\(m \leq 5\)
\(2 x^{2}+2 m x+m+1>0\)
For real roots
\(4 m^{2}-8(m+1)>0 \)
\(4 m^{2} - 8 m-8>0 \)
\(m^{2}-2 m-2>0 \)
\(\text {Then } m \text { can be } 5,4,3\)
\(n(S)=5 ,n(A)=3 ,P(A)=\frac{3}{5} \)

\(\int \frac{x+2}{\sqrt{x^{2}-1}} d x =\int\left(\frac{x}{\sqrt{x^{2}-1}} d x+\frac{2}{\sqrt{x^{2}-1}}\right) d x \)
\(=\sqrt{x^{2}-1}+2 \log \left|x+\sqrt{x^{2}-1}\right|+c \)

\(y=f\left(x^{2}+2\right) \)
\(\frac{d y}{d x} =f^{\prime}\left(x^{2}+2\right)(2 x) \)
\(\text { At } x =1, \frac{d y}{d x}=f^{\prime}(1+2)(2)=f^{\prime}(3)(2) \)
\(=5(2)=10 \)

\(2 X=\left[\begin{array}{ll} 3 & 8 \\ 7 & 2 \end{array}\right]-\left[\begin{array}{ll} 1 & 2 \\ 3 & 4 \end{array}\right]=\left[\begin{array}{cc} 2 & 6 \\ 4 & -2 \end{array}\right]\)
\(X=\left[\begin{array}{cc} 1 & 3 \\ 2 & -1 \end{array}\right]\)

\(\left(a t^{2}, 2 a t\right) \Rightarrow x=a t^{2}, \quad y=2 a t\)
\(y^{2} =4 a^{2} t^{2} \)
\(=4 a^{2}\left(\frac{x}{a}\right) \)
\(y^{2} =4 a x \)

No. of ways of answering = 5³ = 125
Correct answer - 1
.'. Number of incorrect answer = 125 - 1 = 124

\(\frac{1}{\sqrt{3}}, \frac{1}{\sqrt{3}+\sqrt{2}}, \frac{1}{\sqrt{3}+2 \sqrt{2}}, \ldots \ldots \text { form } a \text { HP. }\)

\(\cos 28^{\circ}+\sin 28^{\circ} =\mathrm{k}^{3} \)
\(\cos 28^{\circ}+\sin \left(90^{\circ}-62^{\circ}\right) =\mathrm{k}^{3} \)
\(\cos 28^{\circ}+\cos 62^{\circ} =\mathrm{k}^{3} \)
\(2 \cos 45^{\circ} \cos 17^{\circ} =\mathrm{k}^{3} \)
\(2 \frac{1}{\sqrt{2}} \cos 17^{\circ} =\mathrm{k}^{3} \)
\(\cos 17^{\circ} =\frac{\mathrm{k}^{3}}{\sqrt{2}} \)

It is onto not one-one

\(\text { Since } \sin 30^{\circ}=\frac{1}{2}\)
\(\sin 150^{\circ}=\frac{1}{2}\)