Here you can download a free Algebra questions PDF with answers for MAH MBA CET 2022 by Cracku. These are some tricky questions in the MAH MBA CET 2022 exam that you need to find the solutions for the given Algebra questions. These questions will help you to practice and solve the Algebra questions in the MAH MBA CET exam. Utilize this best PDF practice set<\/strong> which is included answers in detail. Click on the below link to download the Algebra MCQ<\/strong> PDF for MBA-CET 2022 for free.<\/p>\n Download Algebra Questions for MAH-CET<\/a><\/p>\n Enroll to MAH-CET 2022 Crash Course<\/a><\/p>\n Question 1:\u00a0<\/b>If $\\sqrt{x}-\\frac{1}{\\sqrt{x}}=\\sqrt{7}$, then the value of $x^2 + \\frac{1}{x^2}$ is:<\/p>\n a)\u00a081<\/p>\n b)\u00a060<\/p>\n c)\u00a079<\/p>\n d)\u00a075<\/p>\n 1)\u00a0Answer\u00a0(C)<\/strong><\/p>\n Solution:<\/b><\/p>\n $\\sqrt{x}-\\frac{1}{\\sqrt{x}}=\\sqrt{7}$<\/p>\n $\\left(\\sqrt{x}-\\frac{1}{\\sqrt{x}}\\right)^2=\\left(\\sqrt{7}\\right)^2$<\/p>\n $x+\\frac{1}{x}-2=7$<\/p>\n $x+\\frac{1}{x}=9$<\/p>\n $\\left(x+\\frac{1}{x}\\right)^2=9^2$<\/p>\n $x^2+\\frac{1}{x^2}+2=81$<\/p>\n $x^2+\\frac{1}{x^2}=79$<\/p>\n Hence, the correct answer is Option C<\/p>\n Question 2:\u00a0<\/b>If $(56\\sqrt{7}x^3-2\\sqrt{2}y^3)\\div(2\\sqrt{7}x-\\sqrt{2}y)=Ax^2+By^2-Cxy$, then find the value of $A + B – \\sqrt{14}C$.<\/p>\n a)\u00a038<\/p>\n b)\u00a010<\/p>\n c)\u00a019<\/p>\n d)\u00a058<\/p>\n 2)\u00a0Answer\u00a0(D)<\/strong><\/p>\n Solution:<\/b><\/p>\n $(56\\sqrt{7}x^3-2\\sqrt{2}y^3)\\div(2\\sqrt{7}x-\\sqrt{2}y)=Ax^2+By^2-Cxy$<\/p>\n $\\frac{\\left(2\\sqrt{7}x-\\sqrt{2}y\\right)\\left(28x^2+2\\sqrt{14}xy+2y^2\\right)}{\\left(2\\sqrt{7}x-\\sqrt{2}y\\right)}=Ax^2+By^2-Cxy$<\/p>\n $28x^2+2\\sqrt{14}xy+2y^2=Ax^2+By^2-Cxy$<\/p>\n Comparing both sides,<\/p>\n A = 28, B = 2, C =\u00a0$-2\\sqrt{14}$<\/p>\n $A+B-\\sqrt{14}C=28+2-\\sqrt{14}\\left(-2\\sqrt{14}\\right)$<\/p>\n $=30+28$<\/p>\n $=58$<\/p>\n Hence, the correct answer is Option D<\/p>\n Question 3:\u00a0<\/b>If $\\frac{x}{y} + \\frac{y}{x} = 2, (x, y \\neq 0)$, then the value of $(x – y)$ is:<\/p>\n a)\u00a01<\/p>\n b)\u00a00<\/p>\n c)\u00a02<\/p>\n d)\u00a0-2<\/p>\n 3)\u00a0Answer\u00a0(B)<\/strong><\/p>\n Solution:<\/b><\/p>\n $\\frac{x}{y}+\\frac{y}{x}=2$<\/p>\n $\\frac{x^2+y^2}{xy}=2$<\/p>\n $x^2+y^2=2xy$<\/p>\n $x^2+y^2-2xy=0$<\/p>\n $\\left(x-y\\right)^2=0$<\/p>\n $x-y=0$<\/p>\n Hence, the correct answer is Option B<\/p>\n Question 4:\u00a0<\/b>If $\\left(2a+\\frac{3}{a}-1\\right)=11$, what is the value of $\\left(4a^2 + \\frac{9}{a^2}\\right)?