Mathematics Test - 20

Given: \(f(x)=\sqrt{9-x^{2}}\)

Let, \(y^{2}=9-x^{2}\)

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

\(x=\sqrt{9-y^{2}} \quad\quad\ldots\)(1)

We know that for any function \(f(x)=\sqrt{9-x^{2}}, y \geq 0\quad\quad \ldots\). (A)

From equation (1).

\(9-y^{2} \geq 0\)

\(y^{2}-9 \leq 0\)

\(\Rightarrow(y+3)(y-3) \leq 0\)

\(\Rightarrow-3 \leq y \leq 3\)

But from equation (A) y must be positive, then, \(y=[0,3]\).

Range \(=\) the value of \({y}=[0,3]\)

\(\begin{aligned}& f(x)=(a+b-2 c) x^2+(b+c-2 a) x+(c+a-2 b) \\& f(x)=a+b-2 c+b+c-2 a+c+a-2 b=0 \\& f(1)=0 \\& \therefore \alpha \cdot 1=\frac{c+a-2 b}{a+b-2 c} \\& \alpha=\frac{c+a-2 b}{a+b-2 c} \\& \text { If }-1<\alpha<0 \\& -1<\frac{c+a-2 b}{a+b-2 c}<0 \\& b+c<2 a \text { and } b>\frac{a+c}{2}\end{aligned}\)

therefore, b cannot be G.M. between a and c.

If, \(0<\alpha<1\)

\(\begin{aligned}& 0<\frac{c+a-2 b}{a+b-2 c}<1 \\& b>c \text { and } b<\frac{a+c}{2}\end{aligned}\)

Therefore, \(\mathrm{b}\) may be the G.M. between \(\mathrm{a}\) and \(\mathrm{c}\).

Given:

Expansion \(\left(\frac{a}{b}-\frac{b}{2 a^{2}}\right)^{12}\)

To find the \(9^{\text {th }}\) term in the expansion of \(\left(\frac{a}{b}-\frac{b}{2 a^{2}}\right)^{12}\)

Formula used: (i) \({ }^{n} C_{r}=\frac{n !}{(n-r) !(r) !}\) and \(T_{r+1}={ }^{n} C_{r} a^{n-r} b^{r}\)

For \(9^{\text {th }}\) term,

\(r+1=9\)

\(\Rightarrow r=8 \)

In \(\left(\frac{a}{b}-\frac{b}{2 a^{2}}\right)^{12} \)

\(9^{\text {th }}\) term \(=T_{8+1}\)

\({ }^{12} \mathrm{C}_{8}\left(\frac{a}{b}\right)^{12-8}\left(\frac{-b}{2 a^{2}}\right)^{8} \)

\(\Rightarrow \frac{12 !}{8 !(12-8) !}\left(\frac{a}{b}\right)^{4}\left(\frac{-b}{2 a^{2}}\right)^{8} \)

\(\Rightarrow 495\left(\frac{a^{4}}{b^{4}}\right)\left(\frac{b^{8}}{256 a^{16}}\right) \)

\(\Rightarrow\left(\frac{495 b^{4}}{256 a^{12}}\right) \)

As we know,

In an ellipse \(\frac{ x ^{2}}{ a ^{2}}+\frac{ y ^{2}}{ b ^{2}}=1, a >b\)

The length of the latus rectum \(=\frac{2 b ^{2}}{ a }\)

Its eccentricity is given by,

\(e=\sqrt{1-\frac{b^{2}}{a^{2}}}\)

Let the equation of the ellipse be \(\frac{ x ^{2}}{ a ^{2}}+\frac{ y ^{2}}{ b ^{2}}=1\) where \(a >b\).

According to the question,

Length of the latus rectum \(=\) Half of its minor axis

\(\frac{2 b ^{2}}{ a }=b\)

\(\Rightarrow \frac{2 b ^{2}}{b}=a\)

\(\therefore a =2 b\)

Now, the eccentricity of the ellipse \((e)=\sqrt{1-\frac{b^{2}}{a^{2}}}\)

\(=\sqrt{1-\frac{b^{2}}{ (2b)^{2}}}\)

\(=\sqrt{1-\frac{b^{2}}{ 4b^{2}}}\)

\(=\sqrt{1-\frac{1}{4}}\)

\(=\sqrt{\frac{3}{4}}\)

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

P - Q = The set of elements which belong to P but not to Q.

\(P \cup Q=\) The set which consists of all elements of P and all elements of Q, the common elements being taken only once.

\(P \cap Q=\) The set of all elements which are common to both Pand Q.

let us supposeP = {1, 2, 3, 4} and Q = {3, 4, 5, 6}

\(\Rightarrow P-Q=\{1,2\}\)

\(\Rightarrow Q-P=\{5,6\}\)

\(\Rightarrow P \cup Q=\{1,2,3,4,5,6\}\)

\(\Rightarrow P \cap Q=\{3,4\}\)

\(\Rightarrow(P-Q) \cup(Q-P) \cup(P \cap Q)=\{1,2\} \cup\{5,6\} \cup\{3,4\}=\{1,2,3,4,5,6\}=P \cup Q\)

As we know,

Let two points be \(A\left(x_{1}, y_{1}, z_{1}\right)\) and \(B\left(x_{2}, y_{2}, z_{2}\right)\).

