1. scenario-1: g(x)=cot−1x g(x)=\cot ^{-1} x g(x)=cot−1x
scenario-2: y2=2x y^{2}=2 x y2=2x.
Find ∫dx5−3x2 \int \frac{d x}{\sqrt{5-3 x^{2}}} ∫5−3x2dx
Find the ∫3xg(x)dx \int^{\sqrt{3}} \mathrm{xg}(\mathrm{x}) \mathrm{dx} ∫3xg(x)dx with the help of scenario-1.
Find the area bounded by scenario- 2 and the straight line x=3y x=3 y x=3y.
2. P=(1531−1612−5),Q=(690),R=(xyz) P=\left(\begin{array}{rrr} 1 & 5 & 3 \\ 1 & -1 & 6 \\ 1 & 2 & -5 \end{array}\right), Q=\left(\begin{array}{l} 6 \\ 9 \\ 0 \end{array}\right), R=\left(\begin{array}{l} x \\ y \\ z \end{array}\right) P=1115−1236−5,Q=690,R=xyz
Explain the competence of multiplication of two matrices.
If f(x)=x2−3x f(x)=x^{2}-3 x f(x)=x2−3x, then determine f(p) f(\mathrm{p}) f(p).
Solve the system linear equation which are obtained from the PTR=Q \mathrm{P}^{\mathrm{T}} \mathrm{R}=\mathrm{Q} PTR=Q with the help of determinant.
3. L(2,−1),M(−3,3) \mathrm{L}(2,-1), \mathrm{M}(-3,3) L(2,−1),M(−3,3) and 2x−y+1=0 2 \mathrm{x}-\mathrm{y}+1=0 2x−y+1=0
Find the equation of locus of points which ate always at a distance 5 units from the point (1,1) (1,1) (1,1).
Determine the perpendicular bisector of the straight line which is joining by the two points L \mathrm{L} L and M \mathrm{M} M.
Determine the such type of straight line equation (13) \left(\frac{1}{3}\right) (31) with stem straight line.
4. g(x)=x3−3xy+y3−15f(x)=23x3+112x2−6x+5 \begin{array}{l}g(x)=x^{3}-3 x y+y^{3}-15 \\ f(x)=\frac{2}{3} x^{3}+\frac{11}{2} x^{2}-6 x+5\end{array} g(x)=x3−3xy+y3−15f(x)=32x3+211x2−6x+5
Find the value of limx→01+2x−1−3xx \lim _{x \rightarrow 0} \frac{\sqrt{1+2 x}-\sqrt{1-3 x}}{x} limx→0x1+2x−1−3x
Find the equation of tangent which is drawn on the curve line g(x)=0 g(x)=0 g(x)=0 of the stem at point (2,1) (2,1) (2,1).
Find the maximum and minimum value of f(x) f(x) f(x) function of the stem.
5. S=(−112−3),T=(3−5−12)U=(abc2a3+12b3+12c3+1a2b2c2) \begin{array}{l}S=\left(\begin{array}{rr}-1 & 1 \\ 2 & -3\end{array}\right), T=\left(\begin{array}{rr}3 & -5 \\ -1 & 2\end{array}\right) \\ U=\left(\begin{array}{ccc}a & b & c \\ 2 a^{3}+1 & 2 b^{3}+1 & 2 c^{3}+1 \\ a^{2} & b^{2} & c^{2}\end{array}\right)\end{array} S=(−121−3),T=(3−1−52)U=a2a3+1a2b2b3+1b2c2c3+1c2
Prove without expansion [a−xa+xb−yb+yc−zc+z]=0 \left[\begin{array}{lll}a & -x & a+x \\ b & -y & b+y \\ c & -z & c+z\end{array}\right]=0 abc−x−y−za+xb+yc+z=0.
Show that, (ST)−1−T−1 S−1 (\mathrm{ST})^{-1}-\mathrm{T}^{-1} \mathrm{~S}^{-1} (ST)−1−T−1 S−1 is a null matrix.
Prove that, ∣U∣=−(2abc+1)(a−b)(b−c)(c−a) |U|=-(2 a b c+1)(a-b)(b-c)(c-a) ∣U∣=−(2abc+1)(a−b)(b−c)(c−a).
