TS EAMCET MATHEMATICS PREVIOUS PROBLEMS MCQ

1)➤ The 13 th term in the expansion of \({(1-4 x)^{-4}}\)

A.\(15_{C_{4}}4^{12}x^{12}\)
B.\(728x^{12}\)
C.\(15_{C_{3}}4^{12}x^{12}\)
D. \(1092x^{12}\)

Answer: Option C
Solution: $${(1-4 x)^{-4}=1+4(4 x)+\frac{4 \cdot 5}{2 !}(4 x)^{2} +\frac{4 \cdot 5 \cdot 6}{3 !}(4 x)^{3}+\ldots}.$$ $${T_{r+1}=\frac{4 \cdot 5 \cdot 6(r+3)}{r !}(4 x)^{r}}.$$ $${\Rightarrow \quad T_{13}=\left(\frac{4 \cdot 5 \cdot 6 \ldots \cdot 15}{12 !}\right) 4^{12} x^{12}}.$$ When, r=12 $${=\left(\frac{15 !}{12 ! 3 !}\right) 4^{12} x^{12}={ }^{15} C_{3} 4^{12} x^{12}}.$$

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2)➤If \((\sqrt{3}+i)^{8}-(\sqrt{3}-i)^{8}=\alpha+i \beta\) then \(\alpha-\frac{\sqrt{3}}{2}\beta\) is equal to

A. 256
B. \(384\sqrt{3}\)
C. 384
D. \(256\sqrt{3}\)

Answer: Option C
Solution:
\((\sqrt{3}+i)^{8}-(\sqrt{3}-i)^{8}=\alpha+i \beta\) \(\begin{aligned} &\frac{\sqrt{3}}{2}+\frac{i}{2}=\cos \left(\frac{\pi}{6}\right)+i \sin \left(\frac{\pi}{6}\right)=e^{i \pi / 6} \\ &\frac{\sqrt{3}}{2}-\frac{i}{2}=\cos \left(2 \pi-\frac{\pi}{6}\right)+i \sin \left(2 \pi-\frac{\pi}{6}\right) \end{aligned}\) \(=e^{i\left(2 \pi-\frac{\pi}{6}\right)}\) From Eq-1 \(\begin{aligned} &\left(2 e^{i \frac{\pi}{6}}\right)^{8}-\left(2 e^{i \frac{11 \pi}{6}}\right)^{8}=\alpha+i \beta \\ \Rightarrow & 2^{8}\left(e^{i \frac{8 \pi}{6}}-e^{i \frac{88 \pi}{6}}\right)=\alpha+i \beta \\ \Rightarrow & 2^{8}\left[e^{i\left(\pi+\frac{\pi}{3}\right)}-e^{i\left(15 \pi-\frac{\pi}{3}\right)}\right]=\alpha+i \beta \\ \Rightarrow & 2^{8}\left[\left(-\frac{1}{2}-\frac{\sqrt{3} i}{2}\right)-\left(-\frac{1}{2}+\frac{\sqrt{3} i}{2}\right)\right]\newline=\alpha+i \beta \\ \Rightarrow & 2^{8}(-\sqrt{3} i)=\alpha+i \beta \Rightarrow \alpha=0, \beta=-\sqrt{3} \cdot 2^{8} \\ \alpha & -\frac{\sqrt{3}}{2} \beta \Rightarrow 0-\frac{\sqrt{3}}{2}\left(-\sqrt{3} \cdot 2^{8}\right)\ \newline\displaylines{=3 \cdot 2^{7} \\=384} \end{aligned}\)

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3)➤\(n \in N,\left(\frac{1+\cos \theta+i \sin \theta}{1+\cos \theta-i \sin \theta}\right)^{n}\) is equal to

A.\(cos( n \theta)-i \sin( n \theta)\)
B.\(-\cos (n \theta)+i \sin( n \theta)\)
C.\(cos (n \theta)+i \sin( n \theta)\)
D.\(-\cos (n \theta)-i \sin( n \theta)\)

