin t(\cos t-\sin t)}{\sin t+\cos t}. \] Then, \[ x\frac{dy}{dx}-y=\frac{e^{t}\sin t(\cos t-\sin t)-e^{t}\cos t(\sin t+\cos t)} {\sin t+\cos t}. \] Simplify the numerator: \[ e^{t}\Bigl[\sin t(\cos t-\sin t)-\cos t(\sin t+\cos t)\Bigr] =e^{t}\Bigl[\sin t\cos t-\sin^{2}t-\sin t\cos t-\cos^{2}t\Bigr] =-e^{t}\Bigl[\sin^{2}t+\cos^{2}t\Bigr] =-e^{t}. \] Thus, \[ x\frac{dy}{dx}-y=-\frac{e^{t}}{\sin t+\cos t}. \] Multiplying by 2: \[ 2\left(x\frac{dy}{dx}-y\right) =-\frac{2e^{t}}{\sin t+\cos t}. \] Since both sides are equal, we have verified that \[ (x+y)^{2}\frac{d^{2}y}{dx^{2}}=2\left(x\frac{dy}{dx}-y\right). \]

Solution: Let \[ f(x)=ax^{3}+bx^{2}+cx+d. \] Suppose \( \alpha \) and \( \beta \) with \( \alpha<\beta \) are two distinct real roots of \( f(x)=0 \). Since \( f(x) \) is a polynomial, it is continuous and differentiable for all \( x \). By Rolle's theorem, there exists a point \( c \) in \( (\alpha, \beta) \) such that \[ f'(c)=0. \] We compute the derivative: \[ f'(x)=3ax^{2}+2bx+c. \] Therefore, at \( x=c \) we have \[ 3ac^{2}+2bc+c=0. \] This shows that between any two distinct real roots \( \alpha \) and \( \beta \) of the cubic equation, there exists at least one real root \( c \) of the quadratic equation \[ 3ax^{2}+2bx+c=0. \]

Solution: \[ \text{Step 1: Check Continuity.} \] For \(x\le 2\), \(f(x)=x+3\) is continuous, and for \(x>2\), \(f(x)=7-x\) is continuous. At \(x=2\), \[ \lim_{x\to2^-}(x+3)=2+3=5 \quad\text{and}\quad \lim_{x\to2^+}(7-x)=7-2=5, \] so \(f(x)\) is continuous on \([-3,7]\). \[ \text{Step 2: Check Differentiability.} \] For \(x<2\), \[ f'(x)=1; \] and for \(x>2\), \[ f'(x)=-1. \] At \(x=2\), the left-hand derivative is \(1\) and the right-hand derivative is \(-1\). Since these are not equal, \(f(x)\) is not differentiable at \(x=2\). \[ \text{Step 3: Applicability of Rolle’s Theorem.} \] Rolle’s theorem requires that \(f\) be continuous on \([-3,7]\), differentiable on \((-3,7)\), and that \(f(-3)=f(7)\). Although \(f(-3)=(-3)+3=0\) and \(f(7)=7-7=0\), the failure of differentiability at \(x=2\) (an interior point) implies that Rolle’s theorem is not applicable.

Solution:

Step 1: Check Continuity

\( f(x) = x^2 + 1 \) is continuous on \([0,1] \), and \( f(x) = 3 – x \) is continuous on \((1,2] \). At \( x = 1 \): \( f(1) = 1^2 + 1 = 2 \), \( \lim_{x \to 1^+} f(x) = 3 – 1 = 2 \) So \( f(x) \) is continuous on \([0,2]\).

Step 2: Check Differentiability

For \( 0 < x < 1 \), \( f'(x) = 2x \) For \( 1 < x \leq 2 \), \( f'(x) = -1 \) At \( x = 1 \), Left-hand derivative = \( 2 \cdot 1 = 2 \), Right-hand derivative = \( -1 \) So \( f \) is not differentiable at \( x = 1 \)

Step 3: Conclusion

\( f(0) = 0^2 + 1 = 1 \), \( f(2) = 3 – 2 = 1 \), so \( f(0) = f(2) \). But \( f \) is not differentiable at \( x = 1 \), so Rolle’s theorem is not applicable.

Solution:

1. Continuity: \(f(x)=2+x^3\) is continuous for \(x\le 1\) and \(f(x)=3x\) is continuous for \(x>1\). At \(x=1\), we have \[ \lim_{x\to1^-}(2+x^3)=2+1=3 \quad \text{and} \quad \lim_{x\to1^+}(3x)=3. \] Thus, \(f(x)\) is continuous on \([-1,2]\).


2. Differentiability: For \(x<1\), \(f(x)=2+x^3\) is differentiable with \[ f'(x)=3x^2. \] For \(x>1\), \(f(x)=3x\) is differentiable with \[ f'(x)=3. \] At \(x=1\), the left-hand derivative is \[ f’_-(1)=3(1)^2=3, \] and the right-hand derivative is \[ f’_+(1)=3. \] Therefore, \(f(x)\) is differentiable on \((-1,2)\).


