BYJU'S



The sum of coefficients of even powers of x in [latex]{{\left( x+\sqrt{{{x}^{3}}-1} \right)}^{6}}+{{\left( x-\sqrt{{{x}^{3}}-1} \right)}^{6}}[/latex](A) 23(B) 24(C) 18(D) 21Solution:[latex]{{\left( x+y \right)}^{n}}+{{\left( x-y \right)}^{n}}=2\left( n{{c}_{0}}{{x}^{n}}+n{{c}_{2}}{{x}^{n-2}}{{y}^{2}}+n{{c}_{4}}{{x}^{n-4}}.{{y}^{4}}+.... \right)[/latex][latex]{{\left( x+\sqrt{{{x}^{3}}-1} \right)}^{6}}+{{\left( x-\sqrt{{{x}^{3}}-1} \right)}^{6}}=2\left[ 6{{c}_{0}}{{x}^{6}}+6{{c}_{2}}.{{x}^{4}}\left( {{x}^{3}}-1 \right)+6{{c}_{4}}{{x}^{2}}{{\left( {{x}^{3}}-1 \right)}^{2}}+6{{c}_{6}}{{\left( {{x}^{3}}-1 \right)}^{3}} \right][/latex][latex]=2\left[ {{x}^{6}}+15{{x}^{4}}\left( {{x}^{3}}-1 \right)+15{{x}^{2}}\left( {{x}^{6}}-2{{x}^{3}}+1 \right)+{{\left( {{x}^{3}}-1 \right)}^{3}} \right][/latex][latex]=2\left[ {{x}^{6}}+15{{x}^{7}}-15{{x}^{4}}+15{{x}^{8}}-30{{x}^{5}}+15{{x}^{2}}+{{x}^{9}}-3{{x}^{6}}+3\left( {{x}^{3}}-1 \right) \right][/latex]Terms with even power of [latex]x=2\left[ {{x}^{6}}-15{{x}^{4}}+15{{x}^{8}}+15{{x}^{2}}-3{{x}^{6}}-1 \right][/latex][latex]\therefore[/latex] Sum of coefficients [latex]=2\left[ 1-\cancel{15}+\cancel{15}+15-3-1 \right]=24[/latex] Let [latex]\sin \left( \alpha -\beta \right)=\frac{5}{13}[/latex]and [latex]\cos \left( \alpha +\beta \right)=\frac{3}{5}[/latex]where [latex]\alpha ,\beta \in \left( 0,\frac{\pi }{4} \right)[/latex] then [latex]Tan2\alpha =[/latex](A) [latex]\frac{63}{16}[/latex](B) [latex]\frac{61}{16}[/latex](C) [latex]\frac{65}{16}[/latex](D) [latex]\frac{32}{9}[/latex]Solution:[latex]\sin \left( \alpha -\beta \right)=\frac{5}{13}\Rightarrow \cos \left( \alpha -\beta \right)=\frac{12}{13}\Rightarrow Tan\left( \alpha -\beta \right)=\frac{5}{12}[/latex][latex]\cos \left( \alpha +\beta \right)=\frac{3}{5}\Rightarrow \sin \left( \alpha +\beta \right)=\frac{4}{5}\Rightarrow Tan\left( \alpha +\beta \right)=\frac{4}{3}[/latex][latex]Tan2\alpha =Tan\left[ \left( \alpha +\beta \right)+\left( \alpha -\beta \right) \right][/latex][latex]=\frac{Tan\left( \alpha +\beta \right)+Tan\left( \alpha -\beta \right)}{1-Tan\left( \alpha +\beta \right).\,Tan\left( \alpha -\beta \right)}=\frac{\frac{4}{3}+\frac{5}{12}}{1-\frac{4}{3}.\frac{5}{12}}=\frac{16+5}{\frac{36-20}{3}}=\frac{63}{16}[/latex] The line [latex]x+y=n,\,\,n\in N[/latex] makes intercepts with [latex]{{x}^{2}}+{{y}^{2}}=16.[/latex] Then the sum of squares of all possible intercepts(A) [latex]\frac{105}{4}[/latex](B) 105(C) 210(D) [latex]\frac{105}{2}[/latex]Solution:31311858255[latex]{{x}^{2}}+{{y}^{2}}=16[/latex]Centre = (0, 0)OA = PRadius = 4P [latex]=\frac{n}{\sqrt{2}}[/latex]To make the intercepts[latex]\frac{n}{\sqrt{2}}<4\Rightarrow n<4\sqrt{2}[/latex]Length of intercepts [latex]=\sqrt{{{r}^{2}}-{{p}^{2}}}=\sqrt{16-\frac{{{n}^{2}}}{2}}[/latex]Square of intercepts [latex]=16-\frac{{{n}^{2}}}{2},n\in N[/latex]Sum of squares of intercepts [latex]=\left( 16-\frac{1}{2} \right)+\left( 16-\frac{4}{2} \right)+\left( 16-\frac{9}{2} \right)+\left( 16-\frac{16}{2} \right)+\left( 16-\frac{25}{2} \right)[/latex][latex]=80-\frac{55}{2}=\frac{105}{2}[/latex] [latex]\int{\frac{\sin \frac{5x}{2}}{\sin \frac{x}{2}}}.