Maths

29.

Given, f (x) = (x-a₁) (x-a₂)… (x-a_n)

So,  $\underset{x\to a_1}{lim}f(x)=\underset{x\to a_1}{lim}\left(x-a_1\right)\underset{x\to a_1}{lim}\left(x-a_2\right)?\underset{x\to a_1}{lim}(x-a^n)$

= (a₁-a₁) (a₁-a₂) … (a₁-a_n)

= 0 (a₁-a₂) … (a₁-a_n)

 = 0

And $\underset{x\to a}{lim}f(x)=\underset{x\to a}{lim}\left(x-a_1\right)\underset{x\to a}{lim}\left(x-a_2\right)\dots \underset{x\to a}{lim}(x-an)$

= (a-a₁) (a-a₂) … (a-a_n)

17. Kindly go through the solution

$=$$=\underset{x\to 0}{lim}\frac{2si{n}^{2}x}{2si{n}^2\raisebox{1ex}{$x$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}$

$=\frac{\underset{x\to 0}{lim}{\left(\frac{sinx}{x}\right)}^2*x^2}{\underset{x\to 0}{lim}{\left(\frac{si{n}^{}\raisebox{1ex}{$x$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}{\raisebox{1ex}{$x$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}\right)}^2{\raisebox{1ex}{$x$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}^2}$

$=\frac{{(1)}^2*x^2*4}{{(1)}^2*x^2}$

= 4

44. Secx = 2.                      .

We have, Secx = 2, | Secx is (+)ve  the principal solution lies in I^st and IV^thquadrent.

= $\frac{\pi }{3}$ and $\frac{6\pi -\pi }{3}$

= $\frac{\pi }{3}$ and $\frac{\mathrm{5\pi }}{3}$ .

As secx = sec $\frac{\pi }{3}$ . cosx = cos $\frac{\pi }{3}$ $\left[\therefore secx=\frac{1}{cosx}\right]$

The general solution is

x = 2nπ ±$\frac{\pi }{3}$, n ∈ z.

28. Kindly consider the following

30. 

30. For inequality y+8 ≤ 2x, the equation of the line is y+8=2x. We consider the table below to plot of y+8=2x.

$\begin{array}{l}x\\ y\end{array}|\begin{array}{l}0\\ -8\end{array}|\begin{array}{l}4\\ 0\end{array}|$

The line devides the xy-plane into half planer I and II. We select a point (0,0) and check the correctness of the inequality.

i.e., 0+8 ≤ 2 * 0

0 ≤ 0 which is true.

So, the solution region is I which includes the rigin (0,0). The continuous line also indicates that any points on the line also satisfy the given inequality.

29. For inequality 3x+4y≥ 12 the equation of the line is 3x+4y=12

We consider the table below to plot 3x+4y=12.

$\begin{array}{l}x\\ y\end{array}|\begin{array}{l}0\\ 3\end{array}|\begin{array}{l}4\\ 0\end{array}|$

This line devides the xy-plane into half planer I and II.

We select point 0 (0,0) and check the correctness of the inequality.

i.e., 3 * 0+4 * 0 ≤ 12.

0+0 ≤ 12

0 ≤12 which is true.

So, the solution region is I which includes the origin (0,0). The continuous line also indicates that any points in the line also satisfy the given inequality.

28. For inequality, 2x+y≥ 6, the equation of line is 2x+y=6.

We consider the table below to pot 2x+y=6.

$\begin{array}{l}x\\ y\end{array}|\begin{array}{l}0\\ 6\end{array}|\begin{array}{l}3\\ 0\end{array}|$

Graph of 2x+y=6 is given as a continuous line in fig 2.

This line divides xy-plane in two half planes I and II.

We select 0 (0,0) and check the correctness of the inequality.

is 2 * 0+0 ≥ 6.

0 ≥ 6 which is false.

So, the solution region is II where origin (0,0) is included.

The continuous line indicates that any point on the line. also satisfy the given inequality.