Mathematics Test - 13

Given,

\(x^{2}+7 x-10=0\)

Comparing the given equation with \(a x^{2}+b x+c=0\), we get \(a=1, b=7\) and \(c=-10\)

As we know,

Discriminant \(=b^{2}-4 a c\)

\(\therefore\) Discriminant \(=(7)^{2}-4 \times 1 \times(-10)\)

\(=49+40\)

\(=89\)

As we know,

If discriminant \(\left(b^{2}-4 a c\right)>0\) then roots are real.

As we can see, discriminant \(=89>0\)

\(\therefore\) The roots of the given equation are real.

Objective function: \(\mathrm{Z}=-3 x+4 y\)

We have to minimize Z on constraints

\(x+2 y \leq 8 \)

\(3 x+2 y \leq 12\)

\(x \geq 0, y \geq 0\)

After plotting inequalities, we got feasible region as shown in the image.

Now, there are 3 corner points \((0,4),(2,3)\) and \((4,0)\).

At \((0,4)\), value of \(Z=-3 \times 0+4 \times 4=16\)

At \((2,3)\), value of \(Z=-3 \times 2+4 \times 3=6\)

\(\operatorname{At}(4,0)\), value of \(\mathrm{Z}=-3 \times 4+4 \times 0=-12\)

So, minimum value of \(Z=-12\).

Given,

\(y=4-x^{2} \ldots(1) \)

\(y=x^{2} \ldots(2)\)

From equation (1) and (2), we get

\(4-x^{2}=x^{2} \)

\(\Rightarrow 2 x^{2}=4 \)

\(\Rightarrow x=\pm \sqrt{2} \ldots(3)\)

If \(m_{1}\) is the slope of equation (1) then

\(m_{1}=\frac{d y}{d x}=\frac{d}{d x}\left(4-x^{2}\right)=0-2 x \)

\(\Rightarrow m_{1}=-2(\sqrt{2}) \text { [from (3)] }\)

If \(m_{2}\) is the slope of equation (2) then

\(m_{2}=\frac{d y}{d x}=\frac{d}{d x}\left(x^{2}\right)=2 x \)

\(\Rightarrow m_{2}=2 \sqrt{2}\)

Angle of intersection between (1) and (2) will be

\(\alpha=\tan ^{-1}\left|\frac{m_{1}-m_{2}}{1+m_{1} m_{2}}\right| \)

\(\Rightarrow \alpha=\tan ^{-1}\left|\frac{-2 \sqrt{2}-2 \sqrt{2}}{1+2 \sqrt{2} \times -2 \sqrt{2}}\right| \)

\(\Rightarrow \alpha=\tan ^{-1}\left|\frac{-4 \sqrt{2}}{-7}\right| \)

\(\alpha=\tan ^{-1}\left(\frac{4 \sqrt{2}}{7}\right)\)

\(\mathrm{a}=2, \mathrm{~b}=3, \sin \mathrm{A}=\frac{2}{3} \Rightarrow \sin ^{2} \mathrm{~A}=\frac{4}{9}\)

\(\therefore \cos ^{2} \mathrm{~A}=1-\sin ^{2} \mathrm{~A}=\frac{5}{9}\)

\(\Rightarrow \cos \mathrm{A}=\frac{\sqrt{5}}{3}\)

By sine rule, \(\frac{\sin A}{a}=\frac{\sin B}{b}\)

\(\Rightarrow \frac{2}{3} \times \frac{1}{2}=\frac{\sin \mathrm{B}}{3}\)

\(\Rightarrow \sin B=1 \Rightarrow B=90^{\circ}\)

In \(\Delta \mathrm{ABC}, \cos \mathrm{C}=\frac{\mathrm{a}}{\mathrm{b}}=\frac{2}{3}\)

Given,

For line I

\(x=a y+b \)

\(\Rightarrow y=\frac{x-b}{a} \ldots(1) \)

\(z=c y+d \)