$<\/p>\n a)\u00a0121<\/p>\n b)\u00a0148<\/p>\n c)\u00a0132<\/p>\n d)\u00a0110<\/p>\n 4)\u00a0Answer\u00a0(C)<\/strong><\/p>\n Solution:<\/b><\/p>\n $\\left(2a+\\frac{3}{a}-1\\right)=11$<\/p>\n $2a+\\frac{3}{a}=12$<\/p>\n $4a^2+\\frac{9}{a^2}+2.2a.\\frac{3}{a}=144$<\/p>\n $4a^2+\\frac{9}{a^2}+12=144$<\/p>\n $4a^2+\\frac{9}{a^2}=132$<\/p>\n Hence, the correct answer is Option C<\/p>\n Question 5:\u00a0<\/b>If $a^3 – b^3 = 2349$ and $(a – b) = 9$, then $(a + b)^2 – ab$ is equal to:<\/p>\n a)\u00a0280<\/p>\n b)\u00a0244<\/p>\n c)\u00a0261<\/p>\n d)\u00a0229<\/p>\n 5)\u00a0Answer\u00a0(C)<\/strong><\/p>\n Solution:<\/b><\/p>\n $(a-b)=9$………….(1)<\/p>\n $(a-b)^3=729$<\/p>\n $a^3-b^3-3ab\\left(a-b\\right)=729$<\/p>\n $2349-3ab\\left(9\\right)=729$<\/p>\n $27ab=1620$<\/p>\n $ab=60$…………..(2)<\/p>\n $(a-b)=9$<\/p>\n $(a-b)^2=81$<\/p>\n $a^2+b^2-2ab=81$<\/p>\n $a^2+b^2-2\\left(60\\right)=81$<\/p>\n $a^2+b^2-120=81$<\/p>\n $a^2+b^2=201$……….(3)<\/p>\n $(a+b)^2-ab=a^2+b^2+2ab-ab$<\/p>\n $=a^2+b^2+ab$<\/p>\n $=201+60$<\/p>\n $=261$<\/p>\n Hence, the correct answer is Option C<\/p>\n Take Free MAH-CET mock tests here<\/a><\/strong><\/span><\/p>\n Enrol to 5 MAH CET Latest Mocks For Just Rs. 299<\/a><\/strong><\/span><\/p>\n Question 6:\u00a0<\/b>If $x – \\frac{1}{x} = \\sqrt{77}$, then one of the values of $x^3 + \\frac{1}{x^3}$ is:<\/p>\n a)\u00a0$80\\sqrt{77}$<\/p>\n b)\u00a0-702<\/p>\n c)\u00a0$77\\sqrt{77}$<\/p>\n d)\u00a0$3\\sqrt{77}$<\/p>\n 6)\u00a0Answer\u00a0(B)<\/strong><\/p>\n Solution:<\/b><\/p>\n $x – \\frac{1}{x} = \\sqrt{77}$<\/p>\n $\\left(x-\\frac{1}{x}\\right)^2=77$<\/p>\n $x^2+\\frac{1}{x^2}-2=77$<\/p>\n $x^2+\\frac{1}{x^2}=79$<\/p>\n $x^2+\\frac{1}{x^2}+2=81$<\/p>\n $\\left(x+\\frac{1}{x}\\right)^2=81$<\/p>\n $x+\\frac{1}{x}=9$ or\u00a0$x+\\frac{1}{x}=-9$<\/p>\n When\u00a0$x+\\frac{1}{x}=-9$<\/p>\n $x^3+\\frac{1}{x^3}+3.x.