f(x)=\(\left\{\begin{matrix} |4x^2-8x+5|, & if\,8x^2-6x+1\geq0 \\ |4x^2-8x+5|, & if\,8x^2-6x+1<0 \end{matrix}\right.\)
=\(\left\{\begin{matrix} |4x^2-8x+5|, & if\,x\in [-\infty,\frac{1}{4}]\bigcup[\frac{1}{2},\infty] \\ |4x^2-8x+5|, & if\, x\in(\frac{1}{4},\frac{1}{2})\end{matrix}\right.\)
f(x)=\(f(x)=\left\{\begin{matrix} 4x^2-8x+5 &if\,x\in [-\infty,\frac{1}{4}]\bigcup[\frac{1}{2},\infty] & \\ 3&x\in(\frac{1}{4},\frac{2-\sqrt2}{2}) & \\ 2& x\in(\frac{2-\sqrt2}{2},\frac{1}{2}) & \end{matrix}\right.\)
∴ Non-diff at
x=\(\frac{1}{4}\),\(\frac{2-\sqrt2}{2}\),\(\frac{1}{2}\)
Let the vectors \(\mathbf{u}_1 = \hat{i} + \hat{j} + a\hat{k}, \mathbf{u}_2 = \hat{i} + b\hat{j} + \hat{k}\), and \(\mathbf{u}_3 = c\hat{i} + \hat{j} + \hat{k}\) be coplanar. If the vectors \(\mathbf{v}_1 = (a + b)\hat{i} + c\hat{j} + c\hat{k}, \mathbf{v}_2 = a\hat{i} + (b + c)\hat{j} + a\hat{k}, \mathbf{v}_3 = b\hat{i} + b\hat{j} + (c + a)\hat{k}\) are also coplanar, then \(6(a + b + c)\) is equal to:
If the points with position vectors \(a\hat{i} +10\hat{j} +13\hat{k}, 6\hat{i} +11\hat{k} +11\hat{k},\frac{9}{2}\hat{i}+B\hat{j}−8\hat{k}\) are collinear, then (19α-6β)2 is equal to
A physical quantity, represented both in magnitude and direction can be called a vector.
For the supplemental purposes of these vectors, there are two laws that are as follows;
It means that if we have any two vectors a and b, then for them
\(\overrightarrow{a}+\overrightarrow{b}=\overrightarrow{b}+\overrightarrow{a}\)
It means that if we have any three vectors namely a, b and c.
\((\overrightarrow{a}+\overrightarrow{b})+\overrightarrow{c}=\overrightarrow{a}+(\overrightarrow{b}+\overrightarrow{c})\)
Read More: Addition of Vectors