1. f(x)=sinx f(x)=\sin x f(x)=sinx
Find the value of limx→0x21−cosax \lim _{x \rightarrow 0} \frac{x^{2}}{1-\cos a x} limx→01−cosaxx2.
If y=f(f(x)) y=f(f(x)) y=f(f(x)), then show that, d2ydx2+dydxtanx+[1− \frac{d^{2} y}{d x^{2}}+\frac{d y}{d x} \tan x+[1- dx2d2y+dxdytanx+[1− {f(x)}2]y=0 \left.\{f(x)\}^{2}\right] y=0 {f(x)}2]y=0
If 0≤x≤π2 0 \leq x \leq \frac{\pi}{2} 0≤x≤2π, then find the extreme values of 3+2 3+2 3+2 f(x)+3{f(π2−x)}2 f(x)+3\left\{f\left(\frac{\pi}{2}-x\right)\right\}^{2} f(x)+3{f(2π−x)}2.
2. g(x)=ex \mathrm{g}(\mathrm{x})=\mathrm{e}^{\mathrm{x}} g(x)=ex
Find any area enclosed by the curve y=sinx y=\sin x y=sinx and x x x axis.
Find the integral ∫g(2x){g(x)−1}{g(2x)+1}dx \int \frac{g(2 x)}{\{g(x)-1\}\{g(2 x)+1\}} d x ∫{g(x)−1}{g(2x)+1}g(2x)dx.
Find the value of 12∫0πln∣g(x)∣sin2xdx \frac{1}{2} \int_{0}^{\pi} \ln |g(x)| \sin ^{2} x d x 21∫0πln∣g(x)∣sin2xdx.
3. A=[pp+1p+1p+1pp+1p+1p+1p] A=\left[\begin{array}{ccc}p & p+1 & p+1 \\ p+1 & p & p+1 \\ p+1 & p+1 & p\end{array}\right] A=pp+1p+1p+1pp+1p+1p+1p
Without expansion find the value of ∣123456678∣ \left|\begin{array}{lll}1 & 2 & 3 \\ 4 & 5 & 6 \\ 6 & 7 & 8\end{array}\right| 146257368
From the stem determine A2−7A−8I3 A^{2}-7 A-8 I_{3} A2−7A−8I3 when p=2 p=2 p=2.
If AX=B \mathrm{AX}=\mathrm{B} AX=B, then find ' X \mathrm{X} X ' through determinant where p=1, B=[11109] \mathrm{p}=1, \mathrm{~B}=\left[\begin{array}{c}11 \\ 10 \\ 9\end{array}\right] p=1, B=11109
4. φ(x)=cosx \varphi(x)=\cos x φ(x)=cosx
If the length of the sides of a triangle be 3a,5a 3 \mathrm{a}, 5 \mathrm{a} 3a,5a and 7a 7 \mathrm{a} 7a units, then find the value of the greatest angle of it.
Find the value of φ(2x)φ(4x)φ(8x)φ(14x) \varphi(2 x) \varphi(4 x) \varphi(8 x) \varphi(14 x) φ(2x)φ(4x)φ(8x)φ(14x), when x= x= x= π15 \frac{\pi}{15} 15π.
If pφ(x)+qφ(y)=r=pφ(π2−x)+qφ(π2−y) p \varphi(x)+q \varphi(y)=r=p \varphi\left(\frac{\pi}{2}-x\right)+q \varphi\left(\frac{\pi}{2}-y\right) pφ(x)+qφ(y)=r=pφ(2π−x)+qφ(2π−y), then show that, φ(x−y2)=±2r2−(p−q)24pq \varphi\left(\frac{x-y}{2}\right)= \pm \sqrt{\frac{2 r^{2}-(p-q)^{2}}{4 p q}} φ(2x−y)=±4pq2r2−(p−q)2
5. A+B+C=π2 \mathrm{A}+\mathrm{B}+\mathrm{C}=\frac{\pi}{2} A+B+C=2π
Prove that, cos5θ=16cos5θ−20cos3θ+5cosθ \cos 5 \theta=16 \cos ^{5} \theta-20 \cos ^{3} \theta+5 \cos \theta cos5θ=16cos5θ−20cos3θ+5cosθ.
According to the stem, show that, cos(B+C)+cos(C \cos (\mathrm{B}+\mathrm{C})+\cos (\mathrm{C} cos(B+C)+cos(C +A)+cos(A+B)=1+4sinπ−2 A4sinπ−2 B4sin +\mathrm{A})+\cos (\mathrm{A}+\mathrm{B})=1+4 \sin \frac{\pi-2 \mathrm{~A}}{4} \sin \frac{\pi-2 \mathrm{~B}}{4} \sin +A)+cos(A+B)=1+4sin4π−2 Asin4π−2 Bsin π−2C4 \frac{\pi-2 C}{4} 4π−2C
According to the stem, if tanA+tanB+tanC=3 \tan A+\tan B+\tan C=\sqrt{3} tanA+tanB+tanC=3, then prove that, A=B=C \mathrm{A}=\mathrm{B}=\mathrm{C} A=B=C.