Answer: Option C
Solution: \(\left(\frac{1+\cos \theta+i \sin \theta}{1+\cos \theta-i \sin \theta}\right)^{n}=\left(\frac{2 \cos ^{2} \frac{\theta}{2}+i 2 \sin \frac{\theta}{2} \cos \frac{\theta}{2}}{2 \cos ^{2} \frac{\theta}{2}-i 2 \sin \frac{\theta}{2} \cos \frac{\theta}{2}}\right)^{n}\) \(=\left(\frac{\cos \frac{\theta}{2}+i \sin \frac{\theta}{2}}{\cos \frac{\theta}{2}-i \sin \frac{\theta}{2}}\right)^{n}=\left(\frac{e^{i \frac{\theta}{2}}}{e^{-i \frac{\theta}{2}}}\right)^{n}\) \(\begin{aligned} {\left[e^{\left.i\left(\frac{\theta}{2}+\frac{\theta}{2}\right)\right]^{n}}\right.} &=\left(e^{i \theta}\right)^{n} \\ &=e^{i n \theta} \\ &=\cos n \theta+i \sin n \theta \end{aligned}\)

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4)➤If α and β are the roots of the equation \(x^2+x+1=0\) then \((2-\alpha)(2-\beta)\left(2-\alpha^{10}\right)\left(2-\alpha^{20}\right)\) is equal to

A.36
B.64
C.49
D.81

Answer: Option C
Solution: \begin{aligned} &\alpha=\omega \text { and } \beta=\omega^{2} \\ &\alpha^{10}=\omega^{10}=\omega^{9} \cdot \omega=\omega \\ &\alpha^{20}=\omega^{20}=\omega^{18} \cdot \omega^{2}=\omega^{2} \\ &(2-\alpha)(2-\beta)\left(2-\alpha^{10}\right)\left(2-\alpha^{20}\right) \\ &(2-\omega)\left(2-\omega^{2}\right)(2-\omega)\left(2-\omega^{2}\right) \\ &{\left[4-\left(\omega+\omega^{2}\right) \cdot 2+\omega^{3}\right]^{2}} \\ &(4+2+1)^{2}=49 \end{aligned}

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5)➤Locus of Z=x+iy such that \(\operatorname{Im}\left(\frac{z-3 i}{i z+4}\right)=0\) is

A. \(x^2-y^2+7y-12\)
B. \(x^2+y^2-7y-12\)
C. \(x^2+y^2-7y+12=0\) and \((x,y)\neq(0,4)\)
D.\(x^2-y^2+7y-12=0\) and \((x,y)\neq(0,4)\)

Answer: Option C
Solution: \begin{aligned} &\Rightarrow \operatorname{Im}\left[\frac{x+i y-3 i}{i(x+i y)+4}\right]=0 \Rightarrow \operatorname{Im}\left[\frac{x+i(y-3)}{(4-y)+i x}\right]=0 \\ &\Rightarrow \operatorname{Im}\left[\frac{x+i(y-3)}{(4-y)+i x} \times \frac{(4-y)-i x}{(4-y)-i x}\right]=0 \\ &\Rightarrow \operatorname{Im}\left[\frac{x(4-y)-i x^{2}+i(y-3)(4-y)+x(y-3)}{(4-y)^{2}+x^{2}}\right]=0 \\ &\Rightarrow \operatorname{Im}\left[\frac{4 x-x y-i x^{2}+i\left(-y^{2}+7 y-12\right)+x y-3 x}{(4-y)^{2}+x^{2}}\right]=0 \\ &\Rightarrow \operatorname{Im}\left[\frac{x-i\left(x^{2}+y^{2}-7 y+12\right)}{(4-y)^{2}+x^{2}}\right]=0 \\ &\Rightarrow \frac{\left(x^{2}+y^{2}-7 y+12\right)}{(4-y)^{2}+x^{2}}=0 \\ &\Rightarrow 4-y \neq 0 \text { and } x \neq 0 \Rightarrow y \neq 4 \text { and } x \neq 0 \\ &\Rightarrow x^{2}+y^{2}-7 y+12=0 \text { but }(x, y) \neq(0,4) \end{aligned}

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6)➤If the coefficient of \(x^{3}\) in the expansion of \(\left(2 \sqrt{3} x^{2}+\frac{1}{k x}\right)^{12}\) is 880 then k=

