TS EAMCET MATHEMATICS PREVIOUS PROBLEMS MCQ-2

1)➤\(If\:A=\left[\begin{array}{lll}1 & 0 & 0 \\ 0 & 2 & 0 \\ 0 & 0 & 3\end{array}\right] ,then \\adj(adjA)\: is\: equal\: to\)

A.\(A\)
B.\(36A\)
C.\(6A\)
D.\(\frac{1}{6}A\)

Answer: Option C
Explanation: \(\operatorname{adj}(\operatorname{adj} A)=|A|^{n-2} A\) \(\begin{aligned} |A| &=\left|\begin{array}{lll} 1 & 0 & 0 \\ 0 & 2 & 0 \\ 0 & 0 & 3 \end{array}\right|=6 \\ \therefore \quad \operatorname{adj}(\operatorname{adj} A) &=6^{3-2} \cdot A=6 A \end{aligned}\)

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2)➤\(If\:A=\left[\begin{array}{ccc}1 & -1 & 1 \\ 2 & -1 & 0 \\ 1 & 0 & 0\end{array}\right],then\\ A^5 \: is\: equal \: to \)

A.\(A\)
B.\(identity\: matrix\)
C.\(null\: matrix\)
D.\(A^{-1}\)

Answer: Option D
Explanation:\(A=\left(\begin{array}1 & -1 & 1 \\ 2 & -1 & 0 \\ 1 & 0 & 0\end{array}\right)\) \(A^{2}=\left(\begin{array}{ccc} 1 & -1 & 1 \\ 2 & -1 & 0 \\ 1 & 0 & 0 \end{array}\right)\left(\begin{array}{ccc} 1 & -1 & 1 \\ 2 & -1 & 0 \\ 1 & 0 & 0 \end{array}\right)\) \(A^{2}=\left(\begin{array}{ccc} 0 & 0 & 1 \\ 0 & -1 & 2 \\ 1 & -1 & 1 \end{array}\right)\) \(\text { and } \begin{aligned} A^{3}=A \cdot A^{2} &=\left(\begin{array}{rrr} 1 & -1 & 1 \\ 2 & -1 & 0 \\ 1 & 0 & 0 \end{array}\right)\left(\begin{array}{ccc} 0 & 0 & 1 \\ 0 & -1 & 2 \\ 1 & -1 & 1 \end{array}\right) \\ &=\left(\begin{array}{lll} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{array}\right) \\ A^{3} &=I \end{aligned}\) \(\text { and } \begin{aligned} A^{3}=A \cdot A^{2} &=\left(\begin{array}{rrr} 1 & -1 & 1 \\ 2 & -1 & 0 \\ 1 & 0 & 0 \end{array}\right)\left(\begin{array}{ccc} 0 & 0 & 1 \\ 0 & -1 & 2 \\ 1 & -1 & 1 \end{array}\right) \\ &=\left(\begin{array}{lll} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{array}\right) \\ A^{3} &=I \end{aligned}\) \(\begin{aligned} A^{3} &=I \\ \therefore \quad A^{5}=A^{2} \cdot A^{3} &=A^{2} \cdot I \\ &=A^{2}\left(A A^{-1}\right) \\ &=A^{3} \cdot A^{-1} \\ &=I \cdot A^{-1} \\ \therefore \quad A^{5} &=A^{-1} \end{aligned}\)

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3)➤\(B=\left[\begin{array}{ll}1 & 2 \\ 0 & 1\end{array}\right]\) \(and\: A\: be\: 2×2 \: matrix \: satisfying \\ {(A^T)}^{-1}=A.If X=AB{A}^T,\:then \\A^TX^{2021}A \:is \:equal\: to \)

A.\(\left[\begin{array}{ll}1 & 2 \\ 0 & 1\end{array}\right]\)
B.\(\left[\begin{array}{ll}1 & 2021 \\ 0 & 1\end{array}\right]\)
C.\(\left[\begin{array}{ll}1 & 0 \\ 0 & 1\end{array}\right]\)
D.\(\left[\begin{array}{ll}1 & 4042 \\ 0 & 1\end{array}\right]\)

