Question

Let A = R – {2}, B = R – {1}. If f : A → B is a function defined by $f(x)\,=\,\left({\frac{x-1}{x-2}}\right)\!,$ show that f is one-one and onto. Hence find $f^{-1}.$

Using properties of determinants, prove that : $\begin{vmatrix}1+a & 1 & 1 \\ 1 & 1+b & 1 \\ 1 & 1 & 1+c\end{vmatrix}$ = $a b c+b c+c a+a b.$

Prove that the function f : [0, ∞) → R given by f(x) = 9x² + 6x – 5 is not invertible. Modify the Codomain of the function f to make it invertible, and hence find f⁻¹.

Find the general solution of the following differential equation : $y-x{\frac{d y}{d x}}=a\left(y^{2}+{\frac{d y}{d x}}\right).$

If A = $\left(\begin{array}{l l }1 & -1 \\ 2 & -1\end{array}\right)$ , B = $\left(\begin{array}{l l }a & 1 \\ b & -1\end{array}\right)$ and (A + B)² = A²+ B², then find the values of a and b.

Find the particular solution of the following differential equation : $x y{\frac{d y}{d x}}\,=(x+2)\,(y+2);y=-1,{\mathrm{when}}\,x=1.$

Form the differential equation representing family of curves given by (x – a)² + 2y²= a², where a is an arbitrary constant.