If $y = f(x) = \frac{{ax + b}}{{cx - a}}$, then $x$ is equal to
$1/f(x)$
$1/f(y)$
$yf(x)$
$f(y)$
(d) $y = \frac{{ax + b}}{{cx – a}}$
$⇒ x(cy – a) = b + ay$
$⇒ x = \frac{{ay + b}}{{cy – a}} = f(y)$.
Let $P(x)$ be a polynomial with real coefficients such that $P\left(\sin ^2 x\right)=P\left(\cos ^2 x\right)$ for all $x \in[0, \pi / 2)$. Consider the following statements:
$I.$ $P(x)$ is an even function.
$II.$ $P(x)$ can be expressed as a polynomial in $(2 x-1)^2$
$III.$ $P(x)$ is a polynomial of even degree.
Then,
If $f\left( x \right) = {\log _e}\,\left( {\frac{{1 – x}}{{1 + x}}} \right)$, $\left| x \right| < 1$, then $f\left( {\frac{{2x}}{{1 + {x^2}}}} \right)$ is equal to
For a real number $x,\;[x]$ denotes the integral part of $x$. The value of $\left[ {\frac{1}{2}} \right] + \left[ {\frac{1}{2} + \frac{1}{{100}}} \right] + \left[ {\frac{1}{2} + \frac{2}{{100}}} \right] + …. + \left[ {\frac{1}{2} + \frac{{99}}{{100}}} \right]$ is
Which of the following is true
Show that the function $f: R_* \rightarrow R_*$ defined by $f(x)=\frac{1}{x}$ is one-one and onto, where $R_*$ is the set of all non-zero real numbers. Is the result true, if the domain $R_*$ is replaced by $N$ with co-domain being same as $R _*$ ?
Confusing about what to choose? Our team will schedule a demo shortly.