Mathematics Test - 20

\((K-2) x^{2}+8 x+K+4=0\)
\(x^{2}+\frac{8}{(K-2)} x+\frac{K+4}{(K-2)}=0\)
Let \(f(x)=x^{2}+\frac{8}{(K-2)} x+\frac{K+4}{(K-2)}\)
If the equation \(f(x)=0,\) has real distinct and negative roots then:
\(D>0 \& f(0)>0 \&-\frac{B}{2 A}<0\)
(i) \(D>0 \frac{8^{2}}{(\mathrm{K}-2)^{2}}-4 \frac{(\mathrm{K}+4)}{\mathrm{K}-2}>0\)
\(64-4\left(k^{2}+2 K-8\right)>0\)
\(-k^{2}-2 K+24<0 K^{2}+2 K-24<0\)
\((K+6)(K-4)<0\)
(ii)\( \left(f(0)>0 \frac{(K+4)}{(K-2)}>0\right).\)
\(\Rightarrow \mathrm{K}<-4\) or \(\mathrm{K}>2\)
(iii) \(-\frac{B}{2 \Lambda}<0-\frac{8}{2(K-2)}<0 \frac{4}{(K-2)}>0\)
\(\mathrm{K}-2>0\)
\(K>2\)
\(\therefore D>0 \& f(0)>0 \&\)
\(-\frac{B}{2 A}<0\)
\(-6<\mathrm{K}<4 \&(\mathrm{K}<-4\) or \(\mathrm{K}>2) \& \mathrm{K}>2\)
\(-6<\mathrm{K}<48 \mathrm{k}<-48 \mathrm{K}>2)\) or \((\mathrm{K}>28 \mathrm{K}>2\)
-62)
\(2<\mathrm{K}<4\)
\(\mathrm{K} \in(2,4)\)
Hence, the correct option is C.

\(\mathrm{B}=-\mathrm{A}^{-1} \mathrm{BA}\)
\(\mathrm{AB}=\mathrm{A}\left(-\mathrm{A}^{-1} \mathrm{BA}\right)\)
\(\left.\mathrm{AB}=-\mathrm{AA}^{-1} \mid \mathrm{BA}\right)\)
\(\mathrm{AB}=-\mathrm{BA}\)
\(\mathrm{AB}+\mathrm{BA}=0\)
Now, \((\mathrm{A}+\mathrm{B})^{2}=\mathrm{A}^{2}+\mathrm{B}^{2}+\mathrm{AB}+\mathrm{BA}\)
\((\mathrm{A}+\mathrm{B})^{2}=\mathrm{A}^{2}+\mathrm{B}^{2},(\because \mathrm{AB}+\mathrm{BA}=0)\)
Hence, the correct option is B.

Let \(\frac{1+i}{2}=x\) Hence the given series \(S\) is \(S=\left(1+\mathrm{x} / 1+\mathrm{x}^{2}\right)\left(1+\mathrm{x}^{4}\right) \cdot\left(1+\mathrm{x}^{2^{\mathrm{n}}}\right)\)
\(\mathrm{S}(1-\mathrm{x})=1-\mathrm{x}^{2^{\mathrm{n}}-1}\)
\(\mathrm{S}=\frac{1-\mathrm{x}^{2} \mathrm{n}+1}{1-\mathrm{x}}=\frac{1-\mathrm{x}^{2+1}}{1-\frac{1+\mathrm{i}}{2}}\)
\(=\frac{1-\left(\frac{1+i}{\sqrt{2}} \cdot \frac{\sqrt{2}}{2}\right)^{2^{n+1}}}{\left(\frac{1-i}{2}\right)}=\frac{1-\left(\frac{1+i}{\sqrt{2}}\right)^{2^{n+1}}\left(\frac{\sqrt{2}}{2}\right)^{2^{n+1}}}{\left(\frac{1-i}{2}\right)}\)
\(=\frac{2\left(1+i\left\{1-(-1)^{2^{n-1}} 2^{-2^{n}}\right\}\right.}{2}\)
\(=\left(1+i\left(1-2^{-2^{n}}\right)\left(\because n \geq 2 \Rightarrow(-1)^{2^{n-1}}=1\right)\right.\)
Hence, the correct option is A.

