Question

If non-zero real numbers b and c are such that \[min{\text{ }}f\left( x \right) > max{\text{ }}g\left( x \right)\], where \[f\left( x \right) = {\text{ }}{x^2} + 2bx + 2{c^2}\] and\[g\left( x \right) = - {x^2} - 2cx + {b^2}\]\[(x \in R)\], then \[\left| {\dfrac{c}{b}} \right|\;\] lies in the interval:
(a) \[\left( {0,\dfrac{1}{2}} \right)\]
(b) \[\left[ {\dfrac{1}{2},\dfrac{1}{{\sqrt 2 }}} \right)\]
(c) \[\left[ {\dfrac{1}{{\sqrt 2 }},\sqrt 2 } \right]\]
(d) \[\left( {\sqrt 2 ,\infty } \right)\]

Answer 1

Hint: Here first we will find the derivative of f(x) and then find the value of x. Also we will find the double derivative of f(x), if it comes out to be greater than zero then f(x) will have a minimum value at the evaluated value of x. Hence we will find the minimum value of f(x)
Now we will find the derivative of g(x) and then find the value of x. Also we will find the double derivative of g(x), if it comes out to be less than zero then f(x) will have maximum value at the evaluated value of x. Hence we will find the maximum value of g(x) and then finally we will substitute the values in the given expression and find the desired value.

