MCQ
If ${a^2},\,{b^2},\,{c^2}$ be in $A.P.$, then $\frac{a}{{b + c}},\,\frac{b}{{c + a}},\,\frac{c}{{a + b}}$ will be in
- ✓$A.P.$
- B$G.P.$
- C$H.P.$
- DNone of these
Then ${b^2} - {a^2} = {c^2} - {b^2}$
$ \Rightarrow $ $(b - a)(b + a) = (c - b)(c + b)$
==> $\frac{{b - a}}{{c + b}} = \frac{{c - b}}{{b + a}}$
==> $\frac{{(b - a)(a + b + c)}}{{(c + a)(b + c)}} = \frac{{(c - b)(a + b + c)}}{{(a + b)(c + a)}}$
$ \Rightarrow $ $\frac{{{b^2} + bc - ac - {a^2}}}{{(c + a)(b + c)}} = \frac{{{c^2} + ac - ab - {b^2}}}{{(a + b)(c + a)}}$
$ \Rightarrow $ $\frac{b}{{c + a}} - \frac{a}{{b + c}} = \frac{c}{{a + b}} - \frac{b}{{c + a}}$
Hence $\frac{a}{{b + c}},\;\frac{b}{{c + a}},\;\frac{c}{{a + b}}$ be in $A.P.$
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| Column $I$ | Column $II$ |
| $(A)$ The set $\left\{\operatorname{Re}\left(\frac{2 i z}{1-z^2}\right): z\right.$ is a complex number, $\left.|z|=1, z \neq \pm 1\right\}$ is | $(p)$ $(-\infty,-1) \cup(1, \infty)$ |
| $(B)$ The domain of the function $f(x)=\sin ^{-1}\left(\frac{8(3)^{x-2}}{1-3^{2(x-1)}}\right)$ is is | $(q)$ $(-\infty, 0) \cup(0, \infty)$ |
| $(C)$ If $f(\theta)=\left|\begin{array}{ccc}1 & \tan \theta & 1 \\ -\tan \theta & 1 & \tan \theta \\ -1 & -\tan \theta & 1\end{array}\right|$, then the set $\left\{f(\theta): 0 \leq \theta<\frac{\pi}{2}\right\}$ is | $(r)$ $[2, \infty)$ |
| $(D)$ If $f(x)=x^{3 / 2}(3 x-10), x \geq 0$, then $f(x)$ is increasing in | $(s)$ $(-\infty,-1] \cup[1, \infty)$ |
| $(t)$ $(-\infty, 0] \cup[2, \infty)$ |