Question

Show that a function f : R → R given by f(x) =ax + b,a, b ∈ R, a ≠ 0 is a bijective.

Find the general solution of the differential equation : $x d y-y d x={\sqrt{x^{2}+y^{2}}}d x~.$

Form the differential equation of the family of circles in the second quadrant and touching the co-ordinate axes.

Find the equation of tangents to the curve y = x³ + 2x – 4 which are perpendicular to the line x + 14y – 3 = 0.

Find the position vector of a point R, which divides the line joining two points P and Q whose position vectors are $2\stackrel{\rightarrow}{a}+\stackrel{\rightarrow}{b}$ and $\stackrel{\rightarrow}{a}-3\stackrel{\rightarrow}{b}$ respectively, externally in the ratio 1 : 2 Also, show that P is the mid-point, of line segment RQ.

Find the general solution of the differential equation : $(x-y){\frac{d y}{d x}}\,=x+2y.$

Let A = R – {3}, B = R – {1}. Let f : A → B be defined by f(x) =${\frac{x-2}{x-3}},~~\forall~x\in~A.$ Show that f is bijective. Also, find (i) x, if $f^{-1}(x)$ = 4 (ii) $f^{-1}(7)$