Mathematics Test - 28

For the roots to be real, we must have \(\mathrm{D}=\mathrm{b}^{2}-4 \mathrm{ac} \geq 0\)

Also, both the roots will be positive if and only if the sum of the roots and product of the roots are both positive,that is, \(-\frac{b}{a}>0 \& \frac{c}{a}>0\).

In other words, \(a,-b\) and \(c\) are of the same sign.

Since, sin 190° = - sin 10°, sin 200° = - sin 20°,

sin 210° = - sin 30°, sin 360° = sin 180° = 0 etc.

Hence all the terms in the expression cancels out, therefore answer is 0.

Given, \(f(x)=\int \frac{\ln (x)}{x} d x\)

Let, \(z=\ln (x)\)

=>d z=\(\frac{d x}{x}\)

=>\(f(x)=\int z d z=\frac {z^{2}}{2}\)

\(=\frac{\ln ^{2}(x)}{2}\)

It is given that,\(A=\left\{x \in R: x^{2}=x\right\}\) and \(B=\left\{x \in R: x^{2}-3 x+2=0\right\}\).

Let us first consider \(A=\left\{x \in R: x^{2}=x\right\}\)

\(\because{x}^{2}={x}\)

\(\therefore x^{2}-x=0\)

\(\Rightarrow x(x-1)=0\)

\(\Rightarrow {x}=0\) or 1

So, \(\mathrm{A}=\{0,1\}\)

Now, \(B=\left\{x \in R: x^{2}-3 x+2=0\right\}\)

\(\Rightarrow x^{2}-3 x+2=0\)

\(\Rightarrow x^{2}-x-2 x+2=0\)

\(\Rightarrow(x-1)(x-2)=0\)

\(\Rightarrow x=1\) or \(2\)

So, \(B=\{1,2\}\)

As we know that, \(B-A=\{x: x \in B\) and \(x \notin A\}\)

\(\therefore \mathrm{B}-\mathrm{A}=\{2\}\)