\,dx=[/latex](A) [latex]x+2\sin x+\sin 2x+c[/latex](B) [latex]x+2\cos x+\sin 2x+c[/latex](C) [latex]x-2\sin x+\sin 2x+c[/latex](D) [latex]x+2\sin x-\sin 2x+c[/latex]Solution:[latex]\int{\frac{\sin \frac{5x}{2}}{\sin \frac{x}{2}}}.\,dx=\int{\frac{2\sin \frac{5x}{2}.\cos \frac{x}{2}}{2\sin \frac{x}{2}.\cos \frac{x}{2}}}.\,dx=\int{\frac{\sin 3x+\sin 2x}{\sin 2\left( \frac{x}{2} \right)}}.\,dx[/latex][latex]=\int{\frac{\sin 3x}{\sin x}}.\,dx+\int{\frac{\sin 2x}{\sin x}}.\,dx[/latex][latex]=\int{\frac{3\sin x-4{{\sin }^{3}}x}{\sin x}}.\,dx+\int{\frac{2\sin x\cos x}{\sin x}}dx[/latex][latex]=3\int{dx-4\int{{{\sin }^{2}}x.\,dx}}+2\int{\cos x\,dx=3x-4}\int{\frac{1-\cos 2x}{2}}.\,dx+2\sin x[/latex][latex]=3x-2x+\sin 2x+2\sin x+c=x+2\sin x+\sin 2x+c[/latex] The area bounded by the curve [latex]y\le {{x}^{2}}+3x,0\le y\le 4,0\le x\le 3[/latex] is (A) [latex]\frac{59}{6}[/latex](B) [latex]\frac{57}{4}[/latex](C) [latex]\frac{59}{3}[/latex](D) [latex]\frac{57}{6}[/latex]390525024447500Solution:[latex]y={{x}^{2}}+3x[/latex][latex]y=4[/latex][latex]\Rightarrow {{x}^{2}}+3x=4[/latex][latex]\Rightarrow {{x}^{2}}+3x-4=0\Rightarrow x=1\,\,or\,\,x=-4[/latex]Area [latex]=\int\limits_{0}^{1}{\left( {{x}^{2}}+3x \right).\,dx+}[/latex] Area + Rectangle[latex]=\left[ \frac{{{x}^{3}}}{3}+\frac{3{{x}^{2}}}{2} \right]_{0}^{1}+2\left( 4 \right)[/latex][latex]=\frac{1}{3}+\frac{3}{2}+8=\frac{59}{6}[/latex] “If you are born in India then you are citizen of India” contrapositive of this statement is (A) If you are born in India then you are not citizen of India(B) If you are not citizen of India then you are not born in India(C) If you are citizen of India then you are not born in India(D) If you are citizen of India then you are born in IndiaSolution:Statement: [latex]p\Rightarrow q[/latex]Contrapositive: [latex]\tilde{\ }p\Rightarrow \,\tilde{\ }q[/latex][latex]\therefore[/latex] Answer is if you are not citizen of India then you are not born in India. [latex]A=\left[ \begin{matrix} \cos \alpha & -\sin \alpha \\ \sin \alpha & \cos \alpha \\ \end{matrix} \right][/latex] and [latex]{{A}^{32}}=\left[ \begin{matrix} 0 & -1 \\ 1 & 0 \\ \end{matrix} \right][/latex]. then α may be (A) 0(B) [latex]\frac{\pi }{32}[/latex](C) [latex]\frac{\pi }{64}[/latex](D) [latex]\frac{\pi }{16}[/latex]Solution:[latex]A=\left[ \begin{matrix} \cos \alpha & -\sin \alpha \\ \sin \alpha & \cos \alpha \\ \end{matrix} \right]\Rightarrow {{A}^{2}}\left[ \begin{matrix} {{\cos }^{2}}\alpha -{{\sin }^{2}}\alpha & -\cos \alpha \sin \alpha -\cos \alpha \sin \alpha \\ \sin \alpha \cos \alpha +\sin \alpha \cos \alpha & -{{\sin }^{2}}\alpha +{{\cos }^{2}}\alpha \\ \end{matrix} \right][/latex][latex]\Rightarrow {{A}^{2}}=\left[ \begin{matrix} \cos 2\alpha & -\sin 2\alpha \\ \sin 2\alpha & \cos 2\alpha \\ \end{matrix} \right][/latex][latex]\Rightarrow {{A}^{n}}=\left[ \begin{matrix} \cos \,n\alpha & -\sin \,n\alpha \\ \sin \,n\alpha & \cos \,n\alpha \\ \end{matrix} \right][/latex] [latex]\Rightarrow \cos 32\alpha =0[/latex] and [latex]\sin 32\alpha =1[/latex][latex]\Rightarrow 32\alpha =2n\pi +\frac{\pi }{2}[/latex][latex]\alpha =\frac{n\pi }{16}+\frac{\pi }{64},\,n\in I[/latex]Shortest distance between the curves [latex]{{y}^{2}}=x-2[/latex] and [latex]y=x[/latex] is (A) greater than 4(B) less than 2(C) greater than 3(D) greater than 2Solution:[latex]{{y}^{2}}=x-2[/latex]309308584455[latex]2y.