We know that,
\(\sin \alpha-\cos \alpha =\sqrt{2}\left[\frac{1}{\sqrt{2}} \sin \alpha-\frac{1}{\sqrt{2}} \cos \alpha\right] \)
\(=\sqrt{2}\left[\cos \frac{\pi}{4} \sin \alpha-\sin \frac{\pi}{4} \cos \alpha\right] \)
\(=\sqrt{2} \sin \left(\alpha-\frac{\pi}{4}\right) \)
\(\therefore \lim _{\alpha \rightarrow \frac{\pi}{4}} \frac{\sin \alpha-\cos \alpha}{\alpha-\frac{\pi}{4}} =\lim _{\alpha-\frac{\pi}{4} \rightarrow 0} \frac{\sqrt{2} \sin \left(\alpha-\frac{\pi}{4}\right)}{\left(\alpha-\frac{\pi}{4}\right)}=\sqrt{2} \)

\(A \& B \text { are symmetric } \Rightarrow A+B \text { is symmetric }\)

\(\mathrm{AM} \geq \mathrm{GM} \)
\(\Rightarrow a \geq g \)

Let y = 2x + c be the required line.
Given perpendicular distance from origin'to this line is \(\sqrt5\)
\(\frac{c}{\sqrt{1+4}}=\sqrt{5}\)
c  = 5
This required line is y - 2x + 5
2x - y + 5 = 0

\(\log _{\sqrt{x}} 0.25 =4 \)
\((\sqrt{x})^{4} =0.25 \)
\((\sqrt{x})^{4} =\frac{1}{4} \)
\(x^{2} =\frac{1}{4} \)
\(\Rightarrow x=\frac{1}{2}=0.5\)

\(\cos p \theta+\cos q \theta=0 \)
\(2 \cos \left(\frac{p+q}{2}\right) \theta \cdot \cos \left(\frac{p-q}{2}\right) \theta=0 \)
\(\Rightarrow 2 \cos \left(\frac{p+q}{2}\right) \theta=0 \)
\(\cos \left(\frac{\mathrm{p}-\mathrm{q}}{2}\right) \theta=0\)
\(\text { Principal angle is } \pi / 2\)
\(\therefore\left(\frac{p+q}{2}\right) \theta =n \pi \pm \pi / 2 \)
\(\Rightarrow(p+q) \cdot \theta =2 n \pi \pm \pi \)
\(\theta =\frac{\pi(2 n \pm 1)}{p+q} \)
\(\text { Similarly, } \theta=\frac{\pi(2 n \pm 1)}{p-q}\)

\(\mathrm{T}: x-y \text { is an integer } \Rightarrow x \mathrm{R} y\)
\(\text { i) } x-x=0 \text { is an integer }\)
\(\therefore \text { T is reflexive. }\)
\(\text { ii) }(x-y) \text { is an integer } \Rightarrow y-x \text { is also an integer }\)
\(\therefore \mathrm{T} \text { is symmetry }\)
iii) If (x - y)is an integer and y- z is also an integer, by adding
x - z is also an integer.
\(\therefore \mathrm{T} \text { is transitive }\)
Thus, T is an equivalence relation.
\(\mathrm{S}: \mathrm{y}=x+1 \Rightarrow x \mathrm{~S} y\)
\(\text { i) } x=x+1 \Rightarrow x S x \text { is not true. }\)
\(\therefore S \text { is not reflexive. }\)
Hence T is an equivalence relation but S is not an equivalence relation