The number of ways of arranging letters of the word havana so that v and n do not appear together is

Answer

Hint: First we find the total number of arrangements of the letters of the word 'BANANA'.Then take two of the N's together i.e., take ‘NN’ as one block, and find the number of such permutations where the N’s appear together or adjacently. Now, note that the required number of permutations of the letters of the word 'BANANA' in which the two N’s do not appear adjacently is given by\[ = \]Total number of arrangements \[ - \]Number of arrangements where two N’s appear together

Complete step by step Answer:

The number of arrangements of the letters of the word 'BANANA' in which the two N’s do not appear adjacently is 40.

Note: A permutation is an act of arranging the objects or numbers in order while Combinations are the way of selecting the objects or numbers from a group of objects or collection, in such a way that the order of the objects does not matter.

Your browser is no longer supported. Update it to get the best YouTube experience and our latest features. Learn more

• Correct Answer: B

  Solution :

  Given word is HAVANA(3A, 1H, 1N, 1V) Total number of ways arranging the given word \[=\frac{6!}{3!}=120\] Total number of words in which N, V together \[=\frac{5!}{3!}\times 2!=40\] \[\therefore \] Required number of ways \[=120-40=80\]

warning Report Error

Page 2

• Correct Answer: A

  Solution :

  Let E be the event that a six occurs and A be the event that man reports that It is a six. \[\therefore \]\[P(E)=\frac{1}{6},P(E)=\frac{5}{6},P\left( \frac{A}{E} \right)=\frac{3}{4}\] and        \[P\left( \frac{A}{E} \right)=\frac{1}{4}\] Using Bays theorem                 \[P\left( \frac{E}{A} \right)=\frac{P(E).P\left( \frac{A}{E} \right)}{P(E).P\left( \frac{A}{E} \right)+P(E)P\left( \frac{A}{E} \right)}\]       \[=\frac{\frac{1}{6}\times \frac{3}{4}}{\frac{1}{6}\times \frac{3}{4}+\frac{5}{6}\times \frac{1}{4}}=\frac{3}{8}\]

warning Report Error

Page 3

• Correct Answer: B

  Solution :

  \[{{\cos }^{2}}(A-B)+oc{{s}^{2}}B-2\cos (A-B)cos\,A\,cos\,B\] \[={{\cos }^{2}}(A-B)+co{{s}^{2}}B-\cos (A-B)\] \[[cos(A+B)+cos(A-B)]\]                 \[={{\cos }^{2}}B-\cos (A-B)cos(A+B)\]                 \[={{\cos }^{2}}B-\frac{1}{2}[cos2A+cos2B]\] \[={{\cos }^{2}}B-\frac{1}{2}[2co{{s}^{2}}B-1+cos2A]\] \[=\frac{1}{2}-\frac{1}{2}\cos 2A\] \[\frac{1}{2}-\frac{1}{2}(2co{{s}^{2}}A-1)\] \[=1-{{\cos }^{2}}A={{\sin }^{2}}A\] Hence, it is independent of B.

warning Report Error

Page 4

• Correct Answer: A

  Solution :

  The intersection point of \[x+2y=3\]and \[3x+4y=7\]is \[(1,1).\] For consistent, point\[(1,1)\]satisfies the equation \[ax+y=3\]                 \[\therefore \]  \[a+1=3\]                 \[\Rightarrow \]               \[a=2\]

warning Report Error

Page 5

• Correct Answer: D

  Solution :

  Given, \[\left| \begin{matrix}    x-1 & 5x & 7  \\    {{x}^{2}}-1 & x-1 & 8  \\    2x & 3x & 0  \\ \end{matrix} \right|=a{{x}^{2}}+b{{x}^{2}}+cx+d\] LHS\[=(x-1)(0-24x)-5x(0-16x)\] \[+\,7(3{{x}^{2}}-3x-2{{x}^{2}}+2x)\] \[=-24{{x}^{2}}+24x+80{{x}^{2}}+21{{x}^{3}}-14{{x}^{2}}-7x\] \[=21{{x}^{3}}+42{{x}^{2}}+17x\] \[\therefore \]  \[21{{x}^{3}}+42{{x}^{2}}+17x\] \[=a{{x}^{3}}+b{{x}^{2}}+cx+d\]                 \[\Rightarrow \]               \[c=17\]                

warning Report Error

Page 6

• Correct Answer: B

  Solution :