\frac{dy}{dx}=1[/latex][latex]\frac{dy}{dx}=\frac{1}{2y}[/latex]Slope of the line [latex]y=x[/latex] is 1Slope of the tangent to the curve [latex]=\frac{1}{2y}[/latex][latex]\frac{1}{2y}=1\Rightarrow y=\frac{1}{2}\Rightarrow K=\frac{1}{2}[/latex][latex]\therefore \,\,N=\left( \frac{9}{4},\frac{1}{2} \right)[/latex]Minimum distance = MN = perpendicular distance from point N to line [latex]y=x[/latex][latex]\left| \frac{a{{x}_{1}}+b{{y}_{1}}+c}{\sqrt{{{a}^{2}}+{{b}^{2}}}} \right|=\left| \frac{1\left( \frac{9}{4} \right)-1\left( \frac{1}{2} \right)}{\sqrt{1+1}} \right|=\frac{7}{4\sqrt{2}}[/latex] which is less than 2 [latex]\underset{x\to 0}{\mathop{\lim }}\,\frac{\sin 2x}{\sqrt{2}-\sqrt{1+\cos x}}=[/latex](A) [latex]\sqrt{2}[/latex](B) 2(C) 4(D) [latex]4\sqrt{2}[/latex]Solution:Rationalise the denominater[latex]\underset{x\to 0}{\mathop{\lim }}\,\frac{{{\sin }^{2}}x\left[ \sqrt{2}+\sqrt{1+\cos x} \right]}{2-\left( 1+\cos x \right)}=\underset{x\to 0}{\mathop{\lim }}\,\frac{{{\sin }^{2}}x\left( \sqrt{2}+\sqrt{1+\cos x} \right)}{1-\cos x}[/latex][latex]=\underset{x\to 0}{\mathop{Lt}}\,\frac{{{\sin }^{2}}x\left( \sqrt{2}+\sqrt{1+\cos x} \right)}{2{{\sin }^{2}}\frac{x}{2}}=\underset{x\to 0}{\mathop{Lt}}\,\frac{{{\sin }^{2}}x}{{{x}^{2}}}.\frac{1}{2}\frac{1}{{{\left( \frac{\sin \frac{x}{2}}{\frac{x}{2}} \right)}^{2}}\times \frac{1}{4}}\left( \sqrt{2}+\sqrt{1+\cos } \right)[/latex][latex]=\frac{4}{2}\left( \sqrt{2}+\sqrt{1+1} \right)=4\sqrt{2}[/latex] How many 9 digit numbers can be formed by using the digits 1, 1, 2, 2, 2, 2, 3, 4, 4 such that odd numbers occur at even places.(A) 160(B) 175(C) 180(D) 220Solution:2931160247650Even places = 4Odd numbers = 3Even numbers = 6[latex]\therefore[/latex] Number of numbers [latex]=4{{c}_{3}}\times \frac{3!}{2!}\times \frac{6!}{4!2!}=180[/latex] Let [latex]g\left( x \right)=\ell n\left( x \right)[/latex] and [latex]f\left( x \right)=\left( \frac{1-x\cos x}{1+x\cos x} \right)[/latex] then [latex]\int\limits_{-\frac{\pi }{4}}^{\frac{\pi }{4}}{g\left[ f\left( x \right) \right]}.\,dx=[/latex](A) [latex]\ell n\,1[/latex](B) [latex]\ell n\,2[/latex](C) [latex]\ell n\,e[/latex](D) [latex]\ell n\,4[/latex]Solution:[latex]g\left[ f\left( x \right) \right]=g\left[ \frac{1-x\cos x}{1+x\cos x} \right]=\ell n\left( \frac{1-x\cos x}{1+x\cos x} \right).\,dx[/latex][latex]I=\int\limits_{-\frac{\pi }{4}}^{\frac{\pi }{4}}{g\left[ f\left( x \right) \right]dx=\int\limits_{-\frac{\pi }{4}}^{\frac{\pi }{4}}{\ell n\left( \frac{1-x\cos x}{1+x\cos x} \right)}\,dx}[/latex]… (1)[latex]\int\limits_{a}^{b}{f\left( x \right)}=\int\limits_{a}^{b}{f\left( a+b-x \right)}[/latex][latex]I=\int\limits_{-\frac{\pi }{4}}^{\frac{\pi }{4}}{\ell n\left( \frac{1+x\cos x}{1-x\cos x} \right)}[/latex]… (2)Add (1) and (2)[latex]2I=\int\limits_{-\frac{\pi }{4}}^{\frac{\pi }{4}}{\left[ \ell n\left( \frac{1+x\cos x}{1-x\cos x} \right)+\left( \frac{1-x\cos x}{1+x\cos x} \right) \right]}\,dx[/latex][latex]=\int\limits_{-\frac{\pi }{4}}^{\frac{\pi }{4}}{\ell n}\left( \frac{1+x\cos x}{1-x\cos x}.\frac{1-x\cos x}{1+x\cos x} \right)\,dx[/latex][latex]\left[ \because \,\,\log ab=\log a+\log b \right][/latex][latex]2I=\int\limits_{-\frac{\pi }{4}}^{\frac{\pi }{4}}{\ell n}\,1=0\Rightarrow I=0[/latex] Let [latex]2.20{{C}_{0}}+5.20{{C}_{1}}+8.20{{C}_{2}}+...