Mathematics Test - 24

Let the terms of infinite series are \(a, a r, a r^2, a r^3, \ldots\)

So, \(\frac{\mathrm{a}}{1-\mathrm{r}}=3\)

Since, sum of cubes of its terms is \(\frac{27}{19}\) that is sum of \(a^3\), \(a^3 r^3, \ldots \infty\) is \(\frac{27}{19}\)

So, \(\frac{\mathrm{a}^3}{1-\mathrm{r}^3}=\frac{27}{19}\)

\(\Rightarrow \frac{\mathrm{a}}{1-\mathrm{r}} \times \frac{\mathrm{a}^2}{\left(1+\mathrm{r}^2+\mathrm{r}\right)}=\frac{27}{19}\)

\(\Rightarrow \frac{9\left(1+\mathrm{r}^2-2 \mathrm{r}\right) \times 3}{1+\mathrm{r}^2+\mathrm{r}}=\frac{27}{19}\)

\(\Rightarrow 6 \mathrm{r}^2-13 \mathrm{r}+6=0\)

\(\Rightarrow(3 \mathrm{r}-2)(2 \mathrm{r}-3)=0\)

\(\Rightarrow \mathrm{r}=\frac{2}{3}\), or \(\frac{3}{2}\)

As \(|\mathrm{r}|<1\)

So, \(\mathrm{r}=\frac{2}{3}\)

Let, \(\mathrm{I}=\int_{-1}^{1} \frac{5 x^{4}}{\sqrt{x^{5}+3}} d x\)

Let \(x^{5}+3=t\) \(\quad\dots\dots(i)\)

Differentiating w.r.t \(x\), we get

\(5 x^{4} =\frac{d t}{dx}\)

\(5 x^{4} d x=d t\)

The new limits to \(eq^{n}(i)\)

When \(x=-1,=({-1}^{5}+3)\), \(t=2\)

Similarly,

When \(x=1, \mathrm{t}=4\)

\(\therefore \int_{-1}^{1} \frac{5 x^{4}}{\sqrt{x^{5}+3}} d x=\int_{2}^{4} \frac{d t}{\sqrt{t}} \)

\(=[2 \sqrt{t}]_{2}^{4}\)

Put the value of limit,

\(=2(\sqrt{4}-\sqrt{2})\)

\(=4-2 \sqrt{2}\)

If \(A\) be a \(3 \times 3\) matrix given by:

\(A=\left[\begin{array}{lll}a & b & c \\ f & e & d \\ g & h & i\end{array}\right]\)

then the value of \(| A |\) also written as \(\operatorname{det}( A )\)

\(\operatorname{det}(A)=a(e i-d h)-b(f i-d g)+c(f h-e g)\)

\(A=\left[\begin{array}{ccc}4 & 7 & 1 \\ -1 & 3 & 2 \\ -2 & 0 & 5\end{array}\right]\)

\(\therefore \operatorname{det}(A)=4(15-0)-7(-5+4)+1(0+6)\)

\(=4(15)-7(-1)+1(6)\)

\(=60+7+6\)

\(=73\)

Now, If \(A\) be a matrix of order \(n \times n\) and \(\operatorname{det}( A )= k\). Then for a scaler \(c\), the following property holds:

\(\operatorname{det}(c A)=c^{n} \operatorname{det}(A)\)

\(\therefore \operatorname{det}(3 A )=3^{3} \operatorname{det}( A )\)

\(=27 \times 73\)\(=1971\)

Given:

\(\mathrm{f}(\mathrm{x})=2 \mathrm{x}^{3}+\frac{1}{\mathrm{x}}\)

We know that:

If \(f(x)=x^{n}\), then:

\(f'(x)=n x^{n-1}\)

Differentiating with respect to x, we get:

\(f'(x)=6 x^{2}-\frac{1}{x^{2}}\)

Putting \(x=1\) in above, we get:

\(f'(1)=6 \times 1^{2}-\frac{1}{1^{2}}\)

\(f'(1)=6-1=5\)

\(\therefore\) The value of \(f^{\prime}(1)\) is 5.

 \({y}=\sin {x}\)

The graph of \(y=\sin x\) can be drawn as shown in the diagram:

\(\therefore\) Required area \(=\) Area \(O A B O+\) Area \(B C D B\)

\(=\int_{0}^{\pi} \sin x d x+\left|\int_{\pi}^{2 \pi} \sin x d x\right|\)

\(=[-\cos x]_{0}^{\pi}+\left|[-\cos x]_{\pi}^{2 \pi}\right|\)

\(=[-\cos \pi+\cos 0]+\left|-\cos 2 \pi+\cos \pi\right|\)

\(=1+1+|(-1-1)|\)

\(=2+|-2|\) \(=2+2\) \(=4\) sq. units

Given,

The distances of \(\mathrm{P}(x, y)\) from \(\mathrm{A}(4,1)\) and \(\mathrm{B}(-1,4)\) are equal.

The distance ' \(d\) ' between two points \(\left(x_{1}, y_{1}\right)\) and \(\left(x_{2}, y_{2}\right)\) is obtained by using the Pythagoras' Theorem: \(d^{2}=\left(x_{1}-x_{2}\right)^{2}+\left(y_{1}-y_{2}\right)^{2}\).

Using the formula for the distance between two points:

\(\mathrm{AP}^{2}=(x-4)^{2}+(y-1)^{2}\)

\(\mathrm{BP}^{2}=(x+1)^{2}+(y-4)^{2}\)

Since the distances are equal, we have:

\(\mathrm{AP}^{2}=\mathrm{BP}^{2} \)

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

\(\Rightarrow x^{2}-8 x+16+y^{2}-2 y+1=x^{2}+2 x+1+y^{2}-8 y+16 \)

\(\Rightarrow 10 x=6 y\)

\(\Rightarrow 5 x=3 y\)

Given,

\(\tan \left(2 \tan ^{-1}(\cos x)\right)\)

As we know, \(2 \tan ^{-1} x =\tan ^{-1} \frac{2 x }{1- x ^{2}}\)

Therefore,

\(2 \tan ^{-1} \cos x=\tan ^{-1} \frac{2 \cos x}{1-\cos ^{2} x}\)

\(=\tan ^{-1} \frac{2 \cos x}{\sin ^{2} x}\)\(\quad(\because 1 - cos^2{x} =sin^2{x})\)

\(=\tan ^{-1}(2 \cot x \operatorname{cosec} x)\)

\(\tan \left(\tan ^{-1}(2 \cot x \operatorname{cosec} x)\right)=2 \cot x \operatorname{cosec} x \quad\left(\because \tan \left(\tan ^{-1} x\right)=x\right)\)

Here, \(S=\{1,2,3,4, \ldots ., 19,20\}\)

Let \(E=\) event of getting a multiple of \(3\) or \(5\)

\(=\{3,6,9,12,15,18,5,10,20\}\)

\(\therefore P(E)=\frac{n(E)}{n(S)}=\frac{9}{20}\)