Some Random Questions And Answers





Question No 1

If \(-1+p+q = 1 - r \), Where \(p, q \) and \(r\) are constants, then solve the given relation:

\(\frac{\sqrt{x-1}-q-r}{p} + \frac{\sqrt{x-1}-r-p}{q} + \frac{\sqrt{x-1}-p-q}{r} = 3 \)

Solution:


Given,

\( -1+p+q=1-r \)

or, \(p+q+r=1+1\)

or, \( p+q+r = 2 \)

Now

\(\phantom{or,}\frac{\sqrt{x-1}-q-r}{p} + \frac{\sqrt{x-1}-r-p}{q} + \frac{\sqrt{x-1}-p-q}{r} =3\)

or, \(\frac{\sqrt{x-1}-q-r}{p}-1 + \frac{\sqrt{x-1}-r-p}{q} -1+ \frac{\sqrt{x-1}-p-q}{r}-1 =3-1-1-1\)

or, \(\frac{\sqrt{x-1}-q-r-p}{p} + \frac{\sqrt{x-1}-r-p-q}{q} + \frac{\sqrt{x-1}-p-q-r}{r} =0\)

or, \( (\sqrt{x-1}-p-q-r)\left( \frac{1}{p}+\frac{1}{q}+\frac{1}{r}\right)=0\)

or, \( \sqrt{x-1}-p-q-r =0 \)

or, \(\sqrt{x-1}=p+q+r\)

or, \( \sqrt{x-1} =2\)

Squaring on both sides,

\( \left( \sqrt{x-1} \right)^2 = 2^2 \)

or, \(x-1 = 4 \) 

or, \( x = 4+1 \) 

\(\therefore x =5 \) Ans.



Question No 2

\(\frac{1}{x+a}+\frac{1}{x+b}+\frac{1}{x+c}+\frac{ax}{x^3+ax^2}+\frac{bx}{x^3+bx^2}+\frac{cx}{x^3+cx^2}\)

\(=\frac{1}{x+a}+\frac{ax}{x^3+ax^2}+\frac{1}{x+b}+\frac{bx}{x^3+bx^2}+\frac{1}{x+c}+\frac{cx}{x^3+cx^2}\)

\(=\frac{1}{x+a}+\frac{ax}{x^2(x+a)}+\frac{1}{x+b}+\frac{bx}{x^2(x+b)}+\frac{1}{x+c}+\frac{cx}{x^2(x+c)} \)

\(={\color{red}{\frac{1}{x+a}+\frac{a}{x(x+a)}}}+{\color{blue}{\frac{1}{x+b}+\frac{b}{x(x+b)}}}+\frac{1}{x+c}+\frac{c}{x(x+c)}\)

\(={\color{red}{\frac{x+a}{x(x+a)}}}+{\color{blue}{\frac{x+b}{x(x+b)}}}+\frac{x+c}{x(x+c)}\)

\(=\frac{1}{x}+\frac{1}{x}+\frac{1}{x}\)

\(=\frac{3}{x} \)


Question No. 3

Simplify: \( -1+\dfrac{a}{2(a+b)-\dfrac{a+b}{1-\dfrac{b}{a+b}}} \)

Solution:

\( =-1+\dfrac{a}{2(a+b)-\dfrac{a+b}{1-\dfrac{b}{a+b}}} \)

\( =-1+\dfrac{a}{2(a+b)-\dfrac{a+b}{\dfrac{a+b-b}{a+b}}} \)

\( =-1+\dfrac{a}{2(a+b)-\dfrac{a+b}{\dfrac{a}{a+b}}} \)

\( =-1+\dfrac{1}{2(a+b)-\dfrac{(a+b)^2}{a}} \)

\(=-1+\dfrac{a}{\dfrac{2a(a+b)-(a+b)^2}{a}}\)

\(=-1+\dfrac{a^2}{(a+b)\{2a-(a+b)\}}\)

\(=-1+\dfrac{a^2}{(a+b)(2a-a-b)}\)

\(=-1+\dfrac{a^2}{(a+b)(a-b)}\)

\(=-1+\dfrac{a^2}{a^2-b^2}\)

\(=\dfrac{-(a^2-b^2)+a^2}{a^2-b^2}\)

\(=\dfrac{-a^2+b^2+a^2}{a^2-b^2}\)

\(=\dfrac{b^2}{a^2-b^2}\) Ans

By Ambik



Question No 4 

Solve: \( \sqrt{6+\sqrt{6+\sqrt{6+\sqrt{6+\ldots \infty }}}} \)

Solution:

Let \(x=\sqrt{6+\sqrt{6+\sqrt{6+\sqrt{6+\ldots \infty }}}} \)

Squaring on both sides, we get,\\

\(\phantom{or,} x^2 = 6+\sqrt{6+\sqrt{6+\sqrt{6+\sqrt{6+\ldots \infty }}}} \)

or, \( x^2 = 6 +x \)

or, \(x^2 - x -6 =0 \) 

or, \(x^2 -3x+2x -6 = 0 \)

or, \(x(x-3)+2(x-3)=0\)

or, \( (x-3)(x+2)=0\)

\( \implies x = 3 , -2 \)

Since there are only positive sign,

\(x=3\) Ans.


