Question

If ${y^{\dfrac{1}{m}}} + {y^{\dfrac{{ - 1}}{m}}} = 2x$ , then $\left( {{x^2} - 1} \right){y_2} + x{y_1}$ is equal to
$
  \left( a \right){m^2}y \\
  \left( b \right) - {m^2}y \\
  \left( c \right) \pm {m^2}y \\
  \left( d \right){\text{None of these}} \\
 $

Answer 1

Hint-In this question, we use the concept of first and second derivatives. We have to differentiate the given equation with respect to x and calculate the value of first derivative ${y_1} = \dfrac{{dy}}{{dx}}$ and second derivative ${y_2} = \dfrac{{{d^2}y}}{{d{x^2}}}$ then put the value of derivatives in $\left( {{x^2} - 1} \right){y_2} + x{y_1}$ .

Complete step-by-step answer:
Given, ${y^{\dfrac{1}{m}}} + {y^{\dfrac{{ - 1}}{m}}} = 2x$
We can write as
$
   \Rightarrow {y^{\dfrac{1}{m}}} + \dfrac{1}{{{y^{\dfrac{1}{m}}}}} = 2x \\
   \Rightarrow {\left( {{y^{\dfrac{1}{m}}}} \right)^2} + 1 = 2x{y^{\dfrac{1}{m}}} \\
   \Rightarrow {\left( {{y^{\dfrac{1}{m}}}} \right)^2} - 2x{y^{\dfrac{1}{m}}} + 1 = 0..............\left( 1 \right) \\
$
Let $t = {y^{\dfrac{1}{m}}}$ and put in (1) equation.
Now, ${\left( t \right)^2} - 2xt + 1 = 0$
We can see the quadratic equation in t. So, we calculate the roots of the quadratic equation by using the Sridharacharya formula. According to Sridharacharya formula, if a quadratic equation $a{x^2} + bx + c = 0$ so the roots of this quadratic equation is $x = \dfrac{{ - b \pm \sqrt {{b^2} - 4ac} }}{{2a}}$ .
$
   \Rightarrow t = \dfrac{{2x \pm \sqrt {4{x^2} - 4} }}{2} \\
   \Rightarrow t = x \pm \sqrt {{x^2} - 1} \\
 $
Now, ${y^{\dfrac{1}{m}}} = x \pm \sqrt {{x^2} - 1} $
$ \Rightarrow y = {\left( {x \pm \sqrt {{x^2} - 1} } \right)^m}................\left( 2 \right)$
Differentiate (2) equation with respect to x.
\[
  {y_1} = \dfrac{{dy}}{{dx}} = \dfrac{d}{{dx}}\left( {{{\left( {x \pm \sqrt {{x^2} - 1} } \right)}^m}} \right) \\
   \Rightarrow {y_1} = m{\left( {x \pm \sqrt {{x^2} - 1} } \right)^{m - 1}}\left( {1 \pm \dfrac{x}{{\sqrt {{x^2} - 1} }}} \right) \\
   \Rightarrow {y_1} = m{\left( {x \pm \sqrt {{x^2} - 1} } \right)^m}\left( {\dfrac{1}{{\sqrt {{x^2} - 1} }}} \right) \\
\]
We know, $y = {\left( {x \pm \sqrt {{x^2} - 1} } \right)^m}$
\[ \Rightarrow {y_1} = \dfrac{{my}}{{\sqrt {{x^2} - 1} }}\]
Now, squaring both sides
\[
   \Rightarrow {\left( {{y_1}} \right)^2} = \dfrac{{{m^2}{y^2}}}{{{x^2} - 1}} \\
   \Rightarrow \left( {{x^2} - 1} \right){\left( {{y_1}} \right)^2} = {m^2}{y^2}...........\left( 3 \right) \\
\]
Differentiate (3) equation with respect to x.
\[ \Rightarrow \dfrac{d}{{dx}}\left( {\left( {{x^2} - 1} \right){{\left( {{y_1}} \right)}^2}} \right) = \dfrac{{d\left( {{m^2}{y^2}} \right)}}{{dx}}\]
Apply product rule of differentiation and ${y_2} = \dfrac{{{d^2}y}}{{d{x^2}}}$ .
\[
   \Rightarrow \left( {{y_1}^2} \right) \times 2x + \left( {{x^2} - 1} \right) \times 2{y_1} \times {y_2} = {m^2} \times 2y \times {y_1} \\
   \Rightarrow 2{y_1}\left( {\left( {{x^2} - 1} \right){y_2} + x{y_1}} \right) = 2{y_1} \times {m^2}y \\
\]
From above equation \[2{y_1}\] cancel from both sides.
\[\left( {{x^2} - 1} \right){y_2} + x{y_1} = {m^2}y\]
So, the correct option is (a).

Note-Whenever we face such types of problems we use some important points. First we convert given equation into y=f(x) by using quadratic formula then differentiate y=f(x) with respect to x and put the values of first and second derivative in $\left( {{x^2} - 1} \right){y_2} + x{y_1}$ . So, we get the required answer.