If possible let \(z=\) re \(\mathrm{i} \theta\) be a root which lies inside \(|z|=2 / 3 .\) Hence \(0 \leq \mathrm{r}<2 / 3\) Hence \(a_{1}\left(\mathrm{re}^{\mathrm{i} \theta}\right)^{3}+\mathrm{a}_{2}\left(\mathrm{re}^{\mathrm{i} \theta}\right)^{2}+\mathrm{a}_{3}\left(\mathrm{re}^{\mathrm{i} \theta}\right)+\mathrm{a}_{4}=3\)
Comparing real parts we get \(a_{1} t^{3} \cos 3 \theta+a_{2} r^{2} \cos 2 \theta+a_{3} r \cos \theta+a_{4}=3\)
but \(\left(\mathrm{r}^{3} \mathrm{a}_{1} \cos 3 \theta+\mathrm{r}^{2} \mathrm{a}_{2} \cos 2 \theta+\mathrm{ra}_{3} \cos \theta+\mathrm{a}_{4}\right)\)
\(\leq r^{3}+r^{2}+r+1\)
(Qlail \(\leq 1 \&\) so also \(\cos 3 \theta, \cos 2 \theta \& \cos \theta)\)
\(3 \leq r^{3}+r^{2}+r+1\)
\(\frac{r^{4}-1}{r-1} \geq 3\)
\(\frac{1-r^{4}}{1-r} \geq 3\)
\(\frac{1}{1-r} \geq 3\)
\(1-r \leq \frac{1}{3}\)
\(1-\frac{1}{3} \leq\)
\(r \geq 2 / 3\)
Which is a contradiction to the assumption \(0 \leq x<2 / 3\). Hence no root of the given equation can lie inside the circle. \(|z|=\frac{2}{3}\)
Hence, the correct option is A.

The general term of the expansion of \(\left(\frac{3}{2} x^{2}-\frac{1}{3 x}\right)^{9}\) is
\(T_{\mathrm{r}+1}={ }^{9} \mathrm{C}_{\mathrm{r}}\left(\frac{3}{2} \mathrm{x}^{2}\right)^{9-\mathrm{r}}\left(-\frac{1}{3 \mathrm{x}}\right)^{\mathrm{r}}\)
\(={ }^{9} \mathrm{C}_{\mathrm{r}}\left(\frac{3}{2}\right)^{9-\mathrm{t}}\left(-\frac{1}{3 \mathrm{x}}\right)^{\mathrm{r}}\)
The coefficient of independent of \(x\) in the expansion of \(\left(1+x+2 x^{3}\right)\) \(\left(\frac{3}{2} x^{2}-\frac{1}{3 x}\right)^{9}=\) sum of coefficient of \(x^{0}, x^{-1} 2 x^{-3}\) in expansion of \(\left(\frac{3}{2} x^{2}-\frac{1}{3 x}\right)^{9}\)
(i) For coefficient of \(x^{0}: 18-3 r=0,,\) according to (1)\(: r=6 .\) Put in (i):
\(T_{7}={ }^{9} \mathrm{C}_{6}\left(\frac{3}{2}\right)^{9-6}\left(-\frac{1}{3}\right)^{6}={ }^{9} \mathrm{C}_{3}\left(\frac{3}{2}\right)^{3} \cdot\left(\frac{1}{3}\right)^{6}\)
\(={ }^{9} \mathrm{C}_{3} \frac{1}{6^{3}}=\frac{7}{18}\)
(ii) For Coefficient of \(x^{-3}: 18-3 r=-1\) or, \(r=19 / 3=\) fraction. No such term exists.
(iii) For coefficient of \(2 x^{-3}: 18-3 r=-3\) or, \(r=7 .\) Put in (i):
\(\mathrm{T}_{8}=2\left[9 \mathrm{C}_{7}\left(\frac{3}{2}\right)^{9-7}\left(-\frac{1}{3}\right)^{7}\right]\)
\(=\left(9 \mathrm{C}_{2}\right)\left(2\left(\frac{3}{2}\right)^{2}(-1)\left(\frac{1}{3}\right)^{7}=-\frac{2}{27}\right.\)
Sum of coefficients \(=\frac{7}{18}-\frac{2}{27}=\frac{17}{54}\)
Hence, the correct option is C.

