Example- 13

For any real x, show that the value of the expression \(y = 2\left( {k - x} \right)\left( {x + \sqrt {{x^2} + {k^2}} } \right)\) cannot exceed \(2{k^2}\)

Solution: We have upon rearranging:

\[\begin{align}&\qquad x + \sqrt {{x^2} + {k^2}} = \frac{y}{{2\left( {k - x} \right)}} \ldots (1)\\ &\Rightarrow \frac{{{k^2}}}{{\sqrt {{x^2} + {k^2}} - x}} = \frac{y}{{2\left( {k - x} \right)}}\end{align}\]

\[ \Rightarrow \sqrt {{x^2} + {k^2}} - x = \frac{{2{k^2}\left( {k - y} \right)}}{x} \ldots (2)\]

By (1) – (2) and subsequent rearrangement, we can form a quadratic in x:

\[4\left( {{k^2} - y} \right){x^2} - 4k\left( {2{k^2} - y} \right)x + \left( {4{k^4} - {y^2}} \right) = 0 \ldots (3)\]

Now, we impose the condition that the roots of (3) must be real:

\[\begin{array}{l}16{k^2}{\left( {2{k^2} - y} \right)^2} \ge 16\left( {{k^2} - y} \right)\left( {4{k^4} - {y^2}} \right)\\ \Rightarrow {y^2}\left( {y - 2{k^2}} \right) \le 0\\ \Rightarrow y \le 2{k^2}\end{array}\]

Therefore, the maximum value of y is \(2{k^2}\)

Example- 14

Find the real values of a for which the equation

\[{\left( {{{\tan }^2}\theta + 1} \right)^2} + 4a\tan \theta \left( {{{\tan }^2}\theta + 1} \right) + 16{\tan ^2}\theta = 0\]

has four distinct real roots in \(\begin{align}&\left( {0,\;\frac{\pi }{2}} \right)\end{align}\) .

Solution: Dividing throughout by \({\tan ^2}\theta \) , we have

\[{\left( {\tan \theta + \cot \theta } \right)^2} + 4a\left( {\tan \theta + \cot \theta } \right) + 16 = 0\]

Using \(\tan \theta + \cot \theta = y\) ,

\[{y^2} + 4ay + 16 = 0 \Rightarrow y = \tan \theta + \cot \theta = - 2\left( {a \pm \sqrt {{a^2} - 4} } \right) \ldots (1)\]

Now, let \(a \pm \sqrt {{a^2} - 4} \) be represented as b. Thus,

\[y = \tan \theta + \frac{1}{{\tan \theta }} = - 2b \Rightarrow \tan \theta = - b \pm \sqrt {{b^2} - 1} \ldots (2)\]

For real roots, we have

From (1) : \({a^2} > 4 \Rightarrow a < - 2\;{\rm{or}}\;a > 2\)

From (2) : \({b^2} > 1 \Rightarrow {\left( {a \pm \sqrt {{a^2} - 4} } \right)^2} > 1\)

\[ \Rightarrow - \frac{5}{2} < a < \frac{5}{2}\] (Verify!)

Thus, \(\begin{align}&a \in \left( { - \frac{5}{2}, - 2} \right) \cup \left( {2,\;\frac{5}{2}} \right)\end{align}\)

However, we have not paid attention to one particular fact: that we need four distinct roots in \(\begin{align}&\left( {0,\;\frac{\pi }{2}} \right)\end{align}\) . For that, we need to take only negative values of a, that is, \(\begin{align}& a \in \left( { - \frac{5}{2}, - 2} \right)\end{align}\) . The justification of why this should be so is left to the reader as an exercise.

Example- 15

If \({x^2} - 10ax - 11b = 0\) has roots c and d, and \({x^2} - 10cx - 11d = 0\) has roots a and b, find\(a + b + c + d\).

Solution: We have

\[\begin{array}{l}a + b = 10c,\;\;c + d = 10a \Rightarrow \left( {a - c} \right) + \left( {b - d} \right) = 10\left( {c - a} \right)\\ \Rightarrow \left( {b - d} \right) = 11\left( {c - a} \right)\end{array}\]

Now, since c is a root of \({x^2} - 10ax - 11b = 0\), we have

\(\) \[{c^2} - 10ac - 11b = 0 \ldots (1)\]

Similarly

\[{a^2} - 10ac - 11d = 0 \ldots (2)\]

By (1) - (2),

\[\begin{align}&\qquad \left( {{c^2} - {a^2}} \right) = 11\left( {b - d} \right)\\ &\Rightarrow c + a = \frac{{11\left( {b - d} \right)}}{{c - a}} = 121\\ &\Rightarrow a + b + c + d = 10\left( {a + c} \right) = 1210\end{align}\]

Learn math from the experts and clarify doubts instantly

  • Instant doubt clearing (live one on one)
  • Learn from India’s best math teachers
  • Completely personalized curriculum