\\frac{1}{x}\\left(x+\\frac{1}{x}\\right)=-729$<\/p>\n $x^3+\\frac{1}{x^3}+3\\left(-9\\right)=-729$<\/p>\n $x^3+\\frac{1}{x^3}-27=-729$<\/p>\n $x^3+\\frac{1}{x^3}=-702$<\/p>\n Hence, the correct answer is Option B<\/p>\n Question 7:\u00a0<\/b>If $\\sqrt{x}=\\sqrt{3}-\\sqrt{5}$, then the value of $x^2-16x+6$ is:<\/p>\n a)\u00a04<\/p>\n b)\u00a00<\/p>\n c)\u00a02<\/p>\n d)\u00a0-2<\/p>\n 7)\u00a0Answer\u00a0(C)<\/strong><\/p>\n Solution:<\/b><\/p>\n Given,\u00a0 $\\sqrt{x}=\\sqrt{3}-\\sqrt{5}$<\/p>\n $\\Rightarrow$ \u00a0$x=\\left(\\sqrt{3}-\\sqrt{5}\\right)^2$<\/p>\n $\\Rightarrow$ \u00a0$x=3+5-2\\sqrt{15}$<\/p>\n $\\Rightarrow$ \u00a0$x=8-2\\sqrt{15}$ ……………(1)<\/p>\n $\\Rightarrow$ \u00a0$x^2=\\left(8-2\\sqrt{15}\\right)^2$<\/p>\n $\\Rightarrow$ \u00a0$x^2=64+60-32\\sqrt{15}$<\/p>\n $\\Rightarrow$ \u00a0$x^2=124-32\\sqrt{15}$ ………..(2)<\/p>\n $\\therefore\\ $ $x^2-16x+6=124-32\\sqrt{15}-16\\left(8-2\\sqrt{15}\\right)+6$<\/p>\n $=124-32\\sqrt{15}-128+32\\sqrt{15}+6$<\/p>\n $=130-128$<\/p>\n $=2$<\/p>\n Hence, the correct answer is Option C<\/p>\n Question 8:\u00a0<\/b>If $x=\\frac{\\sqrt{3}}{2}$, then the value of $\\frac{\\sqrt{1+x}+\\sqrt{1-x}}{\\sqrt{1+x}-\\sqrt{1-x}}$ is equal to:<\/p>\n a)\u00a0$\\sqrt 2$<\/p>\n b)\u00a0$\\sqrt 3$<\/p>\n c)\u00a03<\/p>\n d)\u00a02<\/p>\n 8)\u00a0Answer\u00a0(B)<\/strong><\/p>\n Solution:<\/b><\/p>\n Given,\u00a0$x=\\frac{\\sqrt{3}}{2}$<\/p>\n $\\frac{\\sqrt{1+x}+\\sqrt{1-x}}{\\sqrt{1+x}-\\sqrt{1-x}}=\\frac{\\sqrt{1+x}+\\sqrt{1-x}}{\\sqrt{1+x}-\\sqrt{1-x}}\\times\\frac{\\sqrt{1+x}+\\sqrt{1-x}}{\\sqrt{1+x}+\\sqrt{1-x}}$<\/p>\n $=\\frac{1+x+1-x+2\\left(\\sqrt{1+x}\\right)\\left(\\sqrt{1-x}\\right)}{1+x-\\left(1-x\\right)}$<\/p>\n $=\\frac{2+2\\left(\\sqrt{1-x^2}\\right)}{2x}$<\/p>\n $=\\frac{1+\\sqrt{1-x^2}}{x}$<\/p>\n $=\\frac{1+\\sqrt{1-\\left(\\frac{\\sqrt{3}}{2}\\right)^2}}{\\frac{\\sqrt{3}}{2}}$<\/p>\n $=\\frac{1+\\sqrt{1-\\frac{3}{4}}}{\\frac{\\sqrt{3}}{2}}$<\/p>\n $=\\frac{1+\\frac{1}{2}}{\\frac{\\sqrt{3}}{2}}$<\/p>\n $=\\frac{3}{2}\\times\\frac{2}{\\sqrt{3}}$<\/p>\n $=\\sqrt{3}$<\/p>\n Hence, the correct answer is Option B<\/p>\n Question 9:\u00a0<\/b>If $a^3+b^3=62$ and a + b = 2, then the value of ab is:<\/p>\n a)\u00a0-6<\/p>\n b)\u00a09<\/p>\n c)\u00a06<\/p>\n d)\u00a0-9<\/p>\n 9)\u00a0Answer\u00a0(D)<\/strong><\/p>\n Solution:<\/b><\/p>\n Given,\u00a0 $a+b=2$<\/p>\n $a^3+b^3=62$<\/p>\n $\\Rightarrow$ \u00a0$\\left(a+b\\right)\\left(a^2-ab+b^2\\right)=62$<\/p>\n $\\Rightarrow$ \u00a0$\\left(2\\right)\\left(a^2+2ab+b^2-3ab\\right)=62$<\/p>\n $\\Rightarrow$ \u00a0$\\left(a+b\\right)^2-3ab=31$<\/p>\n $\\Rightarrow$ \u00a0$\\left(2\\right)^2-3ab=31$<\/p>\n $\\Rightarrow$ \u00a0$4-3ab=31$<\/p>\n $\\Rightarrow$ \u00a0$3ab=-27$<\/p>\n $\\Rightarrow$ \u00a0$ab=-9$<\/p>\n Hence, the correct answer is Option D<\/p>\n Question 10:\u00a0<\/b>If $a-b=18$ and $a^3-b^3=324$, then find ab.<\/p>\n a)\u00a0105<\/p>\n b)\u00a0-102<\/p>\n c)\u00a0-104<\/p>\n d)\u00a0103<\/p>\n 10)\u00a0Answer\u00a0(B)<\/strong><\/p>\n Solution:<\/b><\/p>\n Given,\u00a0 $a-b=18$ and<\/p>\n $a^3-b^3=324$<\/p>\n $\\Rightarrow$ \u00a0$\\left(a-b\\right)\\left(a^2+ab+b^2\\right)=324$<\/p>\n $\\Rightarrow$\u00a0 $\\left(18\\right)\\left(a^2-2ab+b^2+3ab\\right)=324$<\/p>\n $\\Rightarrow$ \u00a0$\\left(a-b\\right)^2+3ab=18$<\/p>\n $\\Rightarrow$ \u00a0$\\left(18\\right)^2+3ab=18$<\/p>\n $\\Rightarrow$ \u00a0$324+3ab=18$<\/p>\n $\\Rightarrow$ \u00a0$3ab=-306$<\/p>\n $\\Rightarrow$ \u00a0$ab=-102$<\/p>\n Hence, the correct answer is Option B<\/p>\n Question 11:\u00a0<\/b>If $x^4+\\frac{1}{x^4}=14159$, then the value of $x+\\frac{1}{x}$ is:<\/p>\n a)\u00a011<\/p>\n b)\u00a012<\/p>\n c)\u00a09<\/p>\n d)\u00a010<\/p>\n 11)\u00a0Answer\u00a0(A)<\/strong><\/p>\n Solution:<\/b><\/p>\n Given, \u00a0$x^4+\\frac{1}{x^4}=14159$<\/p>\n $\\Rightarrow$ \u00a0$x^4+\\frac{1}{x^4}+2=14159+2$<\/p>\n $\\Rightarrow$ \u00a0$\\left(x^2+\\frac{1}{x^2}\\right)^2=14161$<\/p>\n $\\Rightarrow$ \u00a0$\\left(x^2+\\frac{1}{x^2}\\right)^2=\\left(119\\right)^2$<\/p>\n $\\Rightarrow$ \u00a0$x^2+\\frac{1}{x^2}=119$<\/p>\n $\\Rightarrow$ \u00a0$x^2+\\frac{1}{x^2}+2=119+2$<\/p>\n $\\Rightarrow$ \u00a0$\\left(x+\\frac{1}{x}\\right)^2=121$<\/p>\n $\\Rightarrow$ \u00a0 $x+\\frac{1}{x}=11$<\/p>\n Hence, the correct answer is Option A<\/p>\n Question 12:\u00a0<\/b>If $p+\\frac{1}{p}=112$, find $(p-112)^{15}+\\frac{1}{p^{15}}$<\/p>\n a)\u00a01<\/p>\n b)\u00a015<\/p>\n c)\u00a010<\/p>\n d)\u00a00<\/p>\n 12)\u00a0Answer\u00a0(D)<\/strong><\/p>\n Solution:<\/b><\/p>\n Given,\u00a0$p+\\frac{1}{p}=112$<\/p>\n $\\Rightarrow$ \u00a0$p-112=-\\frac{1}{p}$<\/p>\n $(p-112)^{15}+\\frac{1}{p^{15}}=\\left(-\\frac{1}{p}\\right)^{15}+\\frac{1}{p^{15}}$<\/p>\n $=-\\frac{1}{p^{15}}+\\frac{1}{p^{15}}$<\/p>\n $=0$<\/p>\n Hence, the correct answer is Option D<\/p>\n Question 13:\u00a0<\/b>If $\\frac{4}{3}\\left(x^2+\\frac{1}{x^2}\\right)=110\\frac{2}{3}$, find $\\frac{1}{9} \\left(x^3 – \\frac{1}{x^3}\\right)$, where $x$ > 0.