Mathematics Test - 4

$\begin{array}{l}={\sin }^{-1}\left(\sin \frac{2\pi }{3}\right)\\ ={\sin }^{-1}\left(\sin (\pi -\frac{\pi }{3})\right)\\ ={\sin }^{-1}\left(\sin \frac{\pi }{3}\right)\\ =\frac{\pi }{3}\end{array}$

$f\left(x\right)=\left|x\right|+\left|x-1\right|$

$For x\le 0$

$f\left(x\right)=-x-(x-1)=1-2x-----\left(1\right)$

$For 0<x\le 1$

$f\left(x\right)=x-(x-1)=1-----\left(2\right)$

$For x>1$

$f\left(x\right)=x+x-1=2x-1-----\left(3\right)$

$CASE 1:$

$x=0$

$f\left(x\right)=\left|0\right|+\left|0-1\right|=1$

$LHL: \underset{x\to 0^{-}}{\lim }(1-2x)=1$

$RHL: \underset{x\to 0^{+}}{\lim }f\left(x\right)=1$

$LHL=RHL=f\left(0\right)$

$Function is continuous at x=0$

$CASE 2:$

$x=1$

$f\left(1\right)=\left|1\right|+\left|1-1\right|=1$

$LHL: \underset{x\to 1^{-}}{\lim }f\left(x\right)=1$

$RHL: \underset{x\to 1^{+}}{\lim }f\left(x\right)=\underset{x\to 1^{+}}{\lim }(2x-1)=1$

$LHL=RHL=f\left(1\right)$

$Function is continuous at x=1$

Since, intelligency is not defined for students in a class i.e., Not a well defined collection.

Given-

$x+y=10$

$x^2+y^2=68$

$x=8$ and $y=2$ satisfy both the given equations.

$8+2=10$ (True)

$8^2+2^2=68$ (True)

$xy=8\times 2$

$xy=16$

Given expression is,

\(\left\{\frac{1}{2}(a-b)\right\}^{2}+a b=p(a+b)^{2}\)

\(\Rightarrow\left\{\frac{1}{4}\left(a^{2}-2 a b+b^{2}\right\}+a b=p(a+b)^{2}\right.\)

\(\Rightarrow \frac{1}{4}\left(a^{2}-2 a b+b^{2}+4 a b\right)=p(a+b)^{2}\)

\(\Rightarrow \frac{1}{4}\left(a^{2}+2 a b+b^{2}\right)=p(a+b)^{2}\)

\(\Rightarrow \frac{1}{4}(a+b)^{2}=p(a+b)^{2}\)

\(\Rightarrow {p}=\frac{1} {4}\)

Consider r consecutive positive integers 

n + 1,........n + r - 1, n + r

Then $\frac{\left(n+1\right).....\left(n+r\right)}{r!}$

= $\frac{n!\left(n+1\right).....\left(n+r\right)}{n! r!}{=}^{n+r}{C}_r$

Which is a positive integer

$\begin{array}{l}{\tan }^{-1}\left(\frac{1}{4}\right)+{\tan }^{-1}\left(\frac{3}{5}\right)\\ ={\tan }^{-1}\left(\frac{\frac{1}{4}+\frac{3}{5}}{1-\frac{1}{4}\times \frac{3}{5}}\right)\\ ={\tan }^{-1}\left(\frac{\frac{5+12}{20}}{\frac{20-3}{20}}\right)={\tan }^{-1}\left(\frac{17}{17}\right)={\tan }^{-1}\left(1\right)\\ ={\tan }^{-1}\left(\tan \frac{\pi }{4}\right)=\frac{\pi }{4}\end{array}$

It is given that,

$(2x+\frac{1}{3x})=6$

On multiplying by $\frac{3}{2}$ on both the sides,

$\Rightarrow (3x+\frac{1}{2x})=9$

On squaring both the sides,

$\Rightarrow {(3x+\frac{1}{2x})}^2=(9{)}^2$

$\Rightarrow (9x^2+\frac{1}{4x^2}+3)=81 \left[\because {\left(a+b\right)}^2=a^2+b^2+2ab\right]$

$\Rightarrow (9x^2+\frac{1}{4x^2})=78$

Given,

${\left|1-2i\right|}^x=5^x$

$({\sqrt{(1^2+2^2)})}^x=5^x$

${\left(\sqrt{5}\right)}^x=5^x$

$5^{\raisebox{1ex}{$x$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}=5^x$

$\frac{x}{2}=x$

$\frac{x}{2}=0$

$x=0$

There are no non-zero integral solutions.

That square number is a multiple of 4, 5 and 6. 

LCM (4, 5, 6) = 60

60 = 2²×3×5

To make 2²×3×5 a perfect square number, we need to multiply it with 3×5.

Required number = 2²×3²×5² 

=  (2×3×5)² = 900