Mathematics Test - 5

Let \(y=\tan ^{-1}(-\sqrt{3})=-\tan ^{-1}(\sqrt{3})\)

Then, \(\tan y=-\sqrt{3}\)

We know that the range of the principal value branch of \(\tan ^{-1}\) is \(\left(\frac{-\pi}{2}, \frac{\pi}{2}\right)\) and \(\tan \left(\frac{\pi}{3}\right)=\sqrt{3}\)

\(\therefore\) The principal value of \(\tan ^{-1}(\sqrt{3})\) is \(-\frac{\pi}{3}\)     .....(i)

Next,

Let \(\mathscr{x}=\sec ^{-1}\left(\frac{2}{\sqrt{3}}\right)\)

Then, \(\sec x=\frac{2}{\sqrt{3}}\)

We know that the range of the principal value branch of \(\sec ^{-1}\) is \([0, \pi]-\frac{\pi}{2}\)

\(\therefore\) The principal value of \(\sec ^{-1}\left(\frac{2}{\sqrt{3}}\right)\) is \(\frac{\pi}{3}\)     ......(ii)

Adding (i) and (ii), we have

\(\tan ^{-1}(-\sqrt{3})+2 \sec ^{-1}\left(\frac{2}{\sqrt{3}}\right)=-\frac{\pi}{3}+2 \times \frac{\pi}{6}\)

\(=0\)

\(\therefore \tan ^{-1}(-\sqrt{3})+2 \sec ^{-1}\left(\frac{2}{\sqrt{3}}\right)=0\)

Let's say that the two curves are: 

\(f(x, y)=x+y-3=0\) 

and, \(g(x, y)=x^{2}-y-9=0\)

The points of their intersection are the points where \(f(x, y)=g(x, y)\).

\(\Rightarrow x+y-3=x^{2}-y-9=0\)

\(\Rightarrow-x-y+3=x^{2}-y-9=0\)

\(\Rightarrow x^{2}+x-12=0\)

\(\Rightarrow x^{2}+4 x-3 x-12=0\)

\(\Rightarrow x(x+4)-3(x+4)=0\)

\(\Rightarrow(x+4)(x-3)=0\)

\(\Rightarrow x+4=0\) Or \(x-3=0\)

\(\Rightarrow x=-4\) Or \(x=3\)

And, \(y=3-(-4)=7\) Or \(y=3-3=0\)

Therefore, the curves intersect at the points B(3, 0) and C(-4, 7) as shown in the diagram below:

The points where \(y=x^{2}-9\) cuts the \(x\)-axis \((y=0)\) are \(A(-3,0)\) and \(B(3,0)\). 

The required area is the shaded part \(\mathrm{ABC}=\) Area of BDC - Area of ADC

\(=\int_{-4}^{3}(3-\mathrm{x}) \mathrm{dx}-\int_{-4}^{-3}\left(\mathrm{x}^{2}-9\right) \mathrm{dx}\)

\(=3[\mathrm{x}]_{-4}^{3}-\left[\frac{\mathrm{x}^{2}}{2}\right]_{-4}^{3}-\left[\frac{\mathrm{x}^{3}}{3}\right]_{-4}^{-3}+9[\mathrm{x}]_{-4}^{-3}\)

\(=3[3-(-4)]-\frac{1}{2}\left[3^{2}-(-4)^{2}\right]-\frac{1}{3}\left[(-3)^{3}-(-4)^{3}\right]+9[-3-(-4)]\)

\(=3(7)-\frac{1}{2}(-7)-\frac{1}{3}(37)+9(1)\)

\(=21+\frac{7}{2}-\frac{37}{3}+9\)

\(=\frac{127}{6}\)

It is given that \(|z|=1\) and \(\operatorname{Re} z \neq 1\)

Let assume \(z=x+i y \Rightarrow x^{2}+y^{2}=1\)

\(\omega=\frac{1+(1-8 \alpha) z}{1-z}\)

\(\omega=\frac{(1+(1-8 \alpha))(x+i y)}{1-(x+i y)}\)

On solving the above equation ,

\(\operatorname{Re} \omega=\frac{(1+x(1-8 \alpha))(1-x)}{\left.\left(1-x^{2}\right)+y^{2}\right)}=0\) (since it is given that \(\omega\) is purely imaginary)

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

As given \(x^{2}+y^{2}=1\)

\((1-x)+x(1-8) \alpha=(1-8 \alpha)\)

\(\Rightarrow \alpha=0\)

So, \(\alpha \in\{0\}\)

Given: \(\left|\begin{array}{lll}\mathrm{x} & 2 & 3 \\ 1 & \mathrm{x} & 1 \\ 3 & 2 & \mathrm{x}\end{array}\right|=0\)

\(\Rightarrow x\left(x^{2}-2\right)-2(x-3)+3(2-3 x)=0\)

Now, \(x^{3}-2 x-2 x+6+6-9 x=0\)

\(\Rightarrow x^{3}-13 x+12=0\)

\(\because x=3\) is a root of the equation

\(\Rightarrow(x-3)\left(x^{2}+3 x-4\right)=0\)

\(\Rightarrow x^{2}+3 x-4=0\)

\(\Rightarrow(x+4)(x-1)=0\)

\(\Rightarrow x=1,-4\)

Given: \(\vec{a}, \vec{b}\) and \(\vec{c}\) are coplanar i.e., \([\vec{a} \vec{b} \vec{c}]=0\)

\(\Rightarrow(2 \vec{a} \times 3 \vec{b}) \cdot 4 \vec{c}=[2 \vec{a} 3 \vec{b} 4 \vec{c}]\) and \((5 \vec{b} \times 3 \vec{c}) \cdot 6 \vec{a}=[5 \vec{b} 3 \vec{c} 6 \vec{a}]\)