<\/p>\n a)\u00a074<\/p>\n b)\u00a076<\/p>\n c)\u00a084<\/p>\n d)\u00a085<\/p>\n 13)\u00a0Answer\u00a0(C)<\/strong><\/p>\n Solution:<\/b><\/p>\n Given, $\\frac{4}{3}\\left(x^2+\\frac{1}{x^2}\\right)=110\\frac{2}{3}$<\/p>\n $\\Rightarrow$ \u00a0$\\frac{4}{3}\\left(x^2+\\frac{1}{x^2}\\right)=\\frac{332}{3}$<\/p>\n $\\Rightarrow$ \u00a0$x^2+\\frac{1}{x^2}=83$<\/p>\n $\\Rightarrow$ \u00a0$x^2+\\frac{1}{x^2}-2=83-2$<\/p>\n $\\Rightarrow$ \u00a0$\\left(x-\\frac{1}{x}\\right)^2=81$<\/p>\n $\\Rightarrow$ \u00a0$x-\\frac{1}{x}=9$<\/p>\n $\\Rightarrow$ \u00a0$\\left(x-\\frac{1}{x}\\right)^3=9^3$<\/p>\n $\\Rightarrow$ \u00a0$x^3-\\frac{1}{x^3}-3.x.\\frac{1}{x}\\left(x-\\frac{1}{x}\\right)=729$<\/p>\n $\\Rightarrow$ \u00a0$x^3-\\frac{1}{x^3}-3\\left(9\\right)=729$<\/p>\n $\\Rightarrow$ \u00a0$x^3-\\frac{1}{x^3}-27=729$<\/p>\n $\\Rightarrow$ \u00a0$x^3-\\frac{1}{x^3}=756$<\/p>\n $\\Rightarrow$ \u00a0$\\frac{1}{9}\\left(x^3-\\frac{1}{x^3}\\right)=\\frac{1}{9}\\left(756\\right)$<\/p>\n $\\Rightarrow$ \u00a0$\\frac{1}{9}\\left(x^3-\\frac{1}{x^3}\\right)=84$<\/p>\n Hence, the correct answer is Option C<\/p>\n Question 14:\u00a0<\/b>If $x^3+y^3=16$ and $x+y=4$, then the value of $x^4+y^4$ is:<\/p>\n a)\u00a048<\/p>\n b)\u00a032<\/p>\n c)\u00a064<\/p>\n d)\u00a030<\/p>\n 14)\u00a0Answer\u00a0(B)<\/strong><\/p>\n Solution:<\/b><\/p>\n Given,\u00a0 $x^3+y^3=16$ and $x+y=4$<\/p>\n $\\Rightarrow$ \u00a0$\\left(x+y\\right)\\left(x^2-xy+y^2\\right)=16$<\/p>\n $\\Rightarrow$ \u00a0$\\left(4\\right)\\left(x^2+2xy+y^2-3xy\\right)=16$<\/p>\n $\\Rightarrow$ \u00a0$\\left(x+y\\right)^2-3xy=4$<\/p>\n $\\Rightarrow$ \u00a0$\\left(4\\right)^2-3xy=4$<\/p>\n $\\Rightarrow$ \u00a0$3xy=16-4$<\/p>\n $\\Rightarrow$ \u00a0$3xy=12$<\/p>\n $\\Rightarrow$ \u00a0$xy=4$<\/p>\n $\\therefore\\ $ $x^4+y^4=x^4+y^4+2x^2y^2-2x^2y^2$<\/p>\n $=\\left[x^2+y^2\\right]^2-2\\left(xy\\right)^2$<\/p>\n $=\\left[x^2+y^2+2xy-2xy\\right]^2-2\\left(4\\right)^2$<\/p>\n $=\\left[\\left(x+y\\right)^2-2xy\\right]^2-32$<\/p>\n $=\\left[\\left(4\\right)^2-2\\left(4\\right)\\right]^2-32$<\/p>\n $=\\left[16-8\\right]^2-32$<\/p>\n $=64-32$<\/p>\n $=32$<\/p>\n Hence, the correct answer is Option B<\/p>\n Question 15:\u00a0<\/b>If $a^2+\\frac{2}{a^2}=16$, then find the value of $\\frac{72a^2}{a^4+2+8a^2}$<\/p>\n a)\u00a02<\/p>\n b)\u00a04<\/p>\n c)\u00a01<\/p>\n d)\u00a03<\/p>\n 15)\u00a0Answer\u00a0(D)<\/strong><\/p>\n Solution:<\/b><\/p>\n Given, \u00a0$a^2+\\frac{2}{a^2}=16$<\/p>\n $\\frac{72a^2}{a^4+2+8a^2}=\\frac{72a^2}{a^2\\left(a^2+\\frac{2}{a^2}+8\\right)}$<\/p>\n $=\\frac{72}{a^2+\\frac{2}{a^2}+8}$<\/p>\n $=\\frac{72}{16+8}$<\/p>\n $=\\frac{72}{24}$<\/p>\n $=3$<\/p>\n Hence, the correct answer is Option D<\/p>\n Question 16:\u00a0<\/b>If a + 3b = 12 and ab = 9, then the value of (a – 3b) is:<\/p>\n a)\u00a09<\/p>\n b)\u00a08<\/p>\n c)\u00a06<\/p>\n d)\u00a04<\/p>\n 16)\u00a0Answer\u00a0(C)<\/strong><\/p>\n Solution:<\/b><\/p>\n Given,\u00a0$a+3b=13$<\/p>\n $\\Rightarrow$ \u00a0$\\left(a+3b\\right)^2=12^2$<\/p>\n $\\Rightarrow$ \u00a0$a^2+9b^2+6ab=144$<\/p>\n $\\Rightarrow$ \u00a0$a^2+9b^2+6\\left(9\\right)=144$<\/p>\n $\\Rightarrow$ \u00a0$a^2+9b^2+54-6ab+6ab=144$<\/p>\n $\\Rightarrow$ \u00a0$a^2+9b^2-6ab+6ab=90$<\/p>\n $\\Rightarrow$ \u00a0$\\left(a-3b\\right)^2+6ab=90$<\/p>\n $\\Rightarrow$ \u00a0$\\left(a-3b\\right)^2+6\\left(9\\right)=90$<\/p>\n $\\Rightarrow$ \u00a0$\\left(a-3b\\right)^2+54=90$<\/p>\n $\\Rightarrow$ \u00a0$\\left(a-3b\\right)^2=36$<\/p>\n $\\Rightarrow$ \u00a0$a-3b=6$<\/p>\n Hence, the correct answer is Option C<\/p>\n Question 17:\u00a0<\/b>If $1 + 9r^2 + 81r^4 = 256$ and $1 + 3r + 9r^2 = 32$, then find the value of $1 – 3r + 9r^2$.<\/p>\n a)\u00a04<\/p>\n b)\u00a08<\/p>\n c)\u00a016<\/p>\n d)\u00a012<\/p>\n 17)\u00a0Answer\u00a0(B)<\/strong><\/p>\n Solution:<\/b><\/p>\n Given that $1 + 3r + 9r^2 = 32$<\/p>\n and let\u00a0$1 – 3r + 9r^2$ be K<\/p>\n Multiply $1 + 3r + 9r^2$ and $1 – 3r + 9r^2$<\/p>\n we get\u00a0the product as $1 + 9r^2 + 81r^4$ = 32K<\/p>\n but\u00a0$1 + 9r^2 + 81r^4$= 256 256 = 32k<\/p>\n K=8<\/p>\n Therefore, answer is option B<\/p>\n Question 18:\u00a0<\/b>If $x^3 \u2014 6x^2 + ax + b$ is divisible by $(x^2 \u2014 3x + 2)$, then the values of a and b are:<\/p>\n a)\u00a0a = -6 and b = -11<\/p>\n b)\u00a0a = -11 and b = 6<\/p>\n c)\u00a0a = 11 and b = -6<\/p>\n d)\u00a0a = 6 and b = 11<\/p>\n 18)\u00a0Answer\u00a0(C)<\/strong><\/p>\n Solution:<\/b><\/p>\n Given,\u00a0$x^3\u20146x^2+ax+b$ is divisible by $(x^2 \u2014 3x + 2)$<\/p>\n Let the quotient when\u00a0$x^3 \u2014 6x^2 + ax + b$ is divisible by $(x^2 \u2014 3x + 2)$ be $x-p$<\/p>\n $\\Rightarrow$ $(x^2\u20143x+2)\\left(x-p\\right)=x^3\u20146x^2+ax+b$<\/p>\n $\\Rightarrow$\u00a0 $x^3-3x^2+2x-px^2+3px-2p=x^3\u20146x^2+ax+b$<\/p>\n $\\Rightarrow$ \u00a0$x^3-\\left(3+p\\right)x^2+\\left(2+3p\\right)x-2p=x^3\u20146x^2+ax+b$<\/p>\n Comparing both sides,<\/p>\n $-\\left(3+p\\right)=-6$<\/p>\n $\\Rightarrow$ \u00a0$p=3$<\/p>\n $2+3p=a$<\/p>\n $\\Rightarrow$ \u00a0$2+3\\left(3\\right)=a$<\/p>\n $\\Rightarrow$ \u00a0$a=11$<\/p>\n $-2p=b$<\/p>\n $\\Rightarrow$ \u00a0$-2\\left(3\\right)=b$<\/p>\n $\\Rightarrow$ \u00a0$b=-6$<\/p>\n $\\therefore\\ $ a = 11 and b = -6<\/p>\n Hence, the correct answer is Option C<\/p>\n Question 19:\u00a0<\/b>If $x – \\frac{1}{x} = 13,$ then the value of $x^2 + \\frac{1}{x^2}$ is:<\/p>\n a)\u00a0165<\/p>\n b)\u00a0171<\/p>\n c)\u00a0167<\/p>\n d)\u00a0169<\/p>\n 19)\u00a0Answer\u00a0(B)<\/strong><\/p>\n Solution:<\/b><\/p>\n Given,\u00a0$x-\\frac{1}{x}=13$<\/p>\n $\\Rightarrow$\u00a0$\\left(x-\\frac{1}{x}\\right)^2=13^2$<\/p>\n $\\Rightarrow$ $x^2-2.x.\\frac{1}{x}+\\frac{1}{x^2}=169$<\/p>\n $\\Rightarrow$\u00a0 $x^2+\\frac{1}{x^2}-2=169$<\/p>\n $\\Rightarrow$\u00a0 $x^2+\\frac{1}{x^2}=171$<\/p>\n Hence, the correct answer is Option B<\/p>\n Question 20:\u00a0<\/b>If $A = \\frac{x – 1}{x + 1}$, then the value of $A – \\frac{1}{A}$ is:<\/p>\n a)\u00a0$\\frac{-4(2x – 1)}{x^2 – 1}$<\/p>\n b)\u00a0$\\frac{x^2 – 1}{-4(2x – 1)}$<\/p>\n c)\u00a0$\\frac{x^2 – 1}{-4(2x + 1)}$<\/p>\n d)\u00a0$\\frac{-4x}{x^2 – 1}$<\/p>\n 20)\u00a0Answer\u00a0(D)<\/strong><\/p>\n Solution:<\/b><\/p>\n Given,\u00a0$A=\\frac{x-1}{x+1}$<\/p>\n $A-\\frac{1}{A}$ =\u00a0$\\frac{x-1}{x+1}-\\frac{x+1}{x-1}$<\/p>\n $=\\frac{\\left(x-1\\right)^2-\\left(x+1\\right)^2}{x^2-1}$<\/p>\n $=\\frac{x^2-2x+1-\\left(x^2+2x+1\\right)^{ }}{x^2-1}$<\/p>\n $=\\frac{x^2-2x+1-x^2-2x-1}{x^2-1}$<\/p>\n $=\\frac{-4x}{x^2-1}$<\/p>\n Hence, the correct answer is Option D<\/p>\n
\nsubstitute this value in the product<\/p>\n