Maths

$area=2*\left(\frac{1}{2}*1*1\right)=1=k$

  ${\displaystyle \underset{0}{\overset{1.5}{\int }}\left[x^2\right]dx={\displaystyle \underset{0}{\overset{1}{\int }}0dx+{\displaystyle \underset{1}{\overset{\sqrt[]{2}}{\int }}1dx+{\displaystyle \underset{\sqrt[]{2}}{\overset{1.5}{\int }}2dx=2-\sqrt[]{2}}}}}$

  $\underset{x\to 1^{-}}{lim}f\left(x\right)=\underset{x\to 1^{+}}{lim}f\left(x\right)\Rightarrow a+b=2$   

  $\therefore \underset{x\to 3^{-}}{lim}f\left(x\right)=\underset{x\to 3^{+}}{lim}f\left(x\right)\Rightarrow 3a+b=6tan\frac{3\pi }{12}$          

$\therefore$ a = 2, b = 0

  $L=\underset{x\to \infty }{lim}\frac{{\left(2+\frac{1}{x}\right)}^{40}{\left(4-\frac{1}{x}\right)}^5}{{\left(2+\frac{3}{x}\right)}^{45}}=32$

  $\left|\overline{a}*\left(\overline{a}*\overline{c}\right)\right|=\left|3\overline{b}\right|=3\left|\overline{b}\right|$

$\Rightarrow 3=3.2sin\theta \Rightarrow sin\theta =\frac{1}{2}\Rightarrow co{s}^2\theta =\frac{3}{4}$  

$\underset{x\to 0}{lim}{\left[si{n}^2\left(\frac{\pi }{2-3x}\right)\right]}^{se{c}^2\left(\frac{\pi }{2-5x}\right)}$

$e^{\underset{x\to 0}{lim}\left[si{n}^2\left(\frac{\pi }{2-3x}\right)-1\right]se{c}^2\left(\frac{\pi }{2-5x}\right)}$

$=e^{\underset{x\to 0}{lim}\frac{-si{n}^2\left(\frac{-3\pi x}{2\left(2-3x\right)}\right)}{si{n}^2\left(\frac{-5\pi x}{2\left(2-5x\right)}\right)}}=e^{-\frac{9}{25}}$

Any point on x + y = 1 can be taken as (t, 1 – t) The equation of chord with this as mid-point is y (1 – t) -2a (x + t) = (1 – t)² – 4at

It passes through (a, 2a)

So, t² – 2t + 2a² – 2a + 1 = 0

This should have two distinct real roots.

So D > 0

$\begin{array}{l}\Rightarrow a^2-a<0\\ \Rightarrow 0<a<1\end{array}$          

So, length of latus rectum < 4 and 0 < a < 1

$\Rightarrow LR=1,2,3$            

You can use the position vector to find the real displacement of a point from a reference point.

Two vectors which are parallel to each other, irrespective of direction, are called parallel vectors.

Types are given below

• Collinear vectors
• Parallel vectors
• Antiparallel vectors

Out of numbers $\frac{1}{2},\frac{2}{4},\frac{3}{6},\frac{1}{3},\frac{2}{6},\frac{2}{3},\frac{4}{6}$ will result only 3 distinct rational numbers.

Total numbers =