+62.20{{C}_{20}}=[/latex](A) [latex]{{16.2}^{22}}[/latex](B) [latex]{{8.2}^{20}}[/latex](C) [latex]{{8.2}^{21}}[/latex](D) [latex]{{16.2}^{21}}[/latex]Solution:[latex]2.20{{C}_{0}}+5.20{{C}_{1}}+8.20{{C}_{2}}+...+62.20{{C}_{20}}[/latex][latex]=\sum\limits_{r=0}^{20}{\left( 3r+2 \right)20{{C}_{r}}}[/latex][latex]=3\sum\limits_{r=0}^{20}{r.\,20{{C}_{r}}}+2.\sum\limits_{r=0}^{20}{20{{C}_{r}}}.[/latex][latex]=3.20\sum\limits_{r=1}^{20}{19{{C}_{r-1}}+2\left( 20{{C}_{0}}+20{{C}_{1}}+...+20{{C}_{20}} \right)}[/latex][latex]={{60.2}^{19}}+{{2.2}^{20}}={{2}^{21}}\left( 15+1 \right)={{2}^{21}}.16.[/latex] Sum of natural numbers between 100 and 200 whose HCF with 91 should be more than 1.(A) 1121(B) 3210(C) 3121(D) 1520Solution:Given numbers: 101, 102, 103, … 198, 199[latex]91=13\times 7[/latex]HCF of 91 and a number is more than 1 means the number should be either multiple of 7 or 13.[latex]\therefore[/latex] Sum of the numbers = (Numbers divisible by 7) + (Numbers divisible by 13) – (Numbers divisible by 91)[latex]=\left( 105+112+...+196 \right)+\left( 104+117+...195 \right)-\left( 182 \right)[/latex][latex]=\frac{14}{2}\left[ 105+196 \right]+\frac{8}{2}\left[ 104+195 \right]-182[/latex][latex]=7\left( 301 \right)+4\left( 299 \right)-182=2107+1196-182[/latex][latex]=3121[/latex] If mean and variance of 7 variates are 8 and 16 respectively and five of them are 2, 4, 10, 12, 14 then the product of remaining two variates is (A) 49(B) 48(C) 45(D) 40Solution:Let the remaining two variates be x and yGiven mean [latex]=7\Rightarrow \frac{x+y+2+4+10+12+14}{7}=8[/latex][latex]\Rightarrow x+y=14[/latex]… (1)And variance [latex]=16\Rightarrow \frac{{{x}^{2}}+{{y}^{2}}+4+16+100+144+196}{7}-{{\left( 8 \right)}^{2}}=16[/latex][latex]\Rightarrow {{x}^{2}}+{{y}^{2}}=100[/latex]… (2) From (1) [latex]x+y=14\Rightarrow {{\left( x+y \right)}^{2}}={{\left( 14 \right)}^{2}}[/latex][latex]\Rightarrow {{x}^{2}}+{{y}^{2}}+2xy=196[/latex][latex]\Rightarrow 2xy=196-100\Rightarrow xy=48[/latex]If α and β are the roots of [latex]{{x}^{2}}-2x+2=0[/latex] then the minimum value of n such that [latex]{{\left( \frac{\alpha }{\beta } \right)}^{n}}=1[/latex](A) 4(B) 3(C) 2(D) 5Solution:[latex]{{x}^{2}}-2x+2=0\Rightarrow x=\frac{2\pm \sqrt{4-8}}{2}\Rightarrow x=1\pm i[/latex][latex]\alpha =1+i[/latex] and [latex]\beta =1-i[/latex][latex]\frac{\alpha }{\beta }=\frac{1+i}{1-i}=\frac{1+i}{1-i}\times \frac{1+i}{1+i}=\frac{2i}{2}=i[/latex][latex]{{\left( \frac{\alpha }{\beta } \right)}^{n}}={{i}^{n}}[/latex]Given [latex]{{i}^{n}}=1\Rightarrow n=4,8,12,....\,\,\,n\in N[/latex]Minimum value of [latex]n=4[/latex] Solution of differential equation [latex]{{\left( {{x}^{2}}+1 \right)}^{2}}\frac{dy}{dx}=2x\left( {{x}^{2}}+1 \right)y=1[/latex]is (A) [latex]y=\frac{Ta{{n}^{-1}}x}{{{x}^{2}}+1}+c[/latex](B) [latex]y=Ta{{n}^{-1}}x+c[/latex](C) [latex]y\left( {{x}^{2}}+1 \right)=Ta{{n}^{-1}}x+c[/latex](D) [latex]y\left( Ta{{n}^{-1}}x \right)={{x}^{2}}+c[/latex]Solution:[latex]{{\left( {{x}^{2}}+1 \right)}^{2}}\frac{dy}{dx}+2x\left( {{x}^{2}}+1 \right)y=1[/latex][latex]\frac{dy}{dx}+\frac{2x}{{{x}^{2}}+1}.\,y=\frac{1}{{{\left( {{x}^{2}}+1 \right)}^{2}}}[/latex]It is linear differential equation where [latex]P=\frac{2x}{{{x}^{2}}+1},Q=\frac{1}{{{\left( {{x}^{2}}+1 \right)}^{2}}}[/latex]If [latex]={{e}^{\int{P.