  Using \[AM\ge GM\] \[\frac{\frac{a}{b+c}+\frac{b}{c+a}+\frac{c}{a+b}}{3}\ge \sqrt{\frac{abc}{(a+b)(b+c)(c+a)}}\]                                                                 ?(i) Again using \[AM\ge GM\] \[\frac{a+b}{2}\ge \sqrt{ab},\frac{b+c}{2}\ge \sqrt{bc}\frac{c+a}{2}\ge \sqrt{ca}\] \[\Rightarrow \]\[(a+b)(b+c)(c+a)\ge 8abc\] \[\Rightarrow \]\[\sqrt[3]{\frac{abc}{(a+b)(b+c)(c+a)}}\le \frac{1}{2}\] \[\therefore \]From Eq. (i)          \[\frac{a}{b+c}+\frac{b}{c+a}+\frac{c}{a+b}\le \frac{3}{2}\]

warning Report Error

Page 7

• Correct Answer: C

  Solution :

  Given, \[\tan \,{{1}^{o}}\tan {{2}^{o}}...\tan {{89}^{o}}={{x}^{2}}-8\] \[\Rightarrow \]\[\tan {{1}^{o}}\tan {{2}^{o}}...\tan {{44}^{o}}\tan {{45}^{o}}\cot {{44}^{o}}\] \[\Rightarrow \]\[...\,\cot {{2}^{o}}\cot {{1}^{o}}={{x}^{2}}-8\] \[\Rightarrow \]\[1={{x}^{2}}-8\Rightarrow {{x}^{2}}=9\] \[\Rightarrow \]\[x=\pm 3\]

warning Report Error

Page 8

• Correct Answer: C

  Solution :

  Given \[\frac{\sin (x+3\alpha )}{\sin (\alpha -x)}=3\] Applying componendo and dividendo, we get  \[\Rightarrow \]\[\frac{\sin (x+3\alpha )+\sin (\alpha -x)}{\sin (x+3\alpha )-\sin (\alpha -x)}=\frac{3+1}{3-1}\] \[\Rightarrow \]\[\frac{2\sin 2\alpha \cos (\alpha +x)}{2\cos 2\alpha \sin (\alpha +x)}=2\] \[\Rightarrow \]\[\frac{\tan 2\alpha }{\tan (\alpha +x)}=2\] \[\Rightarrow \]\[\frac{2\tan \alpha }{1-{{\tan }^{2}}\alpha }\times \frac{(1-\tan \alpha \tan x)}{(\tan \alpha +\tan x)}=2\] \[\Rightarrow \]\[\tan \alpha -{{\tan }^{2}}\alpha \tan x=\tan \alpha +\tan x\]                 \[-{{\tan }^{3}}\alpha -{{\tan }^{2}}\alpha \tan x\] \[\Rightarrow \]\[\tan x={{\tan }^{3}}\alpha \]

warning Report Error

Page 9

• Correct Answer: D

  Solution :

  \[2{{a}^{2}}+4{{b}^{2}}+{{c}^{2}}=4ab+2ac\] \[\Rightarrow \]\[{{a}^{2}}+{{(2b)}^{2}}-4ab+{{a}^{2}}+{{c}^{2}}-2ac=0\] \[\Rightarrow \]\[{{(a-2b)}^{2}}+{{(a-c)}^{2}}=0\] \[\Rightarrow \]\[a=2b=c\] \[\cos B=\frac{{{a}^{2}}+{{c}^{2}}-{{b}^{2}}}{2ac}\] \[=\frac{{{c}^{2}}+{{c}^{2}}-{{\left( \frac{c}{2} \right)}^{2}}}{2\times c\times c}=\frac{2{{c}^{2}}-\frac{{{c}^{2}}}{4}}{2{{c}^{2}}}\] \[\Rightarrow \]               \[\cos B=\frac{7}{8}\]

warning Report Error

Page 10

• Correct Answer: C

  Solution :

