MCQ
The expression ${\cos ^2}(A - B) + {\cos ^2}B - 2\cos (A - B)\cos A\cos B$ is
- ADependent on $B$
- BDependent on $A$ and $B$
- ✓Dependent on $A$
- DIndependent of $A $ and $B$
$ = {\cos ^2}(A - B) + {\cos ^2}B$
$ - \cos \,(A - B)\,\left\{ {\cos (A - B) + \cos (A + B)} \right\}$
$ = {\cos ^2}B - \cos \,(A - B)\,\,\cos \,\,(A + B)$
$ = {\cos ^2}B - ({\cos ^2}A - {\sin ^2}B) = 1 - {\cos ^2}A$
Hence it depends on $A.$
Trick : Put two different values of $A$.
Let $A = {90^o},$ then the value of expression will be ${\sin ^2}B + {\cos ^2}B = 1$
Now put $A = {0^o}$, then the value of expression will be ${\cos ^2}B + {\cos ^2}B - 2\,\,{\cos ^2}B = 0$
It means that the expression has different values for different $A$
$i.e.$ it depends on $A.$
Now similarly for $B = {90^o},$
the value of expression will be ${\sin ^2}A + 0 - 0$
$ = {\sin ^2}A$ and at $B\,\, = {0^o}$
the value of expression will be ${\cos ^2}A + 1 - 2{\cos ^2}A = {\sin ^2}A$.
Hence, the expression has the same value for different values of $B$,
so it does not depend on $B.$
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$[A]$ $f(x)$ is increasing in $(0, \infty)$
$[B]$ $f(x)$ is decreasing in $(0, \infty)$
$[C]$ $f(x)>e^{2 x}$ in $(0, \infty)$
$[D]$ $f^{\prime}(x) < e^{2 x}$ in $(0, \infty)$