1. Scenario-1: x=0,y=0 x=0, y=0 x=0,y=0 and x=10 x=10 x=10 are the equation of three straight lines.
Scenario-2: The equation of two circles are x2+y2−12x+16y−69=0 x^{2}+y^{2}-12 x+16 y-69=0 x2+y2−12x+16y−69=0 and x2+y2−9x+12y−59=0 x^{2}+y^{2}-9 x+12 y-59=0 x2+y2−9x+12y−59=0
Find the length of the tangent drawn from the point (3,2) (3,2) (3,2) to the circle 2x2+2y2−6x−7=0 2 x^{2}+2 y^{2}-6 x-7=0 2x2+2y2−6x−7=0.
Obtain the equation of the circle which touch the three straight lines of scenario-1.
Find the equation of the circle whose diameter is the common chord of two circles of scenario-2.
2.
Find the co-ordinates of the trisection points of the intercept of the line AB \mathrm{AB} AB made by the axis of coordinate.
Obtain the equation of the straight line which passes through the point (7,9) (7,9) (7,9) and makes an angle 45∘ 45^{\circ} 45∘ with the line AB \mathrm{AB} AB.
Find the co-ordinates of the point D \mathrm{D} D.
3. A=(1−23),X=(xyz),B=(1−231504−21),C= \mathrm{A}=\left(\begin{array}{lll}1 & -2 & 3\end{array}\right), \mathrm{X}=(\mathrm{x} y \mathrm{z}), \mathrm{B}=\left(\begin{array}{rrr}1 & -2 & 3 \\ 1 & 5 & 0 \\ 4 & -2 & 1\end{array}\right), \mathrm{C}= A=(1−23),X=(xyz),B=114−25−2301,C= ((m+n2)l2l2 m2(n+l)2 m2n2n2(l+m)2) \left(\begin{array}{ccc}\left(\mathrm{m}+\mathrm{n}^{2}\right) & l^{2} & l^{2} \\ \mathrm{~m}^{2} & (\mathrm{n}+l)^{2} & \mathrm{~m}^{2} \\ \mathrm{n}^{2} & \mathrm{n}^{2} & (l+\mathrm{m})^{2}\end{array}\right) (m+n2) m2n2l2(n+l)2n2l2 m2(l+m)2
If 3(1−124)+E=I2 3\left(\begin{array}{rr}1 & -1 \\ 2 & 4\end{array}\right)+E=I_{2} 3(12−14)+E=I2, then find E E E.
Solve the system of equation BXT=AT B X^{T}=A^{T} BXT=AT.
Show that, ∣C∣=2lmn(1+m+n)3 |C|=2 \operatorname{lmn}(1+m+n)^{3} ∣C∣=2lmn(1+m+n)3.
4. Scenario-1 : f(x)=1(4+x2)32 f(x)=\frac{1}{\left(4+x^{2}\right)^{\frac{3}{2}}} f(x)=(4+x2)231
Scenario-2: x2+y2=64 x^{2}+y^{2}=64 x2+y2=64
y=5 y=5 y=5
Determine ∫sin9x.sin11xdx \int \sin 9 x . \sin 11 x d x ∫sin9x.sin11xdx.
Using scenario-1, determine ∫04f(x)dx \int_{0}^{4} f(x) d x ∫04f(x)dx.
Find the area of the smaller part bounded by the circle and the line of scenario-2.
5. PQR \mathrm{PQR} PQR is a triangle.
Prove that, 2cosx=2+2+2cos4x 2 \cos x=\sqrt{2+\sqrt{2+2 \cos 4 x}} 2cosx=2+2+2cos4x.
From the stem, prove that, 1+4sinQ+R4⋅sinR+P4⋅sinP+Q4=sinP2+sin 1+4 \sin \frac{\mathrm{Q}+\mathrm{R}}{4} \cdot \sin \frac{\mathrm{R}+\mathrm{P}}{4} \cdot \sin \frac{\mathrm{P}+\mathrm{Q}}{4}=\sin \frac{\mathrm{P}}{2}+\sin 1+4sin4Q+R⋅sin4R+P⋅sin4P+Q=sin2P+sin Q2+sinR2 \frac{\mathrm{Q}}{2}+\sin \frac{\mathrm{R}}{2} 2Q+sin2R.
From the stem, prove that, p3cos(Q−R)+q3cos(R−P)+r3cos(P−Q)=3 \mathrm{p}^{3} \cos (\mathrm{Q}-\mathrm{R})+\mathrm{q}^{3} \cos (\mathrm{R}-\mathrm{P})+\mathrm{r}^{3} \cos (\mathrm{P}-\mathrm{Q})=3 p3cos(Q−R)+q3cos(R−P)+r3cos(P−Q)=3 pqr.