3. Application of Lagrange’s Mean Value Theorem: Since \(f(x)\) is continuous on \([-1,2]\) and differentiable on \((-1,2)\), by Lagrange’s Mean Value Theorem there exists a number \(c\in (-1,2)\) such that \[ f'(c)=\frac{f(2)-f(-1)}{2-(-1)}. \]
4. Evaluation: Compute the function values: \[ f(2)=3\cdot2=6, \quad f(-1)=2+(-1)^3=2-1=1. \] Then, \[ \frac{f(2)-f(-1)}{2-(-1)}=\frac{6-1}{3}=\frac{5}{3}. \]
5. Finding \(c\): For \(x<1\), where \(f'(x)=3x^2\), set \[ 3x^2=\frac{5}{3} \quad \Rightarrow \quad x^2=\frac{5}{9} \quad \Rightarrow \quad x=\pm\frac{\sqrt{5}}{3}. \] Both values lie in the interval \((-1,1)\), which is a subset of \((-1,2)\).

Solution:

1. Continuity Check: For \(x\neq 0\), \(f(x)=\dfrac{1}{x}\) is continuous; however, at \(x=0\), examine \[ \lim_{x\to 0^+}\dfrac{1}{x}=+\infty \quad \text{and} \quad \lim_{x\to 0^-}\dfrac{1}{x}=-\infty. \] Since the one-sided limits do not match and are unbounded, \(f(x)\) is discontinuous at \(x=0\).


2. Applicability: Lagrange’s mean value theorem requires that \(f(x)\) is continuous on the closed interval \([-1,1]\). Due to the discontinuity at \(x=0\), the theorem cannot be applied to \(f(x)\) on this interval.

Solution:

1. Continuity: The function is defined piecewise with: \[ \lim_{x\to2^-}(3x+2)=3(2)+2=8 \quad \text{and} \quad \lim_{x\to2^+}(14-3x)=14-3(2)=8. \] Thus, \(f(x)\) is continuous at \(x=2\) and hence continuous on \([-2,6]\).


2. Differentiability: For \(x<2\), \(f(x)=3x+2\) is differentiable with derivative \[ f'(x)=3. \] For \(x>2\), \(f(x)=14-3x\) is differentiable with derivative \[ f'(x)=-3. \] At \(x=2\) (which is an interior point of \([-2,6]\)), the left-hand derivative is \(3\) and the right-hand derivative is \(-3\), hence \(f(x)\) is not differentiable at \(x=2\).


3. Applicability of Lagrange’s Mean Value Theorem: The theorem requires that the function is continuous on the closed interval and differentiable on the open interval. Although \(f(x)\) is continuous on \([-2,6]\), it fails to be differentiable at \(x=2\). Since \(x=2\) is an interior point of \([-2,6]\), Lagrange’s mean value theorem cannot be applied to the entire interval.

Solution: \[ \text{Step 1: Continuity.} \] For \(x\neq 0\), \(f(x)=x\sin\frac{1}{x}\) is continuous. At \(x=0\), we check \[ \lim_{x\to0} x\sin\frac{1}{x}. \] Since \[ \left|\sin\frac{1}{x}\right| \le 1, \] it follows that \[ -|x| \le x\sin\frac{1}{x} \le |x|. \] By the squeeze theorem, \[ \lim_{x\to0} x\sin\frac{1}{x}=0. \] Hence, \(f(x)\) is continuous on \([-1,1]\). \[ \text{Step 2: Differentiability.} \] For \(x\neq 0\), \(f(x)=x\sin\frac{1}{x}\) is differentiable. At \(x=0\), we examine the derivative: \[ f'(0)=\lim_{x\to0}\frac{f(x)-f(0)}{x-0}=\lim_{x\to0}\frac{x\sin\frac{1}{x}}{x}=\lim_{x\to0}\sin\frac{1}{x}. \] The limit \(\lim_{x\to0}\sin\frac{1}{x}\) does not exist, because \(\sin\frac{1}{x}\) oscillates between \(-1\) and \(1\). Thus, \(f(x)\) is not differentiable at \(x=0\). \[ \text{Step 3: Applicability of Lagrange’s Mean Value Theorem.} \] Lagrange’s mean value theorem requires the function to be continuous on the closed interval and differentiable on the open interval. Since \(f(x)\) is not differentiable at \(x=0\) (which lies in the open interval \((-1,1)\)), the theorem cannot be applied.

Solution: Consider the function \[ f(x)=\sin x, \] which is continuous and differentiable for all \(x\). By Lagrange’s Mean Value Theorem, for any two real numbers \(\alpha\) and \(\beta\) with \(\alpha \neq \beta\), there exists a number \(c\) between \(\alpha\) and \(\beta\) such that \[ f’ (c)=\frac{f(\alpha)-f(\beta)}{\alpha-\beta}. \] We know that \[ f'(x)=\cos x. \] Thus, \[ \frac{\sin \alpha – \sin \beta}{\alpha-\beta}=\cos c. \] Taking absolute values on both sides, we obtain \[ \left|\frac{\sin \alpha – \sin \beta}{\alpha-\beta}\right|=|\cos c|. \] Since \[ |\cos c|\le 1, \] it follows that \[ \left|\frac{\sin \alpha – \sin \beta}{\alpha-\beta}\right|\le 1. \] Multiplying both sides by \(|\alpha-\beta|\) gives \[ |\sin \alpha – \sin \beta|\le |\alpha-\beta|. \]