Question

For real x, the expression $\dfrac{{{{\left( {x + m} \right)}^2} - 4mn}}{{2\left( {x - n} \right)}}$ can have any value except
$\left( a \right)$ Between m and (m + n)
$\left( b \right)$ More than (m + 2n)
$\left( c \right)$ Between 2m and 2n
$\left( d \right)$ All values are possible.

Answer 1

Hint – In this particular question use the concept that if a function attains two minima of the same kind in any interval then in this interval the function does not attain any values, so in this question apply the concept of maxima and minima so use these concepts to reach the solution of the question.

Complete step-by-step answer:
Given expression:
$\dfrac{{{{\left( {x + m} \right)}^2} - 4mn}}{{2\left( {x - n} \right)}}$
Let, f (x) = $\dfrac{{{{\left( {x + m} \right)}^2} - 4mn}}{{2\left( {x - n} \right)}}$
Now differentiate the above function w.r.t x we have,
$ \Rightarrow \dfrac{d}{{dx}}f\left( x \right) = f'\left( x \right) = \dfrac{d}{{dx}}\dfrac{{{{\left( {x + m} \right)}^2} - 4mn}}{{2\left( {x - n} \right)}}$
Now as we know that the differentiation of, $\dfrac{d}{{dx}}\left( {\dfrac{m}{n}} \right) = \dfrac{{n\left( {\dfrac{d}{{dx}}m} \right) - m\left( {\dfrac{d}{{dx}}n} \right)}}{{{n^2}}}$ so use this property in the above equation we have,
$ \Rightarrow \dfrac{d}{{dx}}f\left( x \right) = f'\left( x \right) = \dfrac{1}{2}\left( {\dfrac{{\left( {x - n} \right)\dfrac{d}{{dx}}{{\left( {x + m} \right)}^2} - {{\left( {x + m} \right)}^2}\dfrac{d}{{dx}}\left( {x - n} \right) - \left[ {\left( {x - n} \right)\dfrac{d}{{dx}}4mn - 4mn\dfrac{d}{{dx}}\left( {x - n} \right)} \right]}}{{{{\left( {x - n} \right)}^2}}}} \right)$
Now as we know that the differentiation of constant term (i.e. terms which has no x) is zero and $\dfrac{d}{{dx}}{x^n} = n{x^{n - 1}},\dfrac{d}{{dx}}{\left( {x + a} \right)^2} = 2\left( {x + a} \right)\dfrac{d}{{dx}}\left( {x + a} \right)$ so use this property in the above equation we have,
$ \Rightarrow \dfrac{d}{{dx}}f\left( x \right) = f'\left( x \right) = \dfrac{1}{2}\left( {\dfrac{{\left( {x - n} \right)2\left( {x + m} \right) - {{\left( {x + m} \right)}^2}\left( {1 - 0} \right) - \left[ {\left( {x - n} \right)\left( 0 \right) - 4mn\left( {1 - 0} \right)} \right]}}{{{{\left( {x - n} \right)}^2}}}} \right)$
$ \Rightarrow \dfrac{d}{{dx}}f\left( x \right) = f'\left( x \right) = \dfrac{1}{2}\left( {\dfrac{{2\left( {x - n} \right)\left( {x + m} \right) - {{\left( {x + m} \right)}^2} + 4nm}}{{{{\left( {x - n} \right)}^2}}}} \right)$
$ \Rightarrow \dfrac{d}{{dx}}f\left( x \right) = f'\left( x \right) = \dfrac{1}{2}\left( {\dfrac{{2{x^2} + 2mx - 2nx - 2nm - {x^2} - {m^2} - 2mx + 4nm}}{{{{\left( {x - n} \right)}^2}}}} \right)$
$ \Rightarrow \dfrac{d}{{dx}}f\left( x \right) = f'\left( x \right) = \dfrac{1}{2}\left( {\dfrac{{{x^2} - 2nx - {m^2} + 2nm}}{{{{\left( {x - n} \right)}^2}}}} \right)$
\[ \Rightarrow \dfrac{d}{{dx}}f\left( x \right) = f'\left( x \right) = \dfrac{1}{2}\left( {\dfrac{{{x^2} - {m^2} - 2nx + 2nm}}{{{{\left( {x - n} \right)}^2}}}} \right) = \dfrac{1}{2}\left( {\dfrac{{\left( {x - m} \right)\left( {x + m} \right) - 2n\left( {x - m} \right)}}{{{{\left( {x - n} \right)}^2}}}} \right) = \dfrac{{\left( {x - m} \right)}}{2}\left( {\dfrac{{x + m - 2n}}{{{{\left( {x - n} \right)}^2}}}} \right)\]
Now to check the maxima and minima equate the above value to zero we have,
\[ \Rightarrow \dfrac{d}{{dx}}f\left( x \right) = f'\left( x \right) = \dfrac{{\left( {x - m} \right)}}{2}\left( {\dfrac{{x + m - 2n}}{{{{\left( {x - n} \right)}^2}}}} \right) = 0\]
$ \Rightarrow x = m,\left( { - m + 2n} \right)$
Now double differentiate the function we have,