Answer: Option D
Explanation: \(A^{T}\left(A^{T}\right)^{-1}=A^{T} \cdot A=I A^{T} X A=A^{T}\left(A B A^{T}\right) A\\ =\left(A^{T} A\right) B\left(A^{T} A\right)=B\\ A^{T} X^{2} A=A^{T}\left(A B A^{T}\right)\left(A B A^{T}\right) A\\ =\left(A^{T} A\right) B\left(A^{T} A\right) B\left(A^{T} A\right)=B^{2}\\ Similarly,A^{T} X^{2021} A=B^{2021}\\ B=\left[\begin{array}{ll} 1 & 2 \\ 0 & 1 \end{array}\right] \\B^{2}=\left[\begin{array}{ll} 1 & 2 \\ 0 & 1 \end{array}\right]\left[\begin{array}{ll} 1 & 2 \\ 0 & 1 \end{array}\right]=\left[\begin{array}{ll} 1 & 4 \\ 0 & 1 \end{array}\right]\\ \begin{aligned} B^{3} &=\left[\begin{array}{ll} 1 & 4 \\ 0 & 1 \end{array}\right]\left[\begin{array}{ll} 1 & 2 \\ 0 & 1 \end{array}\right]=\left[\begin{array}{ll} 1 & 6 \\ 0 & 1 \end{array}\right] \\ B^{4} &=\left[\begin{array}{ll} 1 & 6 \\ 0 & 1 \end{array}\right]\left[\begin{array}{ll} 1 & 2 \\ 0 & 1 \end{array}\right]=\left[\begin{array}{ll} 1 & 8 \\ 0 & 1 \end{array}\right] \\ B^{n} &=\left[\begin{array}{cc} 1 & 2 n \\ 0 & 1 \end{array}\right] \\ \therefore B^{2021} &=\left[\begin{array}{cc} 1 & 4042 \\ 0 & 1 \end{array}\right] \end{aligned}\)

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4)➤\(If \:A\: is\: a\: 3 \times 3 \: matrix\: such\: that\\ |A|=27, adj A=k A^{T}, then\\ k^{2}-3 k+5\: is\: equal\: to\)

A.5
B.3
C.0
D.2

Answer: Option A
Explanation: \(\operatorname{adj} A=k A^{T}\\ A\left|=k^{3}\right| A^{T} \mid\\ \Rightarrow \quad|A|^{2}=k^{3}|A|\\ \Rightarrow \quad|A|\left(|A|-k^{3}\right)=0,\\since \quad|A| \neq 0 \Rightarrow \quad k^{3}=|A|=27,\\\quad k=3 k^{2}-3 k+5=9-9+5=5\)

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5)➤\(The \:parametric\: equations\: of \:the\\ parabola\:x^{2}-8 x+12 y+15=0 \:are\)

A.\(x=4+6t,y=\frac{1}{12}-3t^2\)
B. \(x=\frac{1}{12}-3t^2,y=4+6t\)
C. \(x=3t^2,y=6t\)
D. \(x=6t,y=3t^2\)

Answer: Option A
Explanation:\(x^{2}-8 x+12 y+15=0\) \( (x-4)^{2}=-12y+1\) \((x-4)^{2}=-4 \cdot 3 \cdot\left(y-\frac{1}{12}\right)\) \(x-4=6 t, y-\frac{1}{12}=-3 t^{2}\) \(x=4+6 t, y=\frac{1}{12}-3 t^{2}\)

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6)➤\(\lim _{x \rightarrow 0^{-}} \frac{\sqrt{\frac{1}{2}\left(1-\cos ^{2} x\right)}}{x}\)

A.\(\frac{1}{\sqrt2}\)
B.\(\frac{-1}{\sqrt2}\)
C.\(-1\)
D.\(does\: not \:exist\)

