Mathematics Test-1

Given:

$\underset{x\to 0}{lim}\frac{\sin x\log (1-x)}{x^2}$

As we know,

$\underset{x\to a}{lim}[f(x)\cdot g(x)]=\underset{x\to a}{lim}f(x)\cdot \underset{x\to a}{lim}g(x)$

$\therefore \underset{x\to 0}{lim}\frac{\sin x\log (1-x)}{x^2}$

$=\underset{x\to 0}{lim}\frac{\sin x}{x}\times \underset{x\to 0}{lim}\frac{\log (1-x)}{x}$

$=1\times \underset{x\to 0}{lim}\frac{\log (1+(-x))}{x}$ $(\because \underset{x\to 0}{lim}\frac{\sin x}{x}=1)$

$=\underset{x\to 0}{lim}\frac{\log (1+(-x))}{-(-x)}$

$=-1\times \underset{x\to 0}{lim}\frac{\log (1+(-x))}{(-x)}$

$=-1\times 1$ $(\because \underset{x\to 0}{lim}\frac{\log x}{x}=1)$

$=-1$

Given that:

$\underset{x\to 1}{lim}\frac{(2x-3)(\sqrt{x}-1)}{2x^2+x-3}$

On putting the limit in above equation we will get$\frac{0}{0}$ form.

Therefore, we can cancel a factor going to zero out of the numerator and denominator.

We have,

$\underset{x\to 1}{lim}\frac{(2x-3)(\sqrt{x}-1)}{2x^2+x-3}$

$=\underset{x\to 1}{lim}\frac{(2x-3)(\sqrt{x}-1)}{(2x+3)(x-1)}$

$=\underset{x\to 1}{lim}\frac{(2x-3)(\sqrt{x}-1)}{(2x+3)(\sqrt{x}-1)(\sqrt{x}+1)}$

Factor $(\sqrt{x}-1)$ becomes zero at $x$ tends to 1 so, we need to cancel this factor from numerator and denominator.

$=\underset{x\to 1}{lim}\frac{(2x-3)}{(2x+3)(\sqrt{x}+1)}$

$=\frac{2-3}{(2+3)(\sqrt{1}+1)}$

$=\frac{-1}{10}$

As ${\cos }^2\alpha +{\cos }^2\beta +{\cos }^2\gamma =1$

$\Rightarrow \frac{1+\cos 2\alpha }{2}+\frac{1+\cos 2\beta }{2}+\frac{1+\cos 2\gamma }{2}=1$

$\Rightarrow \cos 2\alpha +\cos 2\beta +\cos 2\gamma =-1$

Given,

$4x^2+2x-1=0$

$x=\frac{b\pm \sqrt{b^2-4ac}}{2a}$

$\Rightarrow x=\frac{-2\pm \sqrt{2^2-4(4)(-1)}}{2(4)}$

$\Rightarrow \alpha =\frac{-1+\sqrt{5}}{4}$ और$\beta =\frac{-1-\sqrt{5}}{4}$

As we know that $\sin {18}^{\circ }=\frac{-1+\sqrt{5}}{4}$ and $\sin {54}^{\circ }=\frac{1+\sqrt{5}}{4}$

So, We can say that $\alpha =\sin {18}^{\circ }$ and $\beta =-\sin {54}^{\circ }\dots$ (i)

Now, From the given formula

$\sin 3\theta =3\sin \theta -4{\sin }^3\theta$

On putting $\theta ={18}^{\circ }$ in the above trigonometric formula, we get

$\Rightarrow \sin {54}^{\circ }=3\sin {18}^{\circ }-4{\sin }^3{18}^{\circ }\dots$ (ii)

From (i) and (ii), we get

$\Rightarrow -\beta =3\alpha -4{\alpha }^3$

$\Rightarrow \beta =4{\alpha }^2-3\alpha$

$\therefore$ The correct relation is $\beta =4{\alpha }^2-3\alpha .$

Given,

Numbers are 3, 10, 10, 4, 7, 10, 5

Mean $=\frac{\text{ Sum of all the observations }}{\text{ Total number of observations }}$

$=\frac{3+10+10+4+7+10+5}{7}$

$=\frac{49}{7}=7$

Deviation of all the values from mean:

$7-3=4$

$10-7=3$

$10-7=3$

$7-4=3$

$7-7=0$

$10-7=3$

$7-5=2$

Mean Deviation $=\frac{4+3+3+3+0+3+2}{7}$

$=\frac{18}{7}$

$c=xa+yb+z(a\times b)$

Let $c\cdot a=x,c\cdot b=y,x=y=\cos \theta$

Now, C.C $=|c{|}^2$

So, $[xa+yb+z(a\times b)]\cdot [xa+yb+z(a\times b)]=|c{|}^2$

$\Rightarrow 2x^2+z^2|a\times b{|}^2=1[\because c$ is a unit vector$]$

$\Rightarrow 2x^2+z^2[|a{|}^2|b{|}^2-(a\cdot b{)}^2]=1$

$\Rightarrow 2x^2+z^2[1-0]=1[\because a\perp b\therefore a\cdot b=0]$

$\Rightarrow 2x^2+z^2=1\Rightarrow z^2=1-2x^2$

$=1-2{\cos }^2\theta =-\cos 2\theta$

Given, ${\sin }^{-1}\mathrm{x}+{\cos }^{-1}\mathrm{y}=\frac{2\pi }{5}$

We know that, ${\sin }^{-1}\mathrm{x}+{\cos }^{-1}\mathrm{x}=\frac{\pi }{2}$

${\sin }^{-1}\mathrm{x}=\frac{\pi }{2}-{\cos }^{-1}\mathrm{x}$

Similarly, ${\sin }^{-1}\mathrm{y}+{\cos }^{-1}\mathrm{y}=\frac{\pi }{2}$

${\cos }^{-1}\mathrm{y}=\frac{\pi }{2}-{\sin }^{-1}\mathrm{y}$

Again, ${\sin }^{-1}\mathrm{x}+{\cos }^{-1}\mathrm{y}=\frac{2\pi }{5}$

$\Rightarrow \frac{\pi }{2}-{\cos }^{-1}x+\frac{\pi }{2}-{\sin }^{-1}y=\frac{2\pi }{5}$

$\Rightarrow \pi -{\cos }^{-1}x-{\sin }^{-1}y=\frac{2\pi }{5}$

Therefore, ${\cos }^{-1}x+{\sin }^{-1}y=\frac{3\pi }{5}$

Curve 1: $y=\sin x=f(x)$ (say)

Curve 2: Lines $x=-\frac{\pi }{3}$ and $x=\frac{\pi }{3}$

It can be drawn as follows:

According to the figure the sum of area curve $\mathrm{OAB}$ and curve OCD.

Here, $\mathrm{OAB}$ and $\mathrm{OCD}$ are equal and limit 0 to $\frac{\pi }{3}$.

So,  Area $=2\times$ area of $\mathrm{OAB}$.

The area between the curves $y_1=f(x)$ and $y_2=g(x)$ is given by: 

Area enclosed $=\left|{\int }_{{\mathrm{x}}_1}^{{\mathrm{x}}_2}({\mathrm{y}}_1-{\mathrm{y}}_2)\mathrm{dx}\right|$

Where, ${\mathrm{x}}_1$ and ${\mathrm{x}}_2$ are the intersections of curves ${\mathrm{y}}_1$ and ${\mathrm{y}}_2$

Now, the required area (A) is,

Area of $OAB=|{\int }_{{\mathrm{x}}_1}^{{\mathrm{x}}_2}\mathrm{f}(\mathrm{x})\mathrm{d}\mathrm{x}|$