\,dx}}}={{e}^{\int{\frac{2x}{{{x}^{2}}+1}.\,dx}}}={{e}^{\ell n\left( {{x}^{2}}+1 \right)}}={{x}^{2}}+1[/latex][latex]y.\,IF=\int{Q.\,IF.\,dx}[/latex][latex]y\left( {{x}^{2}}+1 \right)=\int{\frac{1}{{{\left( {{x}^{2}}+1 \right)}^{2}}}}.\left( {{x}^{2}}+1 \right)dx[/latex][latex]y\left( {{x}^{2}}+1 \right)=\int{\frac{1}{{{x}^{2}}+1}}.\,dx[/latex][latex]y\left( {{x}^{2}}+1 \right)=Ta{{n}^{-1}}x+c[/latex] If [latex]f\left( x \right)=\log \left( \frac{1-x}{1+x} \right)[/latex] then [latex]f\left( \frac{2x}{1+{{x}^{2}}} \right)[/latex] is equal to (A) [latex]f\left( x \right)[/latex](B) [latex]2f\left( x \right)[/latex](C) [latex]-2f\left( x \right)[/latex](D) [latex]{{\left[ f\left( x \right) \right]}^{2}}[/latex]Solution:[latex]f\left( x \right)=\log \left( \frac{1-x}{1+x} \right)[/latex][latex]f\left( \frac{2x}{1+{{x}^{2}}} \right)=\log \left[ \frac{1-\frac{2x}{1+{{x}^{2}}}}{1+\frac{2x}{1+{{x}^{2}}}} \right]=\log \left( \frac{1+{{x}^{2}}-2x}{1+{{x}^{2}}+2x} \right)=\log {{\left( \frac{1-x}{1+x} \right)}^{2}}[/latex][latex]=2.log\left( \frac{1-x}{1+x} \right)[/latex][latex]=2.\,f\left( x \right)[/latex] Given that [latex]A\subset B[/latex] then identify the correct statement.(A) [latex]P\left( \frac{A}{B} \right)=P\left( A \right)[/latex](B) [latex]P\left( \frac{A}{B} \right)\le P\left( A \right)[/latex](C) [latex]P\left( \frac{A}{B} \right)\ge P\left( A \right)[/latex](D) [latex]P\left( \frac{A}{B} \right)=P\left( A \right)-P\left( B \right)[/latex]Solution:[latex]P\left( \frac{A}{B} \right)=\frac{P\left( A\cap B \right)}{P\left( B \right)}=\frac{P\left( A \right)}{P\left( B \right)}\ge P\left( A \right)[/latex] Find the value of ‘c’ for which the following equations have non-trivial solutions.[latex]cx-y-z=0,-cx+y-cz=0,x+y-cz=0[/latex](A) [latex]\frac{1}{2}[/latex](B) [latex]-1[/latex](C) 2(D) 0Solution:[latex]\left| \begin{matrix} c & -1 & -1 \\ -c & 1 & -c \\ 1 & 1 & -c \\ \end{matrix} \right|=0[/latex][latex]\Rightarrow c\left( -c+c \right)+1\left( {{c}^{2}}+c \right)-1\left( -c-1 \right)=0[/latex][latex]\Rightarrow {{c}^{2}}+2c+1=0\Rightarrow {{\left( c+1 \right)}^{2}}=0\Rightarrow c=-1[/latex] Let [latex]2y={{\left[ {{\cot }^{-1}}\left( \frac{\sqrt{3}\cos x+\sin x}{\cos x-\sqrt{3}\sin x} \right) \right]}^{2}}[/latex] then [latex]\frac{dy}{dx}=[/latex](A) [latex]x-\frac{\pi }{6}[/latex](B) [latex]x+\frac{\pi }{6}[/latex](C) [latex]2x-\frac{\pi }{6}[/latex](D) [latex]2x-\frac{\pi }{3}[/latex]Solution:[latex]2y={{\left[ {{\cot }^{-1}}\left( \frac{\sqrt{3}\cos x+\sin x}{\cos x-\sqrt{3}\sin x} \right) \right]}^{2}}[/latex][latex]2y={{\left[ {{\cot }^{-1}}\left( \frac{\sqrt{3}+Tan\,x}{1-\sqrt{3}Tan\,x} \right) \right]}^{2}}[/latex][[latex]\because[/latex] Divide Nr & Dr with cos x] [latex]2y={{\left[ {{\cot }^{-1}}\left( \frac{Tan60+Tan\,x}{1-Tan60.\,Tanx} \right) \right]}^{2}}[/latex][latex]2y={{\left[ {{\cot }^{-1}}\left( Tan\frac{\pi }{3}+x \right) \right]}^{2}}[/latex][latex]2y{{\left\{ \frac{\pi }{2}-Ta{{n}^{-1}}\left[ Tan\left( \frac{\pi }{3}+x \right) \right] \right\}}^{2}}[/latex][latex]2y={{\left( \frac{\pi }{2}-\frac{\pi }{3}-x \right)}^{2}}\Rightarrow 2y={{\left( \frac{\pi }{6}-x \right)}^{2}}[/latex][latex]2y=\frac{{{\pi }^{2}}}{36}+{{x}^{2}}-\frac{\pi x}{3}[/latex][latex]2.