Example – 4

For real x, find the condition on a, b, c such that the function

\(f(x) = \begin{align}\frac{{(x - a)(x - b)}}{{(x - c)}}\,\,\,\,\,\,\,\,\,\,\,\,(\mathbb{R} - \{ c\}  \to \mathbb{R})\end{align}\)  is onto

Solution:    Let \(y = f(x) =\begin{align} \frac{{(x - a)(x - b)}}{{(x - c)}}\end{align}\)  . Since the co-domain for this function has been specified has \(\mathbb{R}\), we require the range to be \(\mathbb{R}\) also if f(x) is to be onto. Hence we require y to take on all real values.

Now   \(y(x - c) = (x - a)(x - b)\)

\[ \Rightarrow  {x^2} - (a + b + y)x + (ab + cy) = 0\]

We again follow the method discussed earlier. For x to be real, the discriminant should be non-negative.

\[\Rightarrow {(a + b + y)^2} - 4(ab + cy) \ge 0\qquad\qquad\qquad   ....(*)\]

For the function f to be onto, we require that each y have a real pre-image x. This is only possible if that y satisfies the constraint (*).  Hence, this constraint, or this inequality, should be true for all real y. Rearanging as a quadratic in y

\[\Rightarrow {y^2} + 2(a + b - 2c)y + {(a - b)^2} \ge 0\qquad\qquad\qquad  ...(**)\]

As we saw in Q3, for the LHS of (**) to be always non-negative, we require its graph to lie above the x-axis (or touching it, at the most). If it goes below the axis, the LHS will become negative.

Hence we require the discriminant for (**) to be non-positive. i.e. D < 0

\[\Rightarrow  {\left( {2(a + b - 2c)} \right)^2} - 4{(a - b)^2}\le 0\]

\[\Rightarrow  {(a + b)^2} + 4{c^2} - 4(a + b)c - {(a - b)^2} \le 0\]

\[\Rightarrow  ab + {c^2} - (a + b)c \le 0\]

As in Q3., we can treat the LHS above as a quadratic in ‘c’. LHS < 0 implies that ‘c’ must lie within the roots of this quadratic expression, which are

\[\frac{{(a + b) \pm \sqrt {{{(a + b)}^2} - 4ab} }}{2} = \frac{{(a + b) \pm (a - b)}}{2} = a,b\]

There is no loss of generality in assuming that a > b since the expression for f(x) is symmetric about ‘a’ and ‘b’.

Hence, we get the constraint on a, b, c as

a < c < b

Question: Why cannot c be equal to a or b?  

Example – 5

Plot the graphs for the following:

(a) \(y = \sin \left( {{{\sin }^{ - 1}}x} \right)\)     (b)  \(y = {\sin ^{ - 1}}(\sin x)\)

Solution: By observation, you will have a tendency to say that since y is a composition of two inverse functions, which should cancel out, the output should be y = x, which is a straight line. But we have to be more cautious:

(a)    The inner function, \({\sin ^{ - 1}}x\), is defined only for x \(\in \) [–1, 1]. For these values of x, \(\sin ({\sin ^{ - 1}}x)\)  will give back x again. Hence, the graph is the identity function but only in the interval [–1, 1].

(b)  The inner function sin x, is defined for all x \(\in \mathbb{R}\) and gives an output in [–1, 1]. The outer function, \({\sin ^{ - 1}}(\,\,)\)   , will now give a value in   \(\left[ {\frac{{ - \pi }}{2},\,\frac{\pi }{2}} \right]\) . For example, if x = \(\pi \) , sin x = 0, and  \({\sin ^{ - 1}}(\sin x) = 0\)       and not \(\pi \). Similarly,  if   \(x = \frac{{3\pi }}{2}\) , sin x = – 1 and \({\sin ^{ - 1}}(\sin x) = \frac{{ - \pi }}{2}\)  and not \(\frac{{3\pi }}{2}\)  , and so on

We can state these things concisely as

\[x = {\sin ^{ - 1}}(\sin x) = x\quad\quad\quad if{\rm{  }}\frac{{ - \pi }}{2} \le x < \frac{\pi }{2}\]

\[\pi  - x{\rm{}}\qquad\qquad\qquad if{\rm{}}\frac{{ - \pi }}{2} + \pi {\rm{}} \le x < \frac{\pi }{2} + \pi \]

\[ + (x{\rm{  - }}2\pi ){\qquad\qquad\qquad\rm{ }}if{\rm{ }}\frac{{ - \pi }}{2} + 2\pi {\rm{}} \le x < \frac{\pi }{2} + 2\pi {\rm{ }}\]

\[3\pi  - x{\rm{}}\qquad\qquad\qquad if{\rm{ }}\frac{{ - \pi }}{2} + 3\pi {\rm{}} \le x \le \frac{\pi }{2} + 3\pi \]

and so on.

Convince yourself that the above formulation is correct. The graph is drawn below:

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