$=|{\int }_0^{\frac{\pi }{3}}\sin xdx|$

$=|[-\cos x{]}_0^{\frac{\pi }{3}}|$

$=|-\cos \frac{\pi }{3}+\cos 0^{\circ }|$

$=|1-\frac{1}{2}|$ $=\frac{1}{2}$

$\therefore$ Shaded Area $=2\times \frac{1}{2}=1$

Given,

$5+55+555+\dots \dots$ to $n$ terms

$=5\times [1+11+111+\dots ..$ to $n$ terms $]$

$=\frac{5}{9}\times [9+99+999+\dots$ to $n$ terms $]$

$=\frac{5}{9}\times [(10-1)+({10}^2-1)+({10}^3-1)+\dots$ to $n$ terms $]$

$=\frac{5}{9}\times [(10+{10}^2+{10}^3+\dots .$ to $n$ terms $)-n]$

So, $a=\frac{5}{9},r=10$

As we know that,

$S_n=a\left(\frac{1-r^n}{1-r}\right)$ for $|r|<1$

$S_n=a\left(\frac{r^n-1}{r-1}\right)$ for $|r|>1$

As common ratio $(r)>1$, so the sum of series is given by,

$S_n=a\left(\frac{r^n-1}{r-1}\right)$

$\therefore S_n=\frac{5}{9}\times [\frac{10\times ({10}^n-1)}{(10-1)}-n]$

$\Rightarrow S_n=\frac{5}{9}\times [\frac{10\times ({10}^n-1)}{9}-n]$

$\Rightarrow S_n=\frac{5}{9}\times [\frac{{10}^1\times {10}^n-10}{9}-n]$

$\Rightarrow S_n=\frac{5}{9}\times [\frac{(10{)}^{n+1}-10}{9}-n]$

$\Rightarrow S_n=\frac{5}{9}\times \left[\frac{(10{)}^{n+1}-10-9n}{9}\right]$

$\Rightarrow S_n=\frac{5}{81}\cdot ({10}^{n+1}-9n-10)$

So, the required sum is $\frac{5}{81}\cdot ({10}^{n+1}-9n-10)$.

Given:

$f(x)=x^3-2x-8$

We can start solving by finding the values of $x$, where $f(x)$ is changing sign. $f^{\prime }(x)=3x^2-2$, equating it to zero we get,

$3x^2-2=0$

$x=\pm \sqrt{\frac{2}{3}}$

$f\left(\sqrt{\frac{2}{3}}\right)=-9.089$ and $f(-\sqrt{\frac{2}{3}})=-6.911$

At both these values of $x,f(x)$ is negative so there will be only one real root and rest $2$ roots of cubic equation will be imaginary.

Imaginary roots always exist in pairs.So, to guess the position of real root, we can check values of $f(x)$ at few points,

$f(0)=-8,f(1)=-9,f(2)=-4,f(3)=13,f(4)=48$

From here, we can see that,

The sign of the function does not change between $3$ and $4$, so there is no real roots between $3$ and $4$.

The sign of the function does not change between $1$ and $2$, so there are no real roots between $1$ and $2$.

The sign of the function changes between $2$ and $3$, so there is at least one real root between $2$ and $3$.

Given: $A=\left[\begin{array}{ccc}-1& 2& 3\\ -2& 3& 5\\ 4& -2& 1\end{array}\right]$

Here, we have to find the cofactor matrix for the given matrix A As we know that, cofactor of an element ${\mathrm{a}}_{\mathrm{ij}}$ is given by:

$C_{ij}=\{\begin{array}{c}{M}_{ij},\text{ when }i+j\text{ is even }\\ -M_{ij},\text{ when }i+j\text{ is odd }\end{array}$

$C_{11}=(-1{)}^2\times \left|\begin{array}{cc}3& 5\\ -2& 1\end{array}\right|=13$

$C_{12}=(-1{)}^3\times \left|\begin{array}{cc}-2& 5\\ 4& 1\end{array}\right|=22$

$C_{13}=(-1{)}^4\times \left|\begin{array}{cc}-2& 3\\ 4& -2\end{array}\right|=-8$

Similarly, we can say that ${\mathrm{C}}_{21}=-8,{\mathrm{C}}_{22}=-13$ and ${\mathrm{C}}_{23}=6$

Similarly, we can also say that, ${\mathrm{C}}_{31}=1,{\mathrm{C}}_{32}=-1$ and ${\mathrm{C}}_{33}=1$

So, the required cofactor matrix is $\left[\begin{array}{ccc}13& 22& -8\\ -8& -13& 6\\ 1& -1& 1\end{array}\right]$

Given:

$\mathrm{z}=2\mathrm{x}+5\mathrm{y}$

$2\mathrm{x}+4\mathrm{y}\le 8$

$\Rightarrow \mathrm{x}+2\mathrm{y}\le 4$

$3x+y\le 6,x+y\le 4,x\ge 0,y\ge 0$

Draw the lines $\mathrm{x}+2\mathrm{y}=4$ (passes through (4,0$),(0,2))$

$3\mathrm{x}+\mathrm{y}=6$ (passes through $(2,0),(0,6)$ and

$\mathrm{x}+\mathrm{y}=4$ (passes through $(4,0),(0,4)$

Shade the region satisfied by the given inequations;

The shaded region in the figure gives the feasible region determined by the given inequations.