  Since, \[P(A\cup B)=P(A)+P(B)-P(A)P(B)\] \[[\because \,P(A\cap B)=P(A)P(B)]\]                 \[\Rightarrow \]\[0.8=0.3+P(B)-0.3P(B)\]                 \[\Rightarrow \]\[0.7P(B)=0.5\Rightarrow P(B)=\frac{5}{7}\]

warning Report Error

Page 11

• Correct Answer: D

  Solution :

  \[\therefore \]Required probability \[=\frac{{{\,}^{4}}{{C}_{2}}+{{\,}^{5}}{{C}_{2}}}{{{\,}^{9}}{{C}_{2}}}\]                              \[=\frac{6+10}{36}=\frac{16}{36}\] \[=\frac{4}{9}\]

warning Report Error

Page 12

• Correct Answer: B

  Solution :

  \[\frac{1}{\cos {{290}^{o}}}+\frac{1}{\sqrt{3}\sin {{250}^{o}}}\] \[=\frac{1}{\cos {{70}^{o}}}-\frac{1}{\sqrt{3}\sin {{110}^{o}}}\] \[=\frac{\sqrt{3}\sin {{10}^{o}}-\cos {{70}^{o}}}{\sqrt{3}\sin {{110}^{o}}\cos {{70}^{o}}}\] \[=\frac{\sqrt{3}\sin ({{180}^{o}}-{{70}^{o}})-cos{{70}^{o}}}{\sqrt{3}\sin (180-{{70}^{o}})cos{{70}^{o}}}\] \[=\frac{\frac{\sqrt{3}}{2}\sin {{70}^{o}}-\frac{1}{2}\cos {{70}^{o}}}{\frac{\sqrt{3}}{2}\sin 70\cos {{70}^{o}}}\] \[\frac{\cos {{30}^{o}}\sin {{70}^{o}}-\sin {{30}^{o}}\cos {{70}^{o}}}{\frac{\sqrt{3}}{2}.\frac{1}{2}\sin {{140}^{o}}}\] \[=\frac{\sin ({{70}^{o}}-{{30}^{o}})}{\frac{\sqrt{3}}{4}\sin ({{180}^{o}}-{{40}^{o}})}\] \[=\frac{\sin {{40}^{o}}}{\frac{\sqrt{3}}{4}\sin {{40}^{o}}}=\frac{4}{\sqrt{3}}\]

warning Report Error

Page 13

• Correct Answer: A

  Solution :

  Equation of any line through (1, 2) is \[y-2=m(x-1)\]                 \[\Rightarrow \]               \[y-mx+m-2=0\] The distance of line \[y=2x\]from the point (3, 1) is greatest. \[\therefore \]Required line in \[y=2x.\]

warning Report Error

Page 14

• Correct Answer: B

  Solution :

  Let \[{{S}_{1}}={{x}^{2}}+{{y}^{2}}=9\] \[PA.PB={{(\sqrt{{{S}_{1}}})}^{2}}\] \[={{\left( \sqrt{{{(3)}^{3}}+{{(11)}^{2}}-9} \right)}^{2}}=121\]

warning Report Error

Page 15

• Correct Answer: C

  Solution :

  \[{{x}^{2}}+{{y}^{2}}+6x+6y-2=0\] Centre \[(-3,-3),\]radius \[=\sqrt{9+9+2}\]                 \[=\sqrt{20}\]
  
  Now, \[QC=\sqrt{{{(-3)}^{2}}+{{6}^{2}}}=\sqrt{45}\] In right \[\Delta CPQ\] \[PQ=\sqrt{45-20}=5\]

warning Report Error

Page 16

• Correct Answer: C

  Solution :

