Column I, Column II `sin^(-1)x+x >0,for` ,p. `x<0` `cos^(-1)x-x >0,for` ,q. `x in [0,1]` `tan^(-1)x+x >0,for` ,r. `x in [-1,0]` `cot^(-1)x+x >0,for` ,s. `x >0`

f(x) = sin^(-1)x+x^(2)-3x + (x^(3))/(3),x in[0,1]

Solve sin x tan x -sin x+ tan x-1=0 for x in [0, 2pi] .

solve tan^(-1) ((1-x)/(1+x)) = 1/2 tan^(-1) x "for" x gt0

Each question contains statement given in two columns. Statement a,b in column I have to be matched with statement p,q in column II Column I, Column II cos^(-1)(4x^3-3x)=3cos^(-1)x ,then x can take values , p. [1/2,1] sin^(-1)(3x-4x^3)=3sin^(-1)x ,then x can take values , q. [-1/2,0]

Prove that sin (cot^(-1) (tan (cos^(-1) x))) = x, x gt 0

Solve : tan^(-1) 3x+tan^(-1) 4x = tan^(-1)(1/x), 12 x^(2) lt 1 , x!= 0

Column I: Function, Column II: Value of x for which both the functions in any option of column I are identical f(x)=tan^(-1)((2x)/(1-x^2)),g(x)=2tan^(-1)x , p. x in {-1,1} f(x)=(log)_(x2)25a n dg(x)=(log)_x5 , q. x in (-1,1) f(x)=sec^(-1)x+cos e c^(-1)x ,g(x)=sin^(-1)x+cos^(-1)x , r. x in (0,1)

f(x) is polynomial function of degree 6, which satisfies ("lim")_(xvec0)(1+(f(x))/(x^3))^(1/x)=e^2 and has local maximum at x=1 and local minimum at x=0a n dx=2. Column I, Column II The coefficient of x^6 , p. 0 The coefficient of x^5 , q. 2 The coefficient of x^4 , r. -(12)/5 The coefficient of x^3 , s. 2/3

Prove that: tan^(-1)x+tan^(-1)1/x={pi/2,ifx >0-pi/2,ifx<0