\,\frac{dy}{dx}=2x-\frac{\pi }{3}[/latex][latex]\frac{dy}{dx}=x-\frac{\pi }{6}[/latex] Let [latex]{{S}_{1}}[/latex]is set of minima and [latex]{{S}_{2}}[/latex]is set of maxima for the curve [latex]y=9{{x}^{4}}+12{{x}^{3}}-36{{x}^{2}}-25[/latex] then(A) [latex]{{S}_{1}}=\left\{ -2,-1 \right\},{{S}_{2}}=\left\{ 0 \right\}[/latex](B) [latex]{{S}_{1}}=\left\{ -2,1 \right\},{{S}_{2}}=\left\{ 0 \right\}[/latex](C) [latex]{{S}_{1}}=\left\{ -2,1 \right\},{{S}_{2}}=\left\{ -1 \right\}[/latex](D) [latex]{{S}_{1}}=\left\{ -2,2 \right\},{{S}_{2}}=\left\{ 0 \right\}[/latex]Solution:[latex]y=9{{x}^{4}}+12{{x}^{3}}-36{{x}^{2}}-25[/latex][latex]\frac{dy}{dx}=36{{x}^{3}}+36{{x}^{2}}-72x[/latex][latex]=36x\left( {{x}^{2}}+x-2 \right)[/latex]3864610244475[latex]=36x\left( {{x}^{2}}+2x-x-2 \right)[/latex][latex]=36x\left( x+2 \right)\left( x-1 \right)[/latex]Critical points are 0, -2, 1At [latex]\left\{ -2,1 \right\}\to[/latex] points of minima.[latex]\left\{ 0 \right\}\to[/latex] point of maxima. Let [latex]{{f}^{11}}\left( x \right)>0[/latex] and [latex]\phi \left( x \right)=f\left( x \right)+f\left( 2-x \right),x\in \left( 0,2 \right)[/latex] be a function then the function [latex]\phi \left( x \right)[/latex]is (A) Increasing in (0, 1) and decreasing in (1, 2)(B) Decreasing in (0, 1) and increasing in (1, 2)(C) Increasing in (0, 2) (D) Decreasing in (0, 2)Solution:Let [latex]y=f\left( x \right),x\in \left( 0,2 \right)[/latex][latex]\phi \left( x \right)=f\left( x \right)+f\left( 2-x \right)[/latex][latex]\phi '\left( x \right)=f'\left( x \right)+f'\left( 2-x \right)[/latex]If [latex]\phi '\left( x \right)[/latex] is increasing then [latex]\phi '\left( x \right)>0[/latex] [latex]\Rightarrow f'\left( x \right)-f'\left( 2-x \right)>0[/latex][latex]\Rightarrow f'\left( x \right)>f'\left( 2-x \right)[/latex][latex]\Rightarrow x>2-x[/latex][[latex]f'\left( x \right)[/latex] is increasing in (0, 2)][latex]\Rightarrow x>1[/latex][latex]\Rightarrow x\in \left( 1,\,2 \right)[/latex]It [latex]\phi '\left( x \right)[/latex] is decreasing then [latex]\phi '\left( x \right)<0[/latex][latex]\Rightarrow f'\left( x \right)<f'\left( 2-x \right)[/latex][latex]\Rightarrow x<1\Rightarrow x\in \left( 0,1 \right)[/latex]Let vertices of the triangle ABC is A(0, 0), B(0, 1) and C(x, y) and perimeter is 4 then locus of ‘C’ is (A) [latex]9{{x}^{2}}+8{{y}^{2}}+8y=16[/latex](B) [latex]8{{x}^{2}}+9{{y}^{2}}+9y=16[/latex](C) [latex]9{{x}^{2}}+8{{y}^{2}}-8y=16[/latex](D) [latex]8{{x}^{2}}+9{{y}^{2}}-9x=16[/latex]Solution:Perimeter = 43359785163195[latex]\sqrt{{{x}^{2}}+{{\left( y-1 \right)}^{2}}}+\sqrt{{{x}^{2}}+{{y}^{2}}}+1=4[/latex][latex]\sqrt{{{x}^{2}}+{{y}^{2}}-2y+1}=3-\sqrt{{{x}^{2}}+{{y}^{2}}}[/latex][latex]\cancel{{{x}^{2}}}+\cancel{{{y}^{2}}}-2y+1=9+\cancel{{{x}^{2}}}+\cancel{{{y}^{2}}}-6\sqrt{{{x}^{2}}+{{y}^{2}}}[/latex][latex]6\sqrt{{{x}^{2}}+{{y}^{2}}}=8+2y[/latex][latex]3\sqrt{{{x}^{2}}+{{y}^{2}}}=4+y[/latex][latex]9\left( {{x}^{2}}+{{y}^{2}} \right)=16+{{y}^{2}}+8y[/latex][latex]9{{x}^{2}}+8{{y}^{2}}-8y=16[/latex] Let the equation of a line is [latex]3x+5y=15[/latex] and a point P on this line is equidistant from x and y axis. In which quadrant that P lies.