Complete step-by-step answer:
Let us first consider the function f(x):
\[f\left( x \right) = {\text{ }}{x^2} + 2bx + 2{c^2}\]
Differentiating both the sides we get:-
\[\dfrac{d}{{dx}}f\left( x \right) = {\text{ }}\dfrac{d}{{dx}}\left( {{x^2} + 2bx + 2{c^2}} \right)\]
Solving it further we get:-
\[\dfrac{d}{{dx}}f\left( x \right) = \dfrac{d}{{dx}}\left( {{x^2}} \right) + 2b\dfrac{d}{{dx}}\left( x \right) + \dfrac{d}{{dx}}\left( {2{c^2}} \right)\]
Now we know that:-
\[\dfrac{d}{{dx}}\left( {{x^n}} \right) = n{x^{n - 1}}\]
Also, the derivative of constant term is zero.
Applying these formulas we get:-
\[f'\left( x \right) = 2x + 2b\]
Now we will put \[f'\left( x \right) = 0\]
Hence we get:-
\[2x + 2b = 0\]
Evaluating the value of x we get:-
\[
  2x = - 2b \\
   \Rightarrow x = \dfrac{{ - 2b}}{2} \\
   \Rightarrow x = - b \\
 \]
Now we will again differentiate \[f'\left( x \right)\]to find the double derivative of f(x).
Hence on differentiating we get:-
\[\dfrac{d}{{dx}}f'\left( x \right) = \dfrac{d}{{dx}}\left( {2x + 2b} \right)\]
Solving it further we get:-
\[f''\left( x \right) = 2\dfrac{d}{{dx}}\left( x \right) + \dfrac{d}{{dx}}\left( {2b} \right)\]
Now we know that:-
\[\dfrac{d}{{dx}}\left( {{x^n}} \right) = n{x^{n - 1}}\]
Also, the derivative of constant term is zero.
Applying these formulas we get:-
\[
  f''\left( x \right) = 2 + 0 \\
   \Rightarrow f''\left( x \right) = 2 \\
 \]
Now since,
\[f''\left( x \right) > 0\]
Hence this implies that minimum value of f(x) exists at \[x = - b\]
Therefore putting \[x = - b\]in f(x) we get:-
\[f\left( { - b} \right) = {\left( { - b} \right)^2} + 2b\left( { - b} \right) + 2{c^2}\]
Solving it further we get:
\[
  f\left( { - b} \right) = {b^2} - 2{b^2} + 2{c^2} \\
   \Rightarrow f\left( { - b} \right) = - {b^2} + 2{c^2} \\
 \]
Hence,
\[\min f\left( x \right) = - {b^2} + 2{c^2}\]………………………………(1)
Now we will consider the function g(x):-
Differentiating both the sides we get:-
\[\dfrac{d}{{dx}}g\left( x \right) = \dfrac{d}{{dx}}\left( { - {x^2} - 2cx + {b^2}} \right)\]
Solving it further we get:-
\[\dfrac{d}{{dx}}g\left( x \right) = - \dfrac{d}{{dx}}\left( {{x^2}} \right) - 2c\dfrac{d}{{dx}}\left( x \right) + \dfrac{d}{{dx}}\left( {{b^2}} \right)\]
Now we know that:-
\[\dfrac{d}{{dx}}\left( {{x^n}} \right) = n{x^{n - 1}}\]
Also, the derivative of constant term is zero.
Applying these formulas we get:-
\[g'\left( x \right) = - 2x - 2c\]
Now we will put \[g'\left( x \right) = 0\]
Hence we get:-
\[ - 2x - 2c = 0\]
Evaluating the value of x we get:-
\[
   - 2x = 2c \\
   \Rightarrow x = \dfrac{{2c}}{{ - 2}} \\
   \Rightarrow x = - c \\
 \]
Now we will again differentiate \[g'\left( x \right)\]to find the double derivative of g(x).
Hence on differentiating we get:-
\[\dfrac{d}{{dx}}g'\left( x \right) = \dfrac{d}{{dx}}\left( { - 2x - 2c} \right)\]
Solving it further we get:-
\[g''\left( x \right) = - 2\dfrac{d}{{dx}}\left( x \right) - \dfrac{d}{{dx}}\left( {2c} \right)\]
Now we know that:-
\[\dfrac{d}{{dx}}\left( {{x^n}} \right) = n{x^{n - 1}}\]
Also, the derivative of constant term is zero.
Applying these formulas we get:-
\[
  g''\left( x \right) = - 2 + 0 \\
   \Rightarrow g''\left( x \right) = - 2 \\
 \]
Now since,
\[g''\left( x \right) < 0\]
Hence this implies that maximum value of g(x) exists at \[x = - c\]
Therefore putting \[x = - c\]in g(x) we get:-
\[g\left( { - c} \right) = - {\left( { - c} \right)^2} - 2c\left( { - c} \right) + {b^2}\]
Solving it further we get:
\[
  g\left( { - c} \right) = - {c^2} + 2{c^2} + {b^2} \\
   \Rightarrow g\left( { - c} \right) = {c^2} + {b^2} \\
 \]
Hence,
\[\max g\left( x \right) = {c^2} + {b^2}\]………………………………(2)
Now it is given that:-
\[min{\text{ }}f\left( x \right) > max{\text{ }}g\left( x \right)\]
Hence putting in the values from equation1 and 2 we get:-
\[ - {b^2} + 2{c^2} > {c^2} + {b^2}\]
Solving it further we get:-
\[
  2{c^2} - {c^2} > {b^2} + {b^2} \\
   \Rightarrow {c^2} > 2{b^2} \\
 \]
Simplifying it further we get:-
\[\dfrac{{{c^2}}}{{{b^2}}} > 2\]
Now taking square root of both sides we get:-
\[
  \sqrt {\dfrac{{{c^2}}}{{{b^2}}}} > \sqrt 2 \\
   \Rightarrow \left| {\dfrac{c}{b}} \right| > \sqrt 2 \\
 \]
Hence, \[\left| {\dfrac{c}{b}} \right|\] lies in the interval \[\left( {\sqrt 2 ,\infty } \right)\]

Therefore option D is correct.

Note: Students might make mistakes while observing the values of double derivatives.
They should note that if the double derivative of a function is greater than zero then the minimum value of the function exists at an evaluated value of x.
If the double derivative of a function is less than zero then the maximum value of the function exists at an evaluated value of x.
Also, the modulus value of any quantity is always greater than 0 or positive.
Also,
 \[\dfrac{d}{{dx}}\left( {{x^n}} \right) = n{x^{n - 1}}\]