# Functions Set 3

**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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