(A) 1st (B) 3rd(C) 4th(D) none of these4302760-8890100Solution:[latex]3x+5y=15[/latex][latex]\frac{x}{5}+\frac{y}{3}=1[/latex]If [latex]y=x\Rightarrow 8x=15[/latex][latex]\Rightarrow x=\frac{15}{8},y=\frac{15}{8}[/latex]If [latex]y=-x\Rightarrow -2x=15[/latex][latex]\Rightarrow x=\frac{-15}{2},y=\frac{15}{2}[/latex]Intersection of [latex]3x+5y=15[/latex] with [latex]y=x[/latex] and [latex]y=-x[/latex] will give required P.[latex]\therefore \,\,\,P=\left( \frac{15}{8},\frac{15}{8} \right)[/latex] or [latex]P=\left( \frac{-15}{2},\frac{15}{2} \right)[/latex][latex]\therefore[/latex] P lies in either 1st or 2nd quadrant.[latex]\therefore[/latex] In the given options 2nd quadrant not given[latex]\therefore[/latex] 1st quadrant is the answer. The perpendicular distance of point [latex]\left( 2,-1,4 \right)[/latex] from the line [latex]\frac{x+3}{10}=\frac{y-2}{-7}=\frac{z}{1}[/latex] lies between (A) (2, 3)(B) (3, 4)(C) (4, 5)(D) (1, 2)Solution:Let P be (2, -1, 4)Let Q is a point on line [latex]\frac{x+3}{10}=\frac{y-2}{-7}=\frac{z}{1}=\lambda[/latex]i.e. [latex]Q=\left( 10\lambda -3,-7\lambda +2,\lambda \right)[/latex]Dr’s of PQ [latex]=\left( {{x}_{2}}-{{x}_{1}},{{y}_{2}}-{{y}_{1}},{{z}_{2}}-{{z}_{1}} \right)[/latex][latex]=\left( 10\lambda -5,-7\lambda +3,\lambda -4 \right)[/latex][latex]{{a}_{1}}{{a}_{2}}+{{b}_{1}}{{b}_{2}}+{{c}_{1}}{{c}_{2}}=0[/latex][latex]\Rightarrow 10\left( 10\lambda -5 \right)-7\left( -7\lambda +3 \right)+1\left( \lambda -4 \right)=0[/latex][latex]\Rightarrow 100\lambda -50+49\lambda -21+\lambda -4=0[/latex][latex]\Rightarrow 150\lambda -75=0\Rightarrow \lambda =\frac{1}{2}[/latex][latex]PQ=\sqrt{{{\left( 10\lambda -5 \right)}^{2}}+{{\left( -7\lambda +3 \right)}^{2}}+{{\left( \lambda -4 \right)}^{2}}}[/latex][latex]\sqrt{0+{{\left( \frac{-7}{2}+3 \right)}^{2}}+{{\left( \frac{1}{2}-4 \right)}^{2}}}=\sqrt{\frac{50}{4}}=\sqrt{12.5}[/latex][latex]\sqrt{12.5}[/latex] lies between (3, 4)If a plane passes through intersection of planes [latex]2x-y-4=0[/latex] and [latex]y+2z-4=0[/latex] and also passes through the point (1, 1, 0). Then the equation of the plane is (A) [latex]x-y-z=0[/latex](B) [latex]2x-z=0[/latex](C) [latex]x+2z-1=0[/latex](D) [latex]x-z-1=0[/latex]Solution:Plane passing through intersection two planes is [latex]{{P}_{1}}+\lambda {{P}_{2}}=0[/latex][latex]\left( 2x-y-4 \right)+\lambda \left( y+2z-4 \right)=0[/latex]… (1)(1) passing through (1, 1, 0) [latex]\Rightarrow \left( 2-1-4 \right)+\lambda \left( 1-4 \right)=0[/latex][latex]\Rightarrow \lambda =-1[/latex][latex]\therefore[/latex] equation of plane is [latex]\left( 2x-y-4 \right)-1\left( y+2z-4 \right)=0[/latex][latex]\Rightarrow 2x+2y-2z=0[/latex][latex]\Rightarrow x+y-z=0[/latex]If [latex]\left| \sqrt{x}-2 \right|+\sqrt{x}\left( \sqrt{x}-4 \right)=2[/latex] then sum of roots of equation is (A) 12(B) 8(C) 4(D) 16Solution:[latex]\left| \sqrt{x}-2 \right|+x-4\sqrt{x}=2[/latex][latex]\left| \sqrt{x}-2 \right|+{{\left( \sqrt{x} \right)}^{2}}-4\sqrt{x}+4=2+4[/latex][latex]\left| \sqrt{x}-2 \right|+{{\left( \sqrt{x}-2 \right)}^{2}}=6[/latex]Let [latex]\left| \sqrt{x}-2 \right|=P[/latex][latex]{{P}^{2}}+P-6=0\,\,\Rightarrow {{P}^{2}}+3P-2P-6=0[/latex][latex]\Rightarrow p=-3[/latex] or [latex]P=2[/latex][latex]\left| \sqrt{x}-2 \right|=-3[/latex] which is not possible[latex]\left| \sqrt{x}-2 \right|=2[/latex][latex]\sqrt{x}-2=\pm 2[/latex][latex]\sqrt{x}=2+2[/latex] or [latex]-2+2[/latex][latex]\sqrt{x}=4[/latex] or 0[latex]x=16[/latex] or 0[latex]\therefore[/latex] sum of roots [latex]=16+0=16[/latex] [latex]4{{x}^{2}}+{{y}^{2}}=8,[/latex] Tangent at (1, 2) and another tangent at (a, b) are perpendicular then [latex]{{a}^{2}}=[/latex](A) [latex]\frac{2}{17}[/latex](B) [latex]\frac{1}{17}[/latex](C) [latex]\frac{8}{17}[/latex](D) [latex]\frac{4}{17}[/latex]Solution:[latex]4{{x}^{2}}+{{y}^{2}}=8[/latex][latex]\frac{{{x}^{2}}}{2}+\frac{{{y}^{2}}}{8}=1\Rightarrow \frac{{{x}^{2}}}{{{\left( \sqrt{2} \right)}^{2}}}+\frac{{{y}^{2}}}{{{\left( 2\sqrt{2} \right)}^{2}}}=1[/latex]Let (a, b) [latex]=\left( \sqrt{2}\cos \theta ,2\sqrt{2}\sin \theta \right)[/latex]Tangent at (1, 2) is [latex]4x+2y=8\Rightarrow 2x+y=4\Rightarrow {{m}_{1}}=-2[/latex]Tangent at [latex]\left( \sqrt{2}\cos \theta ,2\sqrt{2}\sin \theta \right)[/latex] is [latex]4\sqrt{2}\cos \theta x+\left( 2\sqrt{2}\sin \theta \right)=y=8\Rightarrow 4\sqrt{2}\cos \theta +2\sqrt{2}\sin \theta y=8[/latex][latex]\Rightarrow {{m}_{2}}=-2\cot \theta[/latex].[latex]{{m}_{1}}{{m}_{2}}=-1[/latex][latex]4\cot \theta =-1\Rightarrow Tan\theta =-4[/latex][latex]\Rightarrow {{\cos }^{2}}\theta =\frac{1}{17}[/latex][latex]\Rightarrow \cos \theta =\pm \frac{1}{\sqrt{17}}[/latex] [latex]a=\sqrt{2}\cos \theta[/latex][latex]{{a}^{2}}=2{{\cos }^{2}}\theta \Rightarrow {{a}^{2}}=\frac{2}{17}[/latex]Find the magnitude of projection of vector [latex]2i+3j+k[/latex] on a vector which is perpendicular to the plane containing vectors [latex]i+j+k[/latex] and [latex]i+2j+3k[/latex](A) [latex]\frac{\sqrt{3}}{\sqrt{2}}[/latex](B) [latex]\frac{\sqrt{2}}{\sqrt{3}}[/latex](C) [latex]\frac{4}{\sqrt{3}}[/latex](D) [latex]\frac{2\sqrt{2}}{\sqrt{3}}[/latex]Solution:Normal vector to the plane containing [latex]i+j+k[/latex] and [latex]i+2j+3k[/latex]is [latex]\overrightarrow{n\,}=\left( i+j+k \right)\times \left( i+2j+3k \right)[/latex][latex]\overrightarrow{n\,}=i-2j+k[/latex]Projection of [latex]\left( 2i+3j+k \right)[/latex] on [latex]\overrightarrow{\,n\,}[/latex][latex]=\left| \frac{\left( 2i+3j+k \right).\left( i-2j+k \right)}{\sqrt{1+4+1}} \right|=\frac{\left| 2-6+1 \right|}{\sqrt{6}}=\frac{3}{\sqrt{6}}=\frac{\sqrt{3}}{\sqrt{2}}[/latex] If [latex]\alpha ={{\cos }^{-1}}\left( \frac{3}{5} \right)[/latex] and [latex]\beta =Ta{{n}^{-1}}\left( \frac{5}{12} \right)[/latex] then [latex]\alpha -\beta =[/latex] (A) [latex]Ta{{n}^{-1}}\left( \frac{33}{56} \right)[/latex](B) [latex]Ta{{n}^{-1}}\left( \frac{31}{66} \right)[/latex](C) [latex]Ta{{n}^{-1}}\left( \frac{11}{56} \right)[/latex](D) [latex]Ta{{n}^{-1}}\left( \frac{9}{56} \right)[/latex]Solution:[latex]\alpha =Ta{{n}^{-1}}\frac{4}{3}[/latex][latex]\beta =Ta{{n}^{-1}}\frac{5}{12}[/latex][latex]\alpha -\beta =Ta{{n}^{-1}}\frac{4}{3}-Ta{{n}^{-1}}\frac{5}{12}=Ta{{n}^{-1}}\left( \frac{\frac{4}{3}-\frac{5}{12}}{1+\frac{4}{3}.\frac{5}{12}} \right)[/latex][latex]=Ta{{n}^{-1}}\left( \frac{16-5}{36+20}.\frac{3}{8} \right)[/latex][latex]=Ta{{n}^{-1}}\left( \frac{33}{56} \right)[/latex] ................
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