A. \(2\sqrt{2}\)
B.\(4\sqrt{3}\)
C. \(2\sqrt{3}\)
D.\(\sqrt{3}\)

Answer: Option D
Solution: \(\begin{aligned} & x^{3}\left[{ }^{12} C_{4}\left(2 \sqrt{3} x^{2}\right)^{4} \cdot\left(\frac{1}{k x}\right)^{8}\right]=880 x^{3} \\ \Rightarrow \quad & \frac{12 !}{4 ! 8 !} \times(16 \times 9) \times \frac{1}{k^{8}}=880 \end{aligned}\) \(\begin{aligned} & x^{3}\left[{ }^{12} C_{4}\left(2 \sqrt{3} x^{2}\right)^{4} \cdot\left(\frac{1}{k x}\right)^{8}\right]=880 x^{3} \\ \Rightarrow \quad & \frac{12 !}{4 ! 8 !} \times(16 \times 9) \times \frac{1}{k^{8}}=880 \end{aligned}\) \(\Rightarrow \quad \frac{12 \times 11 \times 10 \times 9}{4 \times 3 \times 2} \times 9 \times \frac{1}{k^{8}}=55\) \(\Rightarrow 11 \times 5 \times 9 \times 9=55 \times k^{8}\) \(\Rightarrow k^{8}=81 \Rightarrow k=\sqrt{3}\)

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7)➤ The coefficient of \(x^3\) in the expansion of \(\left(1-\frac{3}{4} x\right)^{\frac{1}{2}}\) is

A.\(\frac{27}{1024}\)
B. \(\frac{-27}{1024}\)
C.\(\frac{81}{1024}\)
D. \(\frac{-81}{1024}\)

Answer: Option B
Explanation: \(\left(1-\frac{3}{4} x\right)^{\frac{1}{2}}=1+\frac{1}{2}\left(-\frac{3}{4} x\right)+\frac{\frac{1}{2}\left(-\frac{1}{2}\right)}{2 !}\left(-\frac{3}{4} x\right)^{2}\) \(\begin{gathered} +\frac{\frac{1}{2} \times\left(-\frac{1}{2}\right) \times\left(-\frac{3}{2}\right)}{3 !}\left(-\frac{3}{4} x\right)^{3}+\ldots \\ =1-\frac{3}{8} x-\frac{1}{8} \times \frac{9}{16} x^{2}-\frac{3}{48} \times \frac{27}{64} \times x^{3}+\ldots \end{gathered}\) The coefficient of \(x^{3}=\frac{-27}{1024}\)

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8)➤The sum of the coefficients in the expansion of \((1+\frac{x}{2})^{12}\)

A.0
B. \(2^{11}\)
C.\((\frac{3}{2})^{12}\)
D.\(2^{12}\)

Answer: Option C
Solution: To find the sum of coefficients, replace all the variables with 1 . Here x is the only variable, so put, x=1 Therefore, we get, Sum of coefficients = \((1+\frac{1}{2})^{12}\) =\((\frac{3}{2})^{12}\)

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9)➤If p and q are respectively the coefficient of \(x^{-3}\) and \(x^{-5}\) in the expansion of\((x^{1 / 3}+\frac{1}{2 x^{1 / 3}})\), x>0 then \(\frac{5p}{3q}\) is equal to

A.102
B.408
C.182
D.468

Answer: Option B
Solution:

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10)➤ The term independent of x in the expansion of \(\left(x-\frac{2}{\sqrt{x}}\right)^{21}\) is

A.\(21_{C_{15}}(-2)^{15}\)
B. \(21_{C_{14}}(2)^{14}\)
C. \(-21_{C_7}(2)^{7}\)
D. \(-21_{C_7}(2)^{14}\)

Answer: Option B
Solution: \(\begin{aligned} &T_{r+1}={ }^{21} C_{r} \cdot(x)^{r} \cdot\left(-\frac{2}{\sqrt{x}}\right)^{21-r} \\ &\quad={ }^{21} C_{r} \cdot(x)^{r} \cdot(-2)^{21-r} \cdot(x) \frac{r-21}{2} \\ &\quad={ }^{21} C_{r} \cdot(-2)^{21-r} \cdot(x) \frac{r-21+2 r}{2} \\ &\text { For independent term, } \frac{r-21+2 r}{2}=0 \\ &\Rightarrow \quad 3 r=21 \\ &\Rightarrow \quad r=21 / 3 \\ &\Rightarrow \quad r=7 \\ &\therefore \text { Independent term }=T_{8}={ }^{21} C_{7}(-2)^{21-7} \\ &\quad={ }^{21} C_{14} 2^{14} \quad\left(\because{ }^{n} C_{r}={ }^{n} C_{n-r}\right) \end{aligned}\)

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