\(\Rightarrow y=\frac{z-d}{c} \ldots(2)\)

From (1) and (2),

\(\frac{x-b}{a}=y=\frac{z-d}{c}\)

which can also be written as,

\(\frac{x-a}{a}=\frac{y-0}{1}=\frac{z-d}{c} \ldots(3)\)

Similarly,

For line II

\(\Rightarrow y=\frac{x-b^{\prime}}{a^{\prime}} \ldots(4) \)

\(z=c^{\prime} y+d^{\prime} \)

\(\Rightarrow y=\frac{z-d^{\prime}}{c^{\prime}} \ldots(5)\)

we can write,

\(\frac{x-b^{\prime}}{a^{\prime}}=\frac{y-0}{1}=\frac{z-d^{\prime}}{c^{z}} \text {... (6) }\)

As we know if two lines are perpendicular then \(a_{1} a_{2}+b_{1} b_{2}+c_{1} c_{2}=0\)

From equation (3) and (6), we get \(a a^{\prime}+c c^{\prime}+1=0\)

Given,

\(\tan 20^{\circ}=p\)

So,

\(\frac{\tan 160^{\circ}-\tan 110^{\circ}}{1+\tan 160^{\circ} \tan 110^{\circ}} \)

\(=\frac{\tan \left(180^{\circ}-20^{\circ}\right)-\tan \left(90^{\circ}+20^{\circ}\right)}{1+\tan \left(180^{\circ}-20^{\circ}\right) \tan \left(90^{\circ}+20^{\circ}\right)} \)

\(=\frac{-\tan 20^{\circ}+\cot 20^{\circ}}{1+\tan 20^{\circ} \cot 20^{\circ}} \)

\(=\frac{-p+\frac{1}{p}}{1+1}=\frac{1-p^{2}}{2 p}\)

\(\therefore \frac{\tan 160^{\circ}-\tan 110^{\circ}}{1+\tan 160^{\circ} \tan 110^{\circ}}\) is equal to \(\left(\frac{1-p^{2}}{2 p}\right)\).

The magnitude of a \(3 d\) vector \(x \hat{i}+y \hat{j}+\mathrm{z} \hat{k}\) is given by:

\(\sqrt{x^{2}+y^{2}+z^{2}} \)

\(\vec{c}=4 \hat{i}-5 \hat{j}+11 \hat{k} \)

\(=\sqrt{4^{2}+(-5)^{2}+11^{2}} \)

\(=\sqrt{164} \)

\(=9 \sqrt{2}\)

\(\therefore\) Modulus of vector \(\vec{c}=4 \hat{i}-5 \hat{j}+11 \hat{k}\) is \(9 \sqrt{2}\).

Let \(\mathrm{I}=\int \sqrt{\mathrm{x}} \mathrm{e}^{\sqrt{\mathrm{x}}} \mathrm{dx}\)

Substituting \(\sqrt{\mathrm{x}}=\mathrm{t},\) we get:

\(\frac{1}{2 \sqrt{\mathrm{x}}} \mathrm{dx}=\mathrm{d} \mathrm{t}\)

\(d x=2 t d t\)

\(\therefore \mathrm{I}=\int \mathrm{t} \times \mathbf{e}^{\mathrm{t}} \times 2 \mathrm{t} \mathrm{d} \mathrm{t}\)

Integrating by parts, taking \(\mathrm{t}^{2}\) as the first function and \(\mathrm{e}^{\mathrm{t}}\) as the second function, we get:

\(\mathrm{I}=2\left[\mathrm{t}^{2} \int \mathrm{e}^{\mathrm{t}} \mathrm{d} \mathrm{t}-\int\left(\frac{\mathrm{d}}{\mathrm{dt}} \mathrm{t}^{2} \int \mathrm{e}^{\mathrm{t}} \mathrm{d} \mathrm{t}\right) \mathrm{dt}\right]+\mathrm{C}\)

