1. f(x)=exf(x)=e^xf(x)=ex
Find the derivative of f′(x)log2xf(2x) f^{\prime}(x) \log 2 x f(2 x) f′(x)log2xf(2x).
Find the derivative of f(mx) f(\mathrm{mx}) f(mx) with respect to x x x using first principle.
If y=f(sec−1x) y=f\left(\sec ^{-1} x\right) y=f(sec−1x), then show that, x2(x2−1)y2+ x^{2}\left(x^{2}-1\right) y_{2}+ x2(x2−1)y2+ x(2x2−1)y1−y=0 x\left(2 x^{2}-1\right) y_{1}-y=0 x(2x2−1)y1−y=0.
2. A=[x−112−2x+1320x] and B=[52−10−4] A=\left[\begin{array}{ccc} x-1 & 1 & 2 \\ -2 & x+1 & 3 \\ 2 & 0 & x \end{array}\right] \text { and } B=\left[\begin{array}{rr} 5 & 2 \\ -10 & -4 \end{array}\right] A=x−1−221x+1023x and B=[5−102−4]
Show that, B \mathrm{B} B is an idempotent matrix?
If ∣A∣=0 |A|=0 ∣A∣=0, then find the value of x x x.
Determine (AT)−1 \left(A^{T}\right)^{-1} (AT)−1 where x=0 \mathrm{x}=0 x=0.
3. sinθ=35 \operatorname{sin} \theta=\frac{3}{5} sinθ=53 and A=π12 A=\frac{\pi}{12} A=12π
Show that, cos75∘=14(6−2) \cos 75^{\circ}=\frac{1}{4}(\sqrt{6}-\sqrt{2}) cos75∘=41(6−2).
Find the value of cotθ+cos(−θ)cosec(−θ)+tanθ \frac{\cot \theta+\cos (-\theta)}{\operatorname{cosec}(-\theta)+\tan \theta} cosec(−θ)+tanθcotθ+cos(−θ); where π2<θ< \frac{\pi}{2}<\theta< 2π<θ< π \pi π
Prove that, tanA⋅tan3A⋅tan5A⋅tan7A⋅tan11 A=1 \tan A \cdot \tan 3 A \cdot \tan 5 A \cdot \tan 7 A \cdot \tan 11 \mathrm{~A}=1 tanA⋅tan3A⋅tan5A⋅tan7A⋅tan11 A=1.
4. Scenario-1: f(x,y)=3x−4y−5 f(x, y)=3 x-4 y-5 f(x,y)=3x−4y−5 and g(x,y)=x2+y2−6x g(x, y)=x^{2}+y^{2}-6 x g(x,y)=x2+y2−6x +8y+9 +8 y+9 +8y+9.
Scenario-2: (5,3) (5,3) (5,3) and (−5,7) (-5,7) (−5,7) are the end points of a diameter of a circle.
Find the intercepted portion from the y y y-axis by the circle g(x,y)=0 g(x, y)=0 g(x,y)=0.
Show that, according to the scenario-1, the straight line f(x,y)=0 f(x, y)=0 f(x,y)=0 is a tangent to the circle g(x,y)=0 g(x, y)=0 g(x,y)=0.
Find the equation of circle according to the scenario-2. Also find the equation of the circle which passes through origin and the intersection of the line f(x,y)=0 f(x, y)=0 f(x,y)=0 and the derived circle.
5.
Show that, b+cb−c=sinB+sinCsinB−sinC \frac{b+c}{b-c}=\frac{\sin B+\sin C}{\sin B-\sin C} b−cb+c=sinB−sinCsinB+sinC
Prove that, cosAsinBsinC+cosBsinCsinA+cosCsinAsinB=2 \frac{\cos A}{\sin B \sin C}+\frac{\cos B}{\sin C \sin A}+\frac{\cos C}{\sin A \sin B}=2 sinBsinCcosA+sinCsinAcosB+sinAsinBcosC=2
If α=45∘,β=60∘ \alpha=45^{\circ}, \beta=60^{\circ} α=45∘,β=60∘ and a=(3+1)cm a=(\sqrt{3}+1) \mathrm{cm} a=(3+1)cm then, determine the area of △ABC \triangle \mathrm{ABC} △ABC.