1. P(x)=cosx,Q(x,y)=x2+y2−25,L=x−3 P(x)=\cos x, Q(x, y)=x^{2}+y^{2}-25, L=x-3 P(x)=cosx,Q(x,y)=x2+y2−25,L=x−3.
Integrate: ∫dxex+e−x \int \frac{d x}{e^{x}+e^{-x}} ∫ex+e−xdx
According to the stem, find the value of ∫dθ1−{P(π2−θ)} \int \frac{d \theta}{1-\left\{P\left(\frac{\pi}{2}-\theta\right)\right\}} ∫1−{P(2π−θ)}dθ
According to the stem, find the small area enclosed by Q(x,y)=0 \mathrm{Q}(\mathrm{x}, \mathrm{y})=0 Q(x,y)=0 and L=0 \mathrm{L}=0 L=0.
2. Scenario-II: x2+y2+2x+3y+1=0 x^{2}+y^{2}+2 x+3 y+1=0 x2+y2+2x+3y+1=0 is an equation of the circle.
Determine the centre and radius of the circle x2+y2−12x−8y+34=0 x^{2}+y^{2}-12 x-8 y+34=0 x2+y2−12x−8y+34=0.
Find the equation of the circle in scenario-I.
Find the equation and length of tangent drawn from origin (0,0) (0,0) (0,0) to the circle of scenario-II.
3. f(x)=ln(x),g(x)=(x+1+x2) f(x)=\ln (x), g(x)=\left(x+\sqrt{1+x^{2}}\right) f(x)=ln(x),g(x)=(x+1+x2).
Differentiate with respect to x x x of (cos−1x) \left(\cos ^{-1} \sqrt{x}\right) (cos−1x)
Find the derivative of f(x) f(x) f(x), using first principle according to the stem.
According to the stem prove that, (1+x2)d2zdx2+xdzdx \left(1+x^{2}\right) \frac{d^{2} z}{d x^{2}}+x \frac{d z}{d x} (1+x2)dx2d2z+xdxdz −m2z=0 -m^{2} z=0 −m2z=0 for z={g(x)}m z=\{g(x)\}^{m} z={g(x)}m.
4. M=[a−522a−2],N=[−12−32104−25]P=[−2a+b−c−2b+c−aa+b−cc2ab] \text { } \begin{aligned} \mathrm{M} & =\left[\begin{array}{cc} \mathrm{a}-5 & 2 \\ 2 & \mathrm{a}-2 \end{array}\right], \mathrm{N}=\left[\begin{array}{rrr} -1 & 2 & -3 \\ 2 & 1 & 0 \\ 4 & -2 & 5 \end{array}\right] \\ P & =\left[\begin{array}{ccr} -2 & a+b & -c \\ -2 & b+c & -a \\ a+b-c & c^{2} & a b \end{array}\right] \end{aligned} MP=[a−522a−2],N=−12421−2−305=−2−2a+b−ca+bb+cc2−c−aab
M M M is singular matrix if the value of a a a ?
Determine N2−5 N+4I \mathrm{N}^{2}-5 \mathrm{~N}+4 \mathrm{I} N2−5 N+4I.
Prove that, ∣P∣=(c−a)(a2+b2+c2) |\mathrm{P}|=(\mathrm{c}-\mathrm{a})\left(\mathrm{a}^{2}+\mathrm{b}^{2}+\mathrm{c}^{2}\right) ∣P∣=(c−a)(a2+b2+c2).
5.
Find the value of tan42∘tan78∘cot6∘cot66∘ \frac{\tan 42^{\circ} \tan 78^{\circ}}{\cot 6^{\circ} \cot 66^{\circ}} cot6∘cot66∘tan42∘tan78∘.
Prove that, tanE2=p−qp+qcot(S−T2) \tan \frac{E}{2}=\frac{p-q}{p+q} \cot \left(\frac{S-T}{2}\right) tan2E=p+qp−qcot(2S−T).
If p4+q4+r4=2p2(q2+r2) p^{4}+q^{4}+r^{4}=2 p^{2}\left(q^{2}+r^{2}\right) p4+q4+r4=2p2(q2+r2), then show that, S=45∘ S=45^{\circ} S=45∘ or 135∘ 135^{\circ} 135∘.
Scenario-II: 