Distance between \(A\) and \(B =\sqrt{\left( x _{2}- x _{1}\right)^{2}+\left( y _{2}- y _{1}\right)^{2}+\left( z _{2}- z _{1}\right)^{2}}\)

Midpoint \(=\left(\frac{x_{1}+x_{2}}{2}, \frac{y_{1}+y_{2}}{2}, \frac{z_{1}+z_{2}}{2}\right)\)

The perpendicular distance of the point \((1,2,3)\) from \(x\) axis is \(h\).

\( h =\) distance between \((1,2,3)\) and \((1,0,0)\)

\(\therefore h =\sqrt{(1-1)^{2}+(0-2)^{2}+(0-3)^{2}}\)

\(\Rightarrow h =\sqrt{(0)^{2}+(-2)^{2}+(-3)^{2}}\)

\(\Rightarrow h =\sqrt{0+4+9} \)

\(\Rightarrow h =\sqrt{13}\)

The perpendicular distance of the point \((1,2,3)\) from \(z\) axis is \(k\).

\( k =\) distance between \((1,2,3)\) and \((0,0,3)\)

\(\therefore k =\sqrt{(0-1)^{2}+(0-2)^{2}+(3-3)^{2}}\)

\(\Rightarrow k =\sqrt{(-1)^{2}+(-2)^{2}+(0)^{2}}\)

\(\Rightarrow k =\sqrt{1+4+0} \)

\(\Rightarrow k =\sqrt{5}\)

Now,

\(h k =\sqrt{13} \times \sqrt{5}=\sqrt{65}\)

Form of six digit palindrome number xyzzyx

This will be divisible by 55 Hence, x = 5 and 5yzzy5 will be divisible by 11 .

⇒ (5 + z + y) − (y + z + 5) is divisible by 11 which is true for all values of y and z ⇒y and z can be chosen in 10×10 ways Number of such number =100

Given,

The distance of the point \((k+2,2 k+3)\) from the line \(x+3 y=7\) is \(\frac{4}{\sqrt{10}}\).

Let \(P=(k+2,2 k+3)\)

\(\Rightarrow x_{1}=k+2\) and \(y_{1}=2 k+3\)

Here, \(a=1\) and \(b=3\)

Now substitute \(x_{1}=k+2\) and \(y_{1}=2 k+3\) in the equation \(x+3 y-7=0\) we get,

\(\Rightarrow\left|x_{1}+3 \cdot y_{1}-7\right|=|7 k+4|\)

\(\Rightarrow \sqrt{a^{2}+b^{2}}\)

\(=\sqrt{1^{2}+3^{2}}\)

\(=\sqrt{10}\)

As we know that, the perpendicular distance \(d\) from \(P\left(x_{1}, y_{1}\right)\) to the line \(a x+b y+c\) \(=0\) is given by:

\(d=\left|\frac{a x_{1}+b y_{1}+c}{\sqrt{a^{2}+b^{2}}}\right|\)

\( d=\left|\frac{7 k+4}{\sqrt{1^{2}+3^{2}}}\right|\)

\(=\frac{4}{\sqrt{10}} \)

\(\Rightarrow|7 k+4|\)

\(=4\)

\(\Rightarrow k=0 \text { or }\frac{-8}{ 7}\)

Multiples of 11 such that they are of 3 -digit and less than 500.

121,132,...,495

\(\begin{aligned} & \mathrm{cn}=\frac{495-121}{11}+1=35 \\ & \mathrm{~S}=\frac{35}{2}(121+495)=10780\end{aligned}\)

Again, multiplies of 11 which are 3-digits, less than 500 and having 1 at hundred's place are 121,132,...,198

\(\begin{aligned} & n_1=\left(\frac{198-121}{11}\right)+1=8 \\ & S_1=\frac{8}{2}(121+198)=1276\end{aligned}\)

\(\therefore \mathrm{S}_2=319+418=737\)

The multiple of 11 which are 3-digits, less than 500 and having 1 at unit place are 231,341,451

\(\begin{aligned} & \therefore \mathrm{S}_3=231+341+451=1023 \\ & \therefore \text { Required sum }=\mathrm{S}-\mathrm{S}_1-\mathrm{S}_2-\mathrm{S}_3 \\ & =7744\end{aligned}\)

Let \(\alpha=\omega, b=1+\omega^3+\omega^6+\ldots=101\)

\(\begin{aligned}& a=(1+\omega)\left(1+\omega^2+\omega^4+\ldots \ldots \omega^{198}+\omega^{200}\right) \\& =(1+\omega) \frac{\left(1-\left(\omega^2\right)^{101}\right)}{1-\omega^2}=\frac{(\omega+1)\left(\omega^{2022}-1\right)}{\left(\omega^2-1\right)} \\& \Rightarrow a=\frac{(1+\omega)(1-\omega)}{1-\omega^2}=1\end{aligned}\)

Required equation \(=x^2-(101+1) x+(101) \times 1=0\)

\(\Rightarrow x^2-102 x+101=0\)