Mathematics Test - 6

$\frac{3^x-2^x}{x}$=$\frac{2^x({\left({\displaystyle \frac{3}{2}}\right)}^{0+x}-1)}{x}$

= $\frac{2^x({\left({\displaystyle \frac{3}{2}}\right)}^{0+x}-{\left({\displaystyle \frac{3}{2}}\right)}^x)}{x}$

Now,$\underset{x\to 0}{\lim }\frac{2^x({\left({\displaystyle \frac{3}{2}}\right)}^{0+x}-{\left({\displaystyle \frac{3}{2}}\right)}^x)}{x}$

=$\underset{x\to 0}{\lim }\underset{x\to 0}{2^{x}x\lim }\frac{2^x({\left({\displaystyle \frac{3}{2}}\right)}^{0+x}-{\left({\displaystyle \frac{3}{2}}\right)}^x)}{x}$

=$1{\left[\frac{d{\left({\displaystyle \frac{3}{2}}\right)}^x}{dx}\right]}_{x=0}$

=$={\log }_e\left(\frac{3}{2}\right)$

Given: The line through the points \((2,4,8)\) and \((1,2,4)\) is parallel to the line through the points \((3,6, k)\) and \((1,2,1)\).

Let us consider \({AB}\) be the line joining the points \((2,4,8)\) and \((1,2,4)\) whereas \({CD}\) be the line passing through the points \((3,6, {k})\) and \((1,2,1)\).

Let, the direction ratios of \(A B\) be: \(a_{1}, b_{1}, c_{1}\)

\(a_{1}=(2-1)=1, b_{1}=(4-2)=2\) and \(c_{1}=(8-4)=4\)

Let the direction ratios of \(C D\) be: \(a_{2}, b_{2}, c_{2}\)

\(a_{2}=(3-1)=2, b_{2}=(6-2)=4\) and \(c_{2}=k-1\)

\(\because\) Line \({AB}\) is parallel to \({CD}\).

So,

\( \frac{{a}_{1}}{{a}_{2}}=\frac{{b}_{1}}{{~b}_{2}}=\frac{{c}_{1}}{{c}_{2}}\)

\(\Rightarrow \frac{1}{2}=\frac{2}{4}=\frac{4}{{k}-1}\)

\(\Rightarrow \frac{1}{2}=\frac{4}{{k}-1}\)

\(\Rightarrow {k}-1=8\)

\( \Rightarrow {k}=9\)

Given:

$f\left(x\right)=2x^3–15x^2+36x+1$

$f\text{'}\left(x\right)=6x^2-30x+36$

$=-6(x^2-5x+6)$

$=6(x-2)(x-3)$

f(x) is increasing in [0, 2] and decreasing in [2, 3].f (x) is many one.

f(0) = 1

f(2) = 29

f(3) = 28

Range is [1, 29].

Given:

\(\cos \alpha-\cos \beta=2(\sin \beta+\sin \alpha)\).....(1)

As we know,

\(\cos \alpha-\cos \beta=-2 \sin \left(\frac{\alpha+\beta}{2}\right) \sin \left(\frac{\alpha-\beta}{2}\right)\)

\(\sin \alpha+\sin \beta=2 \sin \left(\frac{\alpha+\beta}{2}\right) \cos \left(\frac{\alpha-\beta}{2}\right)\)

So, from (1),

\(\Rightarrow-2 \sin \left(\frac{\alpha+\beta}{2}\right) \sin \left(\frac{\alpha-\beta}{2}\right)=2 \times 2 \sin \left(\frac{\beta+\alpha}{2}\right) \cos \left(\frac{\beta-\alpha}{2}\right)\)

\(\Rightarrow 4 \sin \left(\frac{\alpha+\beta}{2}\right) \cos \left(\frac{\alpha-\beta}{2}\right)+2 \sin \left(\frac{\alpha+\beta}{2}\right) \sin \left(\frac{\alpha-\beta}{2}\right)=0 \quad(\because \cos (-\theta)=\cos \theta)\)

\(\Rightarrow 2 \sin \left(\frac{\alpha+\beta}{2}\right)\left[2 \cos \left(\frac{\alpha-\beta}{2}\right)+\sin \left(\frac{\alpha-\beta}{2}\right)\right]=0\)

\(\Rightarrow \sin \left(\frac{\alpha+\beta}{2}\right)=0\)

\(\Rightarrow\left(\frac{\alpha+\beta}{2}\right)=0\)

\(\therefore \alpha=-\beta\)

Now,

\(\sin 3 \alpha+\sin 3 \beta\)

\(=\sin 3 \alpha+\sin (-3 \alpha)\)

\(=\sin 3 \alpha-\sin 3 \alpha \quad(\because \sin (-\theta)=-\sin \theta)\)

\(=0\)

Consider a line ……. (1) ( to be a circle)$\alpha x+\beta y=9$

But equation of circle is given by

$x^2+y^2=9$......(2)

Let the midpoint (h,k) at the circle S.

Then by equation (1) and (2) to,

$xh+yk=h^2+k^2=9$ …….(3)

Comparing equation (1) and (3) to, we get

$\alpha =\frac{9h}{h^2+k^2}and\beta =\frac{9h}{h^2+k^2}$

Sine αandβ lies on the given line of Line

$4x-5y=20$

Then,

$4\alpha -5\beta =20$ ……(4)

Put the value of αandβ in equation (4),

$\frac{4\times 9h}{h^2+k^2}-\frac{5\times 9h}{h^2+k^2}=20$

$36h-45k=20(h^2+k^2)$

$20(h^2+k^2)-36h+45k=0$

The locus is a circle

$20(x^2+y^2)-36x+45y=0$