Mathematics Test - 9

since the line \(\frac{x}{\alpha}+\frac{y}{\beta}=1\) touches the circle \(x^{2}+y^{2}=a^{2}\)
\(\therefore\) The perpendicular distance from centre (0, 0) to the tangent = radius of the circle
\(\Rightarrow \quad \frac{|-1|}{\sqrt{\frac{1}{\alpha^{2}}+\frac{1}{\beta^{2}}}}=a\)
\(\Rightarrow \quad \frac{1}{a^{2}}=\frac{1}{\alpha^{2}}+\frac{1}{\beta^{2}}\)
The locus of \(\left(\frac{1}{\alpha}, \frac{1}{\beta}\right)\) is \(\frac{1}{a^{2}}=\frac{1}{x^{2}}+\frac{1}{y^{2}}\)
\(\therefore\) It represents a circle.

Hence option B is the correct answer.

\begin{equation}\begin{aligned}
&\begin{array}{l}
\because f(x)=a x^{2}+b x+c \\
\text { and } \quad g(x)=p x^{2}+q x
\end{array}\\
&\text { since, } g(1)=f(1)\\
&\begin{array}{lr}
\Rightarrow p+q=a+b+c......(i) \\
\text{ and } g(2)-f(2)=1 \\
\Rightarrow 4 p+2 q-4 a-2 b-c=1......(ii) \\
\text { also } g(3)-f(3)=4 \\
\Rightarrow 9 p+3 q-9 a-3 b-c=4.....(iii)
\end{array}
\end{aligned}\end{equation}From Eqs. (i) and (ii)
\[\begin{aligned}
2 p =2 a-c+1 \\
\text { Now, } g(4)-f(4) & \\
= 16 p+4 q-16 a-4 b-c \\
= 12 p+4(p+q)-16 a-4 b-c=6-3 c
\end{aligned}
\]

Hence option D is the correct answer.

Given equation of hyperbola is \(\frac{\mathrm{x}^{2}}{\cos ^{2} \alpha}-\frac{\mathrm{y}^{2}}{\sin ^{2} \alpha}=1\)

Here, \(\mathrm{a}^{2}=\cos ^{2} \alpha\) and \(\mathrm{b}^{2}=\sin ^{2} \alpha\)

(i.e, comparing with standard equation \(\left.\frac{\mathrm{x}^{2}}{\mathrm{a}^{2}}-\frac{\mathrm{y}^{2}}{\mathrm{b}^{2}}=1\right)\)

We know, foci \(=(\pm \mathrm{ae}, 0)\)

where ae \(=\sqrt{\mathrm{a}^{2}+\mathrm{b}^{2}}=\sqrt{\cos ^{2} \alpha+\sin ^{2} \alpha}=1\)

\(\Rightarrow\) foci \(=(\pm 1,0)\)

where as vertices are \((\pm \cos \alpha, 0)\)

Eccentricity ae \(=1 \Rightarrow \mathrm{e}=\frac{1}{\cos \alpha}\)

Hence foci reamins constant with change in \(^{\prime} \alpha^{\prime}\)

Hence option B is the correct answer.

The centre and radius of the first circle \(x^{2}+y^{2}+2 x+8 y-23=0\) are \(C_{1}(-1,-4)\)
and \(r_{1}=\sqrt{40}\)
Similarly, the centre and radius of second circle \(x^{2}+y^{2}-4 x-10 y+9=0\) are \(C_{2}(2,5)\) and
\(\begin{aligned} r_{2}=\sqrt{20} & \\ \text { Now, } C_{1} C_{2} &=\sqrt{(2+1)^{2}+(5+4)^{2}} \\ &=\sqrt{9+81}=\sqrt{90} \end{aligned}\)
and \(\quad r_{1}+r_{2}=\sqrt{40}+\sqrt{20}\)
also \(\quad r_{1}-r_{2}=\sqrt{40}-\sqrt{20}\)
Here, r₁, r₂Two common tangents can be drawn.

Hence option C is the correct answer.