Question No. 5

If \(f(x)=3^x+3^{-x}\) and the functions \(g\) and \(h\) are inverses of each other such that \( goh(x^2+2)=f\left(\frac{-2}{3}\right)\), find the value of \(x\).

Solution:

Here, \(f(x)=3^x+3^{-x}\)

or, \( f\left(\frac{-2}{3}\right) = 3^{\dfrac{-2}{3}}+f^{\dfrac{2}{3}}\)

Also, \(g\) and \(h\) are inverses of each other means 

\(\therefore goh(x^2+2) = x^2 +2 \)  \( [ \because fof^{-1}(x)=x] \)

Now, \( goh(x^2+2)=f\left(\frac{-2}{3}\right)\)

or, \(\phantom{,,,} x^2+2 = 3^{\frac{-2}{3}}+3^{\frac{2}{3}}\)

or, \( x^2 = 3^{\frac{-2}{3}}-2+3^{\frac{2}{3}}\)

or, \( x^2 = (3^{\frac{-1}{3}})^2-2\times 3^{\frac{-1}{3}} \times 3^{\frac{1}{3}}+(3^{\frac{1}{3}})^2\)

or, \( x^2 = \left(3^{\frac{-1}{3}}-3^{\frac{1}{3}} \right)^2 \)

\( \therefore x = 3^{\frac{-1}{3}}-3^{\frac{1}{3}} \)



Question No. 6

Prove that:

\( \frac{1+\sin A + \cos A }{\sin A + \cos A - 1} +\frac{1+\sin A -\cos A}{\sin A - \cos A - 1}\) 

LHS

\( = \frac{1+\sin A + \cos A }{\sin A + \cos A - 1} +\frac{1+\sin A -\cos A}{\sin A - \cos A - 1}\) 

\({\color{blue}{=\frac{(1+\sin A+\cos A)(\sin A -\cos A - 1)+(1+\sin A - \cos A )(\sin A + \cos A -1)}{(\sin A + \cos A -1)(\sin A -\cos A - 1 ) }}} \)

\(=\frac{\{\sin A+(1+\cos A)\}\{\sin A -(1+ \cos A)\}+\{\sin A +(1 - \cos A )\}\{\sin A -(1- \cos A)\}}{\{(\sin A -1)+ \cos A\}\{(\sin A -1)-\cos A ) } \)

\({\color{blue}{=\frac{\sin^2 A - (1+\cos A)^2 + \sin^2 A - (1-\cos A )^2}{(\sin A -1)^2 - \cos^2 A} } }[\because (a-b)(a+b)=a^2-b^2] \)

\(=\frac{\sin^2 A - (1+2\cos A + \cos^2 A) + \sin^2 A - (1 - 2\cos A + \cos^2 A )}{(\sin^2 A -2\sin A + 1) - \cos^2 A } \)

\({\color{blue}{=\frac{\sin^2 A - 1 -2\cos A - \cos^2 A + \sin^2 A - 1 +2\cos A - \cos^2 A }{\sin^2 A - 2\sin A + 1 - \cos^2 A }}} \)

\( = \frac{2\sin^2 A - 2\cos^2A - 2 }{\sin^2A - 2\sin A + 1 - (1-\sin^2 A ) }\)

\( {\color{blue}{= \frac{ 2\sin^2 A - 2(1-\sin^2 A ) - 2 }{\sin^2 A - 2 \sin A + 1 - 1 + \sin^2 A }}}\)

\(=\frac{2\sin^2 A - 2 + 2\sin^2 A - 2 }{2\sin^2 A - 2\sin A}\)

\( = \frac{4\sin^2 A - 4 }{2 \sin A ( \sin A - 1 )} \)

\(=\frac{4(\sin^2 A - 1)}{2\sin A(\sin A -1)}\)

\(=\frac{2(\sin^2 A - 1^2)}{\sin A (\sin A -1)}\)

\(=\frac{2(\sin A -1)(\sin A + 1)}{\sin A (\sin A -1)}\)

\(=\frac{2(\sin A + 1)}{\sin A}\)

\(=2\left(\frac{\sin A}{\sin A}+\frac{1}{\sin A}\right)\)

\(=2(1+\text{ cosec }A )\)

\(=\) RHS


\( \text{Prove that }:\)\(\cos^2 \theta+\sin^2 \theta.\cos2\beta=cos^2\beta+\sin^2\beta\cos2\theta \)


\( \text{Solution:}\)

\(\begin{align*} \text{LHS }&=\cos^2 \theta+\sin^2 \theta.\cos2\beta\\

&=\cos^2 \theta+\sin^2 \theta.(1-2\sin^2\beta)\\

&=\cos^2 \theta+\sin^2 \theta-2\sin^2\theta\sin^2\beta\\

&=1-2\sin^2\theta\sin^2\beta\\

&=\cos^2 \beta+\sin^2 \beta-2\sin^2\theta\sin^2\beta\\

&=\cos^2 \beta+\sin^2 \beta (1-2\sin^2\theta)\\

&=cos^2\beta+\sin^2\beta\cos2\theta\\

&=RHS

\end{align*}\)

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