Answer: Option B
Explanation: \(L=\lim _{x \rightarrow 0^{-}} \frac{\sqrt{\frac{1}{2}\left(1-\cos ^{2} x\right)}}{x}\\ =\lim _{x \rightarrow 0^{-}} \frac{\frac{1}{\sqrt{2}} \sqrt{1-\left(1-\sin ^{2} x\right)}}{x}\\ =\lim _{x \rightarrow 0^{-}} \frac{1}{\sqrt{2}} \frac{|\sin x|}{x}=\frac{1}{\sqrt{2}} \lim _{x \rightarrow 0^{-}} \frac{-\sin x}{x}\\ =-\frac{1}{\sqrt{2}}\)

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7)➤\(\lim _{x \rightarrow 0} \frac{2 \sin x-\sin 2 x}{x^{3}}\)

A.\(1\)
B.\( 0\)
C.\(-1\)
D.\(Does\: not\: exist\)

Answer: Option A
Explanation: \(Let L=\lim _{x \rightarrow 0} \frac{2 \sin x-\sin 2 x}{x^{3}}\left(\text { form } \frac{0}{0}\right)\\ =\lim _{x \rightarrow 0} \frac{2 \cos x-2 \cos 2 x}{3 x^{2}}\left(\text { form } \frac{0}{0}\right)\\(using\: L-Hospital\: rule)\) \(=\lim _{x \rightarrow 0} \frac{-2 \sin x+4 \sin 2 x}{6 x}\left(\text { form } \frac{0}{0}\right)\\(using\: L-Hospital\: rule)\) \(=\lim _{x \rightarrow 0} \frac{-2 \cos x+8 \cos 2 x}{6}=\frac{-2+8}{6}=1\)

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8)➤\(\lim _{x \rightarrow \infty} \frac{3|x|^{3}-x^{2}+2|x|-5}{-5|x|^{3}+3 x^{2}-2|x|+7}\)

A.\(\frac{3}{5}\)
B.\(\frac{-5}{7}\)
C.\(\frac{5}{7}\)
D. \(\frac{-3}{5}\)

Answer: Option D
Explanation: \(\text { } \begin{aligned} &\lim _{x \rightarrow-\infty} \frac{3|x|^{3}-x^{2}+2|x|-5}{-5|x|^{3}+3 x^{2}-2|x|+7} \\ &x \rightarrow-\infty \Rightarrow|x|=-x \\ &\lim _{x \rightarrow-\infty} \frac{-3 x^{3}-x^{2}-2 x-5}{5 x^{3}+3 x^{2}+2 x+7}=-\frac{3}{5} \end{aligned}\)

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9)➤\(If\: the \:angles\: of \:a \triangle A B C \:are\: in\: A.P\\ then\)

A.\(c^2=a^2+b^2-ab\)
B.\(a^2=b^2+c^2-ac\)
C.\(b^2=a^2+c^2-ac\)
D.\(b^2=a^2+c^2-ac\)

Answer: Option D
Explanation: \(Let\: the \:angles\: of\: a \triangle A B C \:be (a-d), a,(a+d)\\ \quad a-d+a+a+d=180^{\circ} \Rightarrow a=60^{\circ}\\ \cos B=\frac{a^{2}+c^{2}-b^{2}}{2 a c}\\ \Rightarrow \frac{1}{2}=\frac{a^{2}+c^{2}-b^{2}}{2 a c} \Rightarrow b^{2}=a^{2}+c^{2}-a c\)

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10)➤\(\text If\: the\: line\: y=mx+c\: is a \:tangent \:to\\ the\: circle\: x^2+y^2=16,\\ then\: m\: is\: equal\: to\)

A.\(\pm \frac{1}{4} \sqrt{C-16}\)
B.\(\pm \frac{1}{c} \sqrt{C^{2}+16}\)
C.\(\pm \frac{1}{4} \sqrt{C^{2}-16}\)
D.\(\pm \frac{1}{16} \sqrt{C^{2}-16}\)

Answer: Option C
Explanation:\(If\: y=mx+c\: is \:a \:tangent\: to\: the \:circle \\x^2+y^2=r^2\: then\\ C^{2}=r^{2}(1+ m^{2})\\ we\: have\: r^2=16,\\ m^{2}=\frac{C^{2}}{16}-1\\ m=\pm \frac{1}{4} \sqrt{C^{2}-16}\)

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