Solving $3\mathrm{x}+\mathrm{y}=6$ and $\mathrm{x}+2\mathrm{y}=4$ simultaneously, we get

$\mathrm{x}=\frac{8}{5}$ and $\mathrm{y}=\frac{6}{5}$

We observe that the feasible region $\mathrm{OABC}$ is a convex polygon and bounded and has corner points.

$\mathrm{O}(0,0),\mathrm{A}(2,0),\mathrm{B}(\frac{8}{5},\frac{6}{5}),\mathrm{C}(0,2)$

The optimal solution occurs at one of the corner points.

At $\mathrm{O}(0,0),\mathrm{z}=2\times 0+5\times 0=0$

At $\mathrm{A}(2,0),\mathrm{z}=2\times 2+5\times 0=4$

At B $(\frac{8}{5},\frac{6}{5}),z=2\times \frac{8}{5}+5\times \frac{6}{5}=\frac{46}{5}$

At $\mathrm{C}(0,2),\mathrm{z}=2\times 0+5\times 2=10$

Therefore, z maximum value at $\mathrm{C}$ and maximum value $=10$

Given,

Equation of line: $\frac{x-1}{3}=\frac{y+1}{-1}=\frac{z-3}{2}$

Equation of plane: $3x+4y+z+5=0$

As we know,

The angle between the line $\frac{x-x_1}{a_1}=\frac{y-y_1}{b_1}=\frac{z-z_1}{c_1}$ and the plane $a_{2}x+b_{2}y+c_{2}z+d=0$ is given by,

$\sin \theta =\frac{a_1{a}_2+b_1{b}_2+c_1{c}_2}{\left(\sqrt{a_1^2+b_1^2+c_1^2}\right)\left(\sqrt{a_2^2+b_2^2+c_2^2}\right)}$

On comparing given equation of line and plane with above equation, we get

$a_1=3,b_1=-1,c_1=2,a_2=3,b_2=4$ and $c_2=1$

$\sin \theta =\frac{a_1{a}_2+b_1{b}_2+c_1{c}_2}{\left(\sqrt{a_1^2+b_1^2+c_1^2}\right)\left(\sqrt{a_2^2+b_2^2+c_2^2}\right)}$

$\Rightarrow \sin \theta =\frac{3\times 3+(-1)\times 4+2\times 1}{\left(\sqrt{(3{)}^2+(-1{)}^2+(2{)}^2}\right)\left(\sqrt{(3{)}^2+(4{)}^2+(1{)}^2}\right)}$

$\Rightarrow \sin \theta =\frac{9-4+2}{\left(\sqrt{9+1+4}\right)\left(\sqrt{9+16+1}\right)}$

$\Rightarrow \sin \theta =\frac{7}{\sqrt{14}\cdot \sqrt{26}}$

$\Rightarrow \sin \theta =\sqrt{\frac{7}{52}}$

$\Rightarrow \theta ={\sin }^{-1}\left(\sqrt{\frac{7}{52}}\right)$

Given that,

$\Rightarrow \frac{xdy}{dx}+3y=4x^3$

Now,

$\Rightarrow \frac{dy}{dx}+\frac{3y}{x}=4x^2$

By comparing with $\frac{dy}{dx}+Py=Q$

$\Rightarrow P=\frac{3}{x}\text{ and }Q=4x^2$

$\Rightarrow \text{ I.F. }=e^{\int Pdx}=e^{\int \frac{3}{x}dx}$

$\Rightarrow \text{ I.F. }=e^{3\ln x}$

$\Rightarrow \text{ I.F. }=e^{\ln x^3}$

$\Rightarrow \text{ I.F. }=x^3(\because e^{\ln x}=x)$

Now general solution will be,

$\Rightarrow y\cdot (I\cdot F\cdot )=\int (Q\cdot (I\cdot F\cdot ))dx+c$

$\Rightarrow y\cdot \left(x^3\right)=\int (4x^2\cdot \left(x^3\right))dx+c$

$\Rightarrow x^3\cdot y=\int 4x^{5}dx+c$

$\Rightarrow x^3\cdot y=4\frac{x^6}{6}+c$

$\Rightarrow x^3\cdot y=\frac{2}{3}\cdot x^6+c$