question_answer

If \[x=\sqrt{{{a}^{{{\sin }^{-1}}t}}}\] and \[y=\sqrt{{{a}^{{{\cos }^{-1}}t}}},\,\,a>0\] and \[-\,1<t<1,\] then prove that \[\frac{dy}{dx}=\frac{-\,y}{x}.\]

OR

In a given function \[f(x)={{x}^{3}}+b{{x}^{2}}+ax,\] \[x\in [1,\,\,3],\] Rolle's theorem holds with \[c=2+\frac{1}{\sqrt{3}}.\] Find the values of a and b.

Answer:

Given,   \[x=\sqrt{{{a}^{{{\sin }^{-1}}t}}}\]                          ?(i)

and       \[y=\sqrt{{{a}^{{{\cos }^{-1}}t}}}\]                           ?(ii)

On differentiating both sides of Eq. (i) w.r.t. t, we get

\[\frac{dx}{dt}=\frac{1}{2}{{({{a}^{{{\sin }^{-1}}t}})}^{\frac{-1}{2}}}\frac{d}{dt}({{a}^{{{\sin }^{-1}}t}})\]

\[=\frac{1}{2}{{({{a}^{{{\sin }^{-1}}t}})}^{\frac{-1}{2}}}({{a}^{{{\sin }^{-1}}t}}{{\log }_{e}}a)\cdot \frac{d}{dt}({{\sin }^{-1}}t)\]

\[=\frac{1}{2}{{({{a}^{{{\sin }^{-1}}t}})}^{-\,\frac{1}{2}}}({{a}^{{{\sin }^{-1}}t}}{{\log }_{e}}a)\frac{1}{\sqrt{1-{{t}^{2}}}}.\]

\[=\frac{1}{2}{{({{a}^{{{\sin }^{-1}}t}})}^{-\,\frac{1}{2}\,+\,1}}{{\log }_{e}}a\frac{1}{\sqrt{1-{{t}^{2}}}}\]

\[=\frac{{{({{a}^{{{\sin }^{-1}}t}})}^{1/2}}{{\log }_{e}}a}{2\sqrt{1-{{t}^{2}}}}\]

\[=\frac{x\,{{\log }_{e}}a}{2\sqrt{1-{{t}^{2}}}}\]        \[[\because x=\sqrt{{{a}^{{{\sin }^{-1}}t}}}]\]

On differentiating both sides of Eq. (ii) w,r.t. t, we get

\[\frac{dy}{dt}=\frac{1}{2}{{({{a}^{{{\cos }^{-1}}t}})}^{\frac{-1}{2}}}\cdot \frac{d}{dt}({{a}^{{{\cos }^{-1}}t}})\]

\[=\frac{1}{2}{{({{a}^{{{\cos }^{-1}}t}})}^{\frac{-1}{2}}}({{a}^{{{\cos }^{-1}}t}}{{\log }_{e}}a)\frac{d}{dt}({{\cos }^{-1}}t)\]

\[=\frac{1}{2}{{({{a}^{{{\cos }^{-1}}t}})}^{\frac{1}{2}}}({{\log }_{e}}a)\frac{d}{dt}\frac{-1}{\sqrt{1-{{t}^{2}}}}\]

\[=\frac{-y{{\log }_{e}}a}{2\sqrt{1-{{t}^{2}}}}\]         \[[\because y=\sqrt{{{a}^{{{\cos }^{-1}}t}}}]\]

Now, \[\frac{dy}{dx}=\frac{dy/dt}{dx/dt}=\frac{-ylo{{g}_{e}}a}{2\sqrt{1-{{t}^{2}}}}\cdot \frac{2\sqrt{1-{{t}^{2}}}}{x\,lo{{g}_{e}}a}=\frac{-y}{x}\]

Hence proved.

OR

Given,   \[f(x)={{x}^{3}}+b{{x}^{2}}+ax\]

and Rolle's theorem holds for \[f(x)\] in {1, 3}.

\[\therefore \]      f (1) = f (3)

\[\Rightarrow \]   \[{{(1)}^{3}}+b\,{{(1)}^{2}}+a\,(1)={{(3)}^{3}}+b\,{{(3)}^{2}}+a\,(3)\]

[putting x = 1 and x = 3 in Eq.    ? (i)

\[\Rightarrow \]   \[1+b\,+a=27+9b\,+3a\]

\[\Rightarrow \]\[2a+8b+26=0\] \[\Rightarrow \] \[a+4b+13=0\]?(ii)

[dividing both sides by 2]

and       f ?(c) = 0

\[\Rightarrow \]   \[3{{c}^{2}}+2bc+a=0\]

\[[\because \,\,f'(x)=3{{x}^{2}}+2bx+a]\]

Given,   \[c=2+\frac{1}{\sqrt{3}}\]

\[\therefore \] \[f'\left( 2+\frac{1}{\sqrt{3}} \right)=0\] \[\Rightarrow \]\[3{{\left( 2+\frac{1}{\sqrt{3}} \right)}^{2}}\]

\[+2b\left( 2+\frac{1}{\sqrt{3}} \right)+a=0\]

\[\Rightarrow \]   \[3\left( 4+\frac{1}{3}+\frac{4}{\sqrt{3}} \right)+4b+\frac{2b}{\sqrt{3}}+a=0\]

\[\Rightarrow \]   \[3\times \frac{13}{3}+\frac{12}{\sqrt{3}}+4b+\frac{2b}{\sqrt{3}}+a=0\]

\[\Rightarrow \]   \[a+4b+13+\frac{12}{\sqrt{3}}+\frac{2b}{\sqrt{3}}=0\]

\[\Rightarrow \]   \[0+\frac{12}{\sqrt{3}}+\frac{2b}{\sqrt{3}}=0\] [from Eq. (i)]

\[\Rightarrow \]   \[b=\frac{-\,\sqrt{3}}{2}\times \frac{12}{\sqrt{3}}=-\,6\]

On putting the value of b in Eq. (ii), we get

\[a+4\,(-\,6)+13=0\]\[\Rightarrow \] \[a-24+13=0\] \[\Rightarrow \]a = 11

Hence, the values of a and b are 11 and \[-\,6\]respectively.