We have,
\(\frac{a+b+\frac{c}{2}+\frac{c}{2}}{4} \geq \sqrt[4]{a b \cdot \frac{c}{2} \cdot \frac{c}{2}}\)
or, \(\frac{a+b+c}{4} \geq \sqrt[4]{\frac{a b c^{2}}{4}}\)
\(\therefore\left(\frac{a+b+c}{4}\right)^{4} \geq \frac{a b c^{2}}{4}\)
or \(a b c^{2} \leq \frac{1}{64}(a+b+c)^{4}\)
\(\therefore\) the greatest value of \(a b c^{2}=\frac{1}{64}(a+b+c)^{4}\)
Also for the greatest value of abc \(^{2}\) the numbers have to be equal, i.e \(a=b=c / 2\) Also given that maximum value \(=1 / 64\) \(\mathrm{s} 0, \mathrm{a}+\mathrm{b}+\mathrm{c}=1\)
i.e. \(a=b=1 / 4, c=1 / 2\)
Hence, the correct option is B.

\(f(x)=x+\frac{1}{x} f\left(x^{3}\right)=x^{3}+\frac{1}{x^{3}}\)
\(f(x)=x+\frac{1}{x}=f\left(\frac{1}{x}\right)=\frac{1}{x}+x f(x)=f\left(\frac{1}{x}\right)\)
\(\therefore f^{3}(x)=\left(x+\frac{1}{x}\right)^{3}\)
\(=\left(x^{3}+\frac{1}{x^{3}}\right)+3 x \cdot \frac{1}{x}\left(x+\frac{1}{x}\right)\)
\(=\left(x^{3}+\frac{1}{x^{3}}\right)+3\left(x+\frac{1}{x}\right)\)
\(=f\left(x^{3}\right)+3 f(x)\)
\(=f\left(x^{3}\right)+3 f\left(\frac{1}{x}\right) \cdot\left[\because f(x)=f\left(\frac{1}{x}\right)\right]\)
Hence, the correct option is A.

Substitute \((x-1)=t .\)
\(\mathrm{x} \rightarrow 1(\mathrm{x}-1) \rightarrow 0 \mathrm{t} \rightarrow 0\)
Here given limit is:
\(\lim _{x \rightarrow 1} \frac{3^{x+1}-9}{4^{2 x+1}-64}=\lim _{t \rightarrow 0} \frac{3^{t+2}-9}{4^{2(t+1)+1}-64}\)
\(=\lim _{t \rightarrow 0} \frac{3^{t} \cdot 3^{2}-3^{2}}{4^{2 t} \cdot 4^{3}-4^{3}}=\lim _{t \rightarrow 0} \frac{9\left(3^{t}-1\right)}{64\left(4^{2 t}-1\right)}\)
\(=\lim _{t \rightarrow 0} \frac{9}{64} \frac{\left(\frac{3^{t}-1}{t}\right) \cdot t}{\left(\frac{4^{2 t}-1}{2 t}\right) \cdot 2 t}=\lim _{t \rightarrow 0} \frac{9}{64} \cdot \frac{\ln 3}{\ln 4} \cdot \frac{1}{2}=\frac{9 \ln 3}{128 \ln 4}\)

Hence, the correct option is B.

\(y=1+a^{2} x-x^{3}\)
\(y^{\prime}=a^{2}-3 x^{2}\)
\(y^{\prime \prime}=-6 x\)
\(y^{\prime}=0 a^{2}-3 x^{2}=0 x=\pm \frac{a}{\sqrt{3}}\)
(i) Let \(a>0:\) Hence \(y^{/ /}>0\) for \(x=-\frac{a}{\sqrt{3}} .\) Hence \(\mathrm{y}\) has minima at \(x=-\frac{a}{\sqrt{3}}\) for \(\mathrm{a}>0 .\) \(\therefore-3<-\frac{a}{\sqrt{3}}<-2 \quad 2 \sqrt{3}<2<3 \sqrt{3}\)
(ii) Let \(a<0\) :
Hence \(y^{/ /}>0\) for \(x=\frac{a}{\sqrt{3}} .\) Hence \(y\) has minima at \(x=\frac{a}{\sqrt{3}}\) for \(a<0\). \(\therefore-3<\frac{a}{\sqrt{3}}<-2-3 \sqrt{3}<2<-2 \sqrt{3}\)
From
(i) \(8(\) ii \()\) a \(\in(-3 \sqrt{3},-2 \sqrt{3}) \cup(2 \sqrt{3}, 3 \sqrt{3})\)
Hence, the correct option is \(D\).