  Let sides of the triangle are \[4x,5x,6x.\] \[s=\frac{4x+5x+6x}{2}=\frac{15}{2}x\] \[\Delta =\sqrt{\frac{15}{2}x\left( \frac{15}{2}x-4x \right)\left( \frac{15}{2}x-5x \right)\left( \frac{15}{2}x-6x \right)}\] \[=\sqrt{\frac{15}{2}x\times \frac{7}{2}x\times \frac{5}{2}x\times \frac{3}{2}x}\] \[=\frac{15\sqrt{7}{{x}^{2}}}{4}\]                 Circumradius,  \[R=\frac{4x\times 5x\times 6x}{4\times \frac{15\sqrt{7}{{x}^{2}}}{4}}\]                                                 \[=\frac{8}{\sqrt{7}}x\]                 Inradius, \[r=\frac{\frac{15\sqrt{7}}{4}{{x}^{2}}}{\frac{15}{2}x}\]                                 \[=\frac{\sqrt{7}}{2}xax\]                                 \[\frac{R}{r}=\frac{\frac{8x}{\sqrt{7}}}{\frac{\sqrt{7x}}{2}}=\frac{16}{7}\]             

warning Report Error

Page 17

• Correct Answer: C

  Solution :

  \[4{{\sin }^{-1}}x+{{\cos }^{-1}}x=\pi \] \[\Rightarrow \]\[4{{\sin }^{-1}}x+\left( \frac{\pi }{2}-{{\sin }^{-1}}x \right)=\pi \] \[\Rightarrow \]\[3{{\sin }^{-1}}x=\pi -\frac{\pi }{2}=\frac{\pi }{2}\] \[\Rightarrow \]\[{{\sin }^{-1}}x=\frac{\pi }{6}\] \[\Rightarrow \]\[x=\sin \frac{\pi }{6}=\frac{1}{2}\]

warning Report Error

Page 18

• Correct Answer: A

  Solution :

  \[\cos \left( \frac{1}{2}{{\cos }^{-1}}\frac{1}{8} \right)=x\]            (let) \[\Rightarrow \]\[{{\cos }^{-1}}\frac{1}{8}=2{{\cos }^{-1}}x\] \[={{\cos }^{-1}}(2{{x}^{2}}-1)\] \[\Rightarrow \]\[\frac{1}{8}=2{{x}^{2}}-1\] \[\Rightarrow \]\[2{{x}^{2}}=\frac{1}{8}+1=\frac{9}{8}\] \[\Rightarrow \]\[{{x}^{2}}=\frac{9}{16}\] \[\Rightarrow \]\[x=\sqrt{\frac{9}{16}}=\frac{3}{4}\]       \[[\because \,0\le x\le 1]\]

warning Report Error

Page 19

• Correct Answer: D

  Solution :

  Given vectors are coplanar, if                 \[\left| \begin{matrix}    1 & 1 & m  \\    1 & 1 & m+1  \\    1 & -1 & m  \\ \end{matrix} \right|=0\] \[\Rightarrow \]\[\left| \begin{matrix}    0 & 0 & -1  \\    1 & 1 & m+1  \\    1 & -1 & m  \\ \end{matrix} \right|=0\]                              \[[{{R}_{1}}\to {{R}_{1}}-{{R}_{2}}]\] \[\Rightarrow \]\[-1(-1-1)=0\] \[\Rightarrow \]\[2\ne 0\] \[\therefore \]No value of m for which vectors are coplanar.

warning Report Error

Page 20

• Correct Answer: D

  Solution :

  Given, \[\vec{a}+\vec{b}+\vec{c}=0\] \[\Rightarrow \]\[{{(\vec{a}+\vec{b}+\vec{c})}^{2}}=0\] \[\Rightarrow \]\[|\vec{a}{{|}^{2}}+|\vec{b}{{|}^{2}}+|\vec{c}{{|}^{2}}\] \[+\,2(\vec{a}.\vec{b}+\vec{b}.\vec{c}+\vec{c}.\vec{a})=0\]                 \[\Rightarrow \]\[4+9+16+2(\vec{a}.\vec{b}+\vec{b}.\vec{c}+\vec{c}.\vec{a})=0\] \[\Rightarrow \]\[\vec{a}.\vec{b}+\vec{b}.\vec{c}+\vec{c}.\vec{a}=\frac{-29}{2}\]

warning Report Error

Page 21

• Correct Answer: C

  Solution :

