Mathematics Test - 1

It is given that the vertex of the parabola is at the origin and its axis lies along the \(X\)-axis.

So, its equation is:

\({y}^{2}=4 {ax}\) OR \({y}^{2}=-4 {ax}\)

Since it passes through the point \({P}(3,4)\), so it lies in the first quadrant.

\(\therefore\) Its equation is \({y}^{2}=4 {ax}\)

Now, \({P}(3,4)\) lies on it, so,

\(4^{2}=4 {a}(3)\)

\(\Rightarrow 16=12 {a}\)

\(\Rightarrow {a}=\frac{4}{3}\)

\(\therefore\) The required equation is \({y}^{2}=4\left(\frac{4}{3}\right) {x}\)

\(\Rightarrow {y}^{2}=\frac{16}{3} {x}\)

\(\frac{\cot 54^{\circ}}{\tan 36^{\circ}}+\frac{\tan 20^{\circ}}{\cot 70^{\circ}}\)

\(=\frac{\tan \left(90^{\circ}-36^{\circ}\right)}{\tan 36^{\circ}}+\frac{\cot \left(90^{\circ}-70^{\circ}\right)}{\cot 70^{\circ}}\)

\(=\frac{\tan 36^{\circ}}{\tan 36^{\circ}}+\frac{\cot 70^{\circ}}{\cot 70^{\circ}}\)

\(=2\)

Total number of times dice was rolled =80

Number of times 3 is obtained =14

$Probability =\frac{Favourable Outcomes}{Total Outcomes}$

⇒ Probability of obtaining 3$=\frac{14}{80}=\frac{7}{40}$

Given: Ellipse=$\frac{x^2}{16}+\frac{y^2}{4}=1$

$e=\sqrt{1-\frac{b^2}{a^2}}=\frac{\sqrt{3}}{2}$

∵P is a point on the ellipse 

So,$P=(4\cos \theta ,2\sin \theta )$

Equation of normal to the ellipse $\frac{x^2}{16}+\frac{y^2}{4}=1$ at point $(x_{1 }, y_1)=(4\cos \theta ,2\sin \theta )$is given by

$a^2{y}_1(x-x_1)=b^2{x}_1(y-y_1)$

$⟹16\times 2\sin \theta (x-4\cos \theta )=4\times 4\cos \theta (y-2\sin \theta )$

$⟹2x\sin \theta -8\sin \theta \cos \theta =y\cos \theta -2\sin \theta \cos \theta$

$⟹2x\sin \theta =y\cos \theta +6\sin \theta \cos \theta$

$⟹\frac{2x}{\cos \theta }=\frac{y}{\sin \theta }+6$

$⟹2xsec\theta -y\cos ec\theta =6$

It meet the x-axis at $Q(3\cos \theta ,0)$

$\therefore M=(\frac{7}{2}\cos \theta ,\sin \theta )=(x,y)$

Locus of M is

$\frac{x^2}{{\left({\displaystyle \frac{7}{2}}\right)}^2}+\frac{y^2}{1}=1$

Latus rectum of the given ellipse is

$x=\pm ae=\pm \sqrt{16-4}=\pm 2\sqrt{3}$

So locus of M meets the latus rectum at points for which

$y^2=1-\frac{12\times 4}{49}=491 \Rightarrow y=\pm \frac{1}{7}$

So, the required point is $(\pm 2\sqrt{3},\pm \frac{1}{7}).$

Given that:

$\frac{{\sin }^{4}x}{2}+\frac{{\cos }^{4}x}{3}=\frac{1}{5}$

$\frac{{\left({\sin }^{2}x\right)}^2}{2}+\frac{{(1-{\sin }^{2}x)}^2}{3}=\frac{1}{5}$

Let $Si{n}^{2}x=t$

$t\in [0,1]$

$\frac{t^2}{2}+\frac{{(1-t)}^2}{3}=\frac{1}{5}$

$\frac{t^2}{2}+\frac{1-2t+t^2}{3}=\frac{1}{5}$

$25t^2-20t+10=6$

$25t^2-20t+4=0$

$25t^2-10t-10t+4=0$

$(5t-2)(5t-2)=0$

$t=\frac{2}{5}$

$\therefore {\sin }^{2}x=\frac{2}{5}$

${\cos }^{2}x=1-{\sin }^{2}x=1-\frac{2}{5}=\frac{3}{5}$

${\tan }^{2}x=\frac{{\sin }^{2}x}{{\cos }^{2}x}=\frac{2}{3}$

Let p be the probability of success and q be the probability of failure.

Given that, the probability of success is twice the probability of failure. Thus,

=>p = 2q

We know that, p+q= 1

So, 2q+q=1

=>3q=1

=>q =⅓ and p=⅔

∴  Required probability =P(Four success)+P(Five success)+P(Six success)

\(\Rightarrow{ }^{6} C_{4}(p)^{4}(q)^{6-4}+{ }^{6} C_{5}(p)^{5}(q)^{6-5}+{ }^{6} C_{6}(p)^{6}(q)^{6-6}\)

\(\Rightarrow \frac{6 \times 5}{2} \times\left(\frac{2}{3}\right)^{4} \times\left(\frac{1}{3}\right)^{2}+6 \times\left(\frac{2}{3}\right)^{5} \times\left(\frac{1}{3}\right)^{1}+1 \times\left(\frac{2}{3}\right)^{6} \times\left(\frac{1}{3}\right)^{0}\)

\(\Rightarrow \frac{240}{729}+\frac{192}{729}+\frac{64}{729}\)

\(\Rightarrow \frac{496}{729}\)