\(\frac{1}{x-\sin \alpha}+\frac{1}{x-\sin \beta}+\frac{1}{x-\sin \gamma}=0\)
\(\Rightarrow 3 x^{2}+2(\sin \alpha+\sin \beta+\sin \gamma) x+\sin \alpha \sin \beta+\sin \beta \sin \gamma+\sin \gamma \sin \alpha=0\)
\(\Delta=b^{2}-4 a c\)
\(=\sin ^{2} \alpha+\sin ^{2} \beta+\sin ^{2} \alpha-\sin \alpha \sin \beta-\sin \beta \sin \gamma-\sin \gamma \sin \alpha=0\)
\(=\frac{1}{2}\left[(\sin \alpha-\sin \beta)^{2}+(\sin \beta-\sin \gamma)^{2}+(\sin \gamma-\sin \alpha)^{2}\right]\)
\(\Delta>0\)
\(\therefore\) Roots are real and unequal.

Hence option C is the correct answer.

Given equation is \(x^{2}-a x+2=0\)
For roots to lie in the interval (0,3) following conditions should hold:
I) \(D \geq 0\) \(\Rightarrow a^{2}-8 \geq 0\)
\(\Rightarrow(a+2 \sqrt{2})(a-2 \sqrt{2}) \geq 0\)
\(\Rightarrow a \in(-\infty,-2 \sqrt{2}) \cup(2 \sqrt{2}, \infty)\)
2) \(0<-\frac{b}{2 a}<3\)
\(\Rightarrow 0<\frac{a}{2}<3\)
\(\Rightarrow a>0\) and \(a<6\)
3) \(a . f(0)>0\) which is true for all \(a>0\)
4) a. \(f(3)>0\) \(9-3 a+2>0\)
\(a<\frac{11}{3}\)
From equations ( 1 ), ( 2 ) and ( 3 ),
we get \(a \in\left(2 \sqrt{2}, \frac{11}{3}\right)\)

Hence option A is the correct answer.

Dividing by the coefficient of \(x^{2}\) we get
\(x^{2}+\frac{k x}{2}-2=0\)
\(\left(x+\frac{k}{4}\right)^{2}-\frac{k^{2}}{4}-2=0\)
\(\left(x+\frac{k}{4}\right)^{2}=\frac{k^{2}+8}{4}\)
Hence for rational roots
\(\frac{k^{2}+8}{4}\) has to be a perfect square.
We get a perfect square at \(x=\pm 1\)
Hence number of integral values of \(k\) is \(2 .\)

Hence option B is the correct answer.

\(a x^{2}+b x+c=0 \ldots \ldots . .(i)\)
\(\alpha, \beta\) are the roots of above equation \(\therefore \alpha+\beta=\frac{-b}{a} \ldots . .(i i)\)
\(\alpha \beta=\frac{c}{a} \ldots \ldots . .(i i i)\)
\(\alpha^{4}\) and \(\beta^{4}\) are the roots of equation \(p x^{2}+q x+r=0\)
\(\therefore \alpha^{4}+\beta^{4}=\frac{-q}{p} \ldots .(i v)\)
\(\alpha^{4} \beta^{4}=\frac{r}{p} \ldots \ldots .(v)\)
\(a^{2} p x^{2}-4 a c p x+2 c^{2} p+a^{2} q=0\)
Let \(r\) and \(\delta\) be roots of above equation
\(\therefore \quad r+\delta=\frac{4 a c p}{a^{2} p}=\frac{4 c}{a}\)
\(r \delta=\frac{2 c^{2} p+a^{2} q}{a^{2} p}\)
\(=2\left(\frac{c}{a}\right)^{2}+\frac{q}{p}\)
\(=2 \alpha^{2} \beta^{2}-\left(\alpha^{4}+\beta^{4}\right) \ldots \ldots\) [From equation \((i i i)\) and \(\left.(i v)\right]\)
\(=-\left(\alpha^{4}+\beta^{4}-2 \alpha^{2} \beta^{2}\right)\)
\(=-\left(\alpha^{2}-\beta^{2}\right)^{2}\)
\(\left(\alpha^{2}-\beta^{2}\right)^{2}>0\)
\(\therefore-\left(\alpha^{2}-\beta^{2}\right)^{2}<0\)
\(\therefore r \delta<0\)
\(\Rightarrow(r<0 \quad \& \quad \delta>0)\) or \((r>0 \quad \& \quad \delta<0)\)
\(\therefore\) Roots of given equation are opposite in sign.

Hence option C is the correct answer.