\[ \Rightarrow \dfrac{{{d^2}}}{{d{x^2}}}f\left( x \right) = \dfrac{d}{{dx}}f'\left( x \right) = \dfrac{d}{{dx}}\dfrac{{\left( {x - m} \right)}}{2}\left( {\dfrac{{x + m - 2n}}{{{{\left( {x - n} \right)}^2}}}} \right)\]
\[ \Rightarrow \dfrac{{{d^2}}}{{d{x^2}}}f\left( x \right) = \dfrac{d}{{dx}}f'\left( x \right) = \dfrac{1}{2}\left( {\dfrac{{{{\left( {x - n} \right)}^2}\dfrac{d}{{dx}}\left[ {\left( {x - m} \right)\left( {x + m - 2n} \right)} \right] - \left( {x - m} \right)\left( {x + m - 2n} \right)\dfrac{d}{{dx}}{{\left( {x - n} \right)}^2}}}{{{{\left( {x - n} \right)}^4}}}} \right)\]
\[ \Rightarrow f''\left( x \right) = \dfrac{1}{2}\left( {\dfrac{{{{\left( {x - n} \right)}^2}\left[ {\left( {x - m} \right)\dfrac{d}{{dx}}\left( {x + m - 2n} \right) + \left( {x + m - 2n} \right)\dfrac{d}{{dx}}\left( {x - m} \right)} \right] - \left( {x - m} \right)\left( {x + m - 2n} \right)\dfrac{d}{{dx}}{{\left( {x - n} \right)}^2}}}{{{{\left( {x - n} \right)}^4}}}} \right)\]
\[ \Rightarrow f''\left( x \right) = \dfrac{1}{2}\left( {\dfrac{{{{\left( {x - n} \right)}^2}\left[ {\left( {x - m} \right)\left( 1 \right) + \left( {x + m - 2n} \right)\left( 1 \right)} \right] - \left( {x - m} \right)\left( {x + m - 2n} \right)2\left( {x - n} \right)}}{{{{\left( {x - n} \right)}^4}}}} \right)\]
\[ \Rightarrow f''\left( x \right) = \dfrac{1}{2}\left( {\dfrac{{{{\left( {x - n} \right)}^2}\left( {2x - 2n} \right) - 2\left( {x - m} \right)\left( {x - n} \right)\left( {x + m - 2n} \right)}}{{{{\left( {x - n} \right)}^4}}}} \right)\]
\[ \Rightarrow f''\left( x \right) = \dfrac{1}{2}\left( {\dfrac{{\left( {x - n} \right)\left( {2x - 2n} \right) - 2\left( {x - m} \right)\left( {x + m - 2n} \right)}}{{{{\left( {x - n} \right)}^3}}}} \right) = \left( {\dfrac{{{{\left( {x - n} \right)}^2} - \left( {x - m} \right)\left( {x + m - 2n} \right)}}{{{{\left( {x - n} \right)}^2}}}} \right)\]
Now for, x = m
\[ \Rightarrow f''\left( m \right) = \dfrac{{{{\left( {m - n} \right)}^2} - 0}}{{{{\left( {m - n} \right)}^2}}} = 1\] = positive, so it is a minima.
Now for, x = -m + 2n
\[ \Rightarrow f''\left( { - m + 2n} \right) = \left( {\dfrac{{{{\left( { - m + 2n - n} \right)}^2} - \left( { - m + 2n - m} \right)\left( { - m + 2n + m - 2n} \right)}}{{{{\left( { - m + 2n - n} \right)}^2}}}} \right)\]
\[ \Rightarrow f''\left( { - m + 2n} \right) = \left( {\dfrac{{{{\left( {n - m} \right)}^2} - 0}}{{{{\left( {n - m} \right)}^2}}}} \right) = 1\]= positive, so it is also a minima of the same kind.
So for both the value of x function attain minima of the same kind.
So the function f(x) at these values of x does not attain any maximum values.
So for, x = m
f (x) = $\dfrac{{{{\left( {x + m} \right)}^2} - 4mn}}{{2\left( {x - n} \right)}} = \dfrac{{{{\left( {2m} \right)}^2} - 4mn}}{{2\left( {m - n} \right)}} = \dfrac{{4m\left( {m - n} \right)}}{{2\left( {m - n} \right)}} = 2m$
And for, x = -m + 2n
f (x) = $\dfrac{{{{\left( {x + m} \right)}^2} - 4mn}}{{2\left( {x - n} \right)}} = \dfrac{{{{\left( {2n} \right)}^2} - 4mn}}{{2\left( { - m + n} \right)}} = \dfrac{{4{n^2} - 4mn}}{{2\left( { - m + n} \right)}} = \dfrac{{4n\left( {n - m} \right)}}{{2\left( {n - m} \right)}} = 2n$
Therefore, function does not attain any value between 2m and 2n.
So this is the required answer.
Hence option (C) is the correct answer.

Note – Whenever we face such types of questions the key concept we have to remember is that always recall the properties of maxima and minima i.e. first differentiate the given function and equate to zero and solve for x then double differentiate the given function and check at the value of x which we calculated earlier if the double differentiation is positive at this value of x, then the function attain minima and if the double differentiation is negative at this value of x, then the function attain the maxima.