\(\Rightarrow(2 \vec{a} \times 3 \vec{b}) \cdot 4 \vec{c}+(5 \vec{b} \times 3 \vec{c}) \cdot 6 \vec{a}=[2 \vec{a} 3 \vec{b} 4 \vec{c}]+[5 \vec{b} 3 \vec{c} 6 \vec{a}]\)

As we know that, \([a b c]=[b c a]=[c a b]\)

\(\Rightarrow(2 \vec{a} \times 3 \vec{b}) \cdot 4 \vec{c}+(5 \vec{b} \times 3 \vec{c}) \cdot 6 \vec{a}=[2 \vec{a} 3 \vec{b} 4 \vec{c}]+[6 \vec{a} 5 \vec{b} 3 \vec{c}]\)

As we know that, \([\lambda a, b c]=\lambda[a b c]\)

\(\Rightarrow[2 \vec{a} 3 \vec{b} 4 \vec{c}]+[6 \vec{a} 5 \vec{b} 3 \vec{c}]=24[\vec{a} \vec{b} \vec{c}]+90[\vec{a} \vec{b} \vec{c}]\)

As we know that, vectors \(\vec{a}, \vec{b}\) and \(\vec{c}\) are coplanar if and only if \([ a b c ]=0\)

\(\Rightarrow[2 \vec{a} 3 \vec{b} 4 \vec{c}]+[6 \vec{a} 5 \vec{b} 3 \vec{c}]=0\)

\(\Rightarrow(2 \vec{a} \times 3 \vec{b}) \cdot 4 \vec{c}+(5 \vec{b} \times 3 \vec{c}) \cdot 6 \vec{a}=0\)

Given,

\(n^{\text {th }}\) terms of two A.P's are \(3 n+8\) and \(7 n+15\).

So, \(12^{\text {th }}\) term in \(1^{\text {st }} \mathrm{AP}\): 

a_n = 3 n + 8

For n = 12,

\(a _{12}= 3 \times 12+8\)

\(=36+8=44\)

Similarly, \(12^{\text {th }}\) term of \(2^{\text {nd }}\) AP: 

a_n = 7n + 15

For n = 12,

\(a_{12}=7 \times 12+15\)

\(=84+15=99\)

\(\therefore\) The required ratio \(=\frac{44 }{ 99}\)

\(=\frac{4 } 9=4: 9\)

As we know,

Straight Lines:

The general equation of a line is \(y=m x+c\), where \(m\) is the slope of the line.

Parallel Lines: If two lines are parallel, then their slopes are equal.

Let the equation of the line passing through \((3,4)\) and \((4,5)\) be \(y=m x+c\).

\(\therefore\) We must have:

\(4=3 m+c\)     .....(i)

And, \(5=4 m+c\)      (ii)

Subtracting equation (i) from equation (ii), we get:

\(1=\mathrm{m}\)

Let the equation of the line passing through the point \((1,2)\) be \(y=n x+d\).

Since this line is parallel to the above line, we must have \({n}={m}=1\).

Also, \(2=1(1)+\mathrm{d}\)

\(\Rightarrow \mathrm{d}=1\)

The required equation is:

\(y=x+1\)

Let, two lines having direction ratios \(a_{1}, b_{1}, c_{1}\), and \(a_{2}, b_{2}, c_{2}\) respectively.

We know that:

Condition for perpendicular lines is: 

\(a _{1} a _{2}+ b _{1} b _{2}+ c _{1}  c_{2}=0\)

Condition for parallel lines is: 

\(\frac{ a _{1}}{ a _{2}}=\frac{ b _{1}}{ b _{2}}=\frac{ c _{1}}{ c _{2}}\)

The equation of the plane passing through the given point is:

\(a(x-1)+b(y-0)+c(z-1)=0\)

Given perpendicular planes are: 

\(2 x+3 y-z=2\) and \(x-y+2 z=1\)

\(\therefore 2 a+3 b-c=0 \quad \quad \ldots\)(i)

Also,

\(a-b+2 c=0\quad \quad \ldots\)(ii)

On substracting \(2 \times\) (ii) from (i),

\(5 b-5 c=0\)

\(b=c\)

Putting it in equation (ii),

\(a-c+2 c=0\)

\(a=-c\)

Now, putting the values of \(a\) and \(b\) in the equation of the plane,

\(-c(x-1)+c(y-0)+c(z-1)=0\)

\(-x+1+y+z-1=0\)

\(x-y-z=0\)

Given,

\(P(n)=2  \times 4^{2 n+1}+3^{3 n+1}\)

By putting \(n=0\), we get

\(P(0)=2  \times 4^{2 \times 0+1}+3^{3 \times 0+1}\)

\(=8+3\)

\(=11\)

As we can say that \(11\) is divisible by \(11\).

By putting \(n=1\), we get

\(P(1)=2  \times 4^{2 \times 1+1}+3^{3 \times 1+1}\)

\(=2  \times 4^{3}+3^{4}=209\)

\(=11 \times 19\)

As we can say that \(209\) is divisible by \(11\).

By putting \(n=2\), we get

\(P(2)=2  \times 4^{2 \times 2+1}+3^{3 \times 2+1}\)

\(=2 \times 4^{5}+3^{7}=209\)

\(=2,048+2,187\)

\(=4235\)

\(=11 \times 11 \times 35\)

As we can say that \(4235\) is divisible by \(11\).

Therefore we can say that \(P(n)\) is divisible by \(11\).

Two digit numbers which are divisible by 4 are 12, 16, 20, ..., 96 forms an AP with first term a = 12, common difference d = 4 and n^th term a_n = 96.

⇒ a_n = a + (n - 1) × d 

⇒ 12 + (n - 1) × 4 = 96

⇒ n = 22