Mathematics Test-1

Given:

\(\lim _{x \rightarrow 0} \frac{\sin x \log (1-x)}{x^{2}}\)

As we know,

\(\lim _{x \rightarrow a}[f(x) \cdot g(x)]=\lim _{x \rightarrow a} f(x) \cdot \lim _{x \rightarrow a} g(x)\)

\(\therefore \lim _{x \rightarrow 0} \frac{\sin x \log (1-x)}{x^{2}}\)

\(=\lim _{x \rightarrow 0} \frac{\sin x}{x} \times \lim _{x \rightarrow 0} \frac{\log (1-x)}{x}\)

\(=1 \times \lim _{x \rightarrow 0} \frac{\log (1+(-x))}{x} \quad\) \( (\because\lim _{x \rightarrow 0} \frac{\sin x}{x}=1)\)

\(=\lim _{x \rightarrow 0} \frac{\log (1+(-x))}{-(-x)}\)

\(=-1 \times \lim _{x \rightarrow 0} \frac{\log (1+(-x))}{(-x)}\)

\(=-1 \times 1 \quad\) \((\because\lim _{x \rightarrow 0} \frac{\log x}{x}=1)\)

\(=-1\)

Given that:

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

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,

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

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

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

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

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

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

\(=\frac{-1}{10}\)

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,

\(4 x^{2}+2 x-1=0\)

\(x=\frac{b \pm \sqrt{b^{2}-4 a c}}{2 a}\)

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

\(\Rightarrow \alpha=\frac{-1+\sqrt{5}}{4}\) 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 \(\beta=-\sin 54^{\circ} \quad \ldots\) (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

\(\Rightarrow \sin 54^{\circ}=3 \sin 18^{\circ}-4 \sin ^{3} 18^{\circ}\quad \ldots\) (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 \(=\frac{4+3+3+3+0+3+2}{7}\)

\(=\frac{18}{7}\)

\(c=x a+y b+z(a \times b)\)

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

Now, C.C \(=|c|^2\)

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

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

\(\Rightarrow 2 x ^2+ z ^2\left[| a |^2| b |^2-( a \cdot b )^2\right]=1\)

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

\(\Rightarrow 2 x ^2+ z ^2=1\Rightarrow z ^2=1-2 x ^2\)

\(=1-2 \cos ^2 \theta=-\cos 2 \theta\)

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 \(=\left|\int_{\mathrm{x}_{1}}^{\mathrm{x}_{2}}\left(\mathrm{y}_{1}-\mathrm{y}_{2}\right) \mathrm{dx}\right|\)

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 \(O A B=\left|\int_{\mathrm{x}_{1}}^{\mathrm{x}_{2}} \mathrm{f}(\mathrm{x}) \mathrm{d} \mathrm{x}\right|\)

\(=\left|\int_{0}^{\frac{\pi}{3}} \sin x ~d x\right|\)

\(=\left|[-\cos x]_{0}^{\frac{\pi}{3}}\right|\)

\(=\left|-\cos \frac{\pi}{3}+\cos 0^{\circ}\right|\)

\(=\left|1-\frac{1}{2}\right|\) \(=\frac{1}{2}\)

\(\therefore\) Shaded Area \(=2 \times \frac{1}{2}=1\)

Given,

\(5+55+555+\ldots \ldots\) to \(n\) terms

\(=5 \times[1+11+111+\ldots . .\) to \(n\) terms \(]\)

\(=\frac{5}{9} \times [9+99+999+\ldots\) to \(n\) terms \(]\)

\(=\frac{5}{9} \times\left[(10-1)+\left(10^{2}-1\right)+\left(10^{3}-1\right)+\ldots\right.\) to \(n\) terms \(]\)

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

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,

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

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

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

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

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

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

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

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

Given:

\(f(x)=x^{3}-2 x-8\)

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

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

\(x=\pm \sqrt{\frac{2}{3}}\)

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

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:

\(C_{i j}=\left\{\begin{array}{c}M_{i j}, \text { when } i+j \text { is even } \\ -M_{i j}, \text { when } i+j \text { is odd }\end{array}\right.\)

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

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

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

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:

\(\mathrm{z}=2 \mathrm{x}+5 \mathrm{y}\)

\(2 \mathrm{x}+4 \mathrm{y} \leq 8 \)

\(\Rightarrow \mathrm{x}+2 \mathrm{y} \leq 4\)

\(3 x+y \leq 6, x+y \leq 4, x \geq 0, y \geq 0\)

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.

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

The optimal solution occurs at one of the corner points.

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

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

At B \(\left(\frac{8}{5}, ~~\frac{6}{5}\right), z=2 × \frac{8}{5}+5 × \frac{6}{5}=\frac{46}{5}\)

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

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: \(3 x+4 y+z+5=0\)

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,

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

Now,

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

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

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

\( \Rightarrow \text { I.F. }=e^{\int P d x}=e^{\int \frac{3}{x} d x} \)

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

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

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

Now general solution will be,

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

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

\( \Rightarrow x ^3 \cdot y =\int 4 x ^5 dx + c \)

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

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