\(\mathrm{I}=2 t^{2} e^{t}-4 \int t e^{t} d t+C\)

Integrating \(\int \mathrm{te}^{\mathrm{t}} \mathrm{dt}\) by parts, we get:

\(\mathrm{I}=2 \mathrm{t}^{2} \mathrm{e}^{\mathrm{t}}-4\left[\mathrm{t} \int \mathrm{e}^{\mathrm{t}} \mathrm{d} \mathrm{t}-\int\left(\frac{\mathrm{d}}{\mathrm{dt}} \mathrm{t} \int \mathrm{e}^{\mathrm{t}} \mathrm{d} \mathrm{t}\right) \mathrm{dt}\right]+\mathrm{C}\)

\(\mathrm{I}=2 \mathrm{t}^{2} \mathrm{e}^{\mathrm{t}}-4\left(\mathrm{te}^{\mathrm{t}}-\mathrm{e}^{\mathrm{t}}\right)+\mathrm{C}\)

\(\mathrm{I}=\left(2 t^{2}-4 t+4\right) e^{t}+C\)

Back substituting \(\sqrt{\mathrm{x}}=\mathrm{t},\) we get:

\(\mathrm{I}=(2 \mathrm{x}-4 \sqrt{\mathrm{x}}+4) \mathrm{e}^{\sqrt{\mathrm{x}}}+\mathrm{C}\)

If a line \(y=m x+c .\) (1)

Touches the ellipse

\(\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1 \text { then } \)

\(c^{2}=a^{2} m^{2}+b^{2} \ldots(2)\)

Given line \(x \cos \alpha+y \sin \alpha=p\)

Let,

\(x \cos \alpha+y \sin \alpha=0 \)

\(\Rightarrow y \sin \alpha=-x \cos \alpha+b \ldots(3) \)

\(\Rightarrow y=-x \frac{\cos \alpha}{\sin \alpha}+\frac{b}{\sin \alpha}\)

On comparing with (1), we get

\(m=-\frac{\cos \alpha}{\sin \alpha} \text { and } c=\frac{b}{\sin \alpha}\)

Then from equation (2)

\(c^{2}=a^{2} m^{2}+b^{2} \)

\(\Rightarrow \frac{b^{2}}{\sin ^{2} \alpha}=a^{2}\left(\frac{\cos ^{2} \alpha}{\sin ^{2} \alpha}\right)+b^{2} \)

\(\Rightarrow b^{2}=a^{2} \cos ^{2} \alpha+b^{2} \sin ^{2} \alpha\)

Comparing it with given line, we get

\(p^{2}=a^{2} \cos ^{2} \alpha+b^{2} \sin ^{2} \alpha\)

Given,

\(\cot A+\operatorname{cosec} A\)

\(\Rightarrow \cot A+\operatorname{cosec} A=\frac{\cos A}{\sin A}+\frac{1}{\sin A}=\frac{1+\cos A}{\sin A}\)

As we know that,

\(\cos 2 \mathrm{~A}=\cos ^{2} \mathrm{~A}-\sin ^{2} \mathrm{~A}=2 \cos ^{2} \mathrm{~A}-1\)

\(\Rightarrow \cot A+\operatorname{cosec} A=\frac{1+\cos A}{\sin A}=\frac{2 \times \cos ^{2}\left(\frac{A}{2}\right)}{\sin A}\)

As we know that, \(\sin 2 \mathrm{~A}=2 \sin \mathrm{A} \times \cos \mathrm{A}\)

\(\Rightarrow \cot A+\operatorname{cosec} A=\frac{1+\cos A}{\sin A}=\frac{2 \times \cos ^{2}\left(\frac{A}{2}\right)}{\sin A} \)

\(=\frac{2 \times \cos ^{2}\left(\frac{A}{2}\right)}{2 \times \cos \left(\frac{A}{2}\right) \times \sin \left(\frac{A}{2}\right)} \)

\(=\cot \left(\frac{A}{2}\right)\)