  Here \[a=2,m=-1\] \[\therefore \]Required point is \[(a{{m}^{2}},-2am)=(2,4)\]

warning Report Error

Page 22

• Correct Answer: C

  Solution :

  Given curve is \[16{{x}^{2}}+25{{y}^{2}}=400\] \[\Rightarrow \]\[\frac{{{x}^{2}}}{25}+\frac{{{y}^{2}}}{16}=1\] Here \[a=5,b=4\] \[{{F}_{1}}\]and \[{{F}_{2}}\]are forcus. \[\therefore \]\[P{{F}_{1}}+P{{F}_{2}}=2a=10\]

warning Report Error

Page 23

• Correct Answer: C

  Solution :

  Let the hyperbola is \[\frac{{{x}^{2}}}{{{a}^{2}}}-\frac{{{y}^{2}}}{{{b}^{2}}}=1\] Eccentricity, \[e=\sqrt{\frac{{{a}^{2}}+{{b}^{2}}}{{{a}^{2}}}}\] \[\Rightarrow \]               \[\frac{1}{{{e}^{2}}}=\frac{{{a}^{2}}}{({{a}^{2}}+{{b}^{2}})}\] Conjugate hyperbola is \[\frac{{{x}^{2}}}{{{a}^{2}}}-\frac{{{y}^{2}}}{{{b}^{2}}}=-1\] Eccentricity, \[{{e}_{1}}=\sqrt{\frac{{{a}^{2}}+{{b}^{2}}}{{{b}^{2}}}}\] \[\Rightarrow \]               \[\frac{1}{e_{1}^{2}}=\frac{{{b}^{2}}}{{{a}^{2}}+{{b}^{2}}}\] Now, \[\frac{1}{{{e}^{2}}}+\frac{1}{e_{1}^{2}}=\frac{{{a}^{2}}}{{{a}^{2}}+{{b}^{2}}}+\frac{{{b}^{2}}}{{{a}^{2}}+{{b}^{2}}}=1\]

warning Report Error

Page 24

• Correct Answer: C

  Solution :

  \[\underset{x\to 0}{\mathop{\lim }}\,\frac{1-{{\cos }^{3}}x}{x\sin x\cos x}=\underset{x\to 0}{\mathop{\lim }}\,\frac{2(1-{{\cos }^{3}}x)}{x\sin 2x}\] [Using LHospital Rule] \[=\underset{x\to 0}{\mathop{\lim }}\,\frac{2[-3{{\cos }^{2}}x(-\sin x)]}{\sin 2x+x+\cos 2x.2}\] \[=\underset{x\to 0}{\mathop{\lim }}\,\frac{6{{\cos }^{2}}x\sin x}{\sin 2x+2x\cos 2x}\] \[=\underset{x\to 0}{\mathop{\lim }}\,\frac{6[-2\cos x{{\sin }^{2}}x+{{\cos }^{3}}x]}{2\cos 2x+2[-x\sin 2x.2+\cos 2x]}\] \[=\underset{x\to 0}{\mathop{\lim }}\,\frac{6[-2\cos x{{\sin }^{2}}x+{{\cos }^{3}}x]}{2\cos 2x-4x\sin 2x+2\cos 2x}\] \[=\frac{6}{2+2}\] \[=\frac{3}{2}\]

warning Report Error

Page 25

• Correct Answer: A

  Solution :

  \[y={{\cos }^{-1}}(\cos x)\] \[y=\frac{-1}{\sqrt{1-{{\cos }^{2}}x}}(-\sin x)\] \[=\frac{\sin x}{\sqrt{{{\sin }^{2}}x}}\] \[=1\forall x\]

warning Report Error

Page 26

• Correct Answer: B

  Solution :

  Position vector of median \[\overrightarrow{AD}=\frac{\overrightarrow{AB}+\overrightarrow{AC}}{2}\] \[=\frac{(-3+5)\hat{i}(0-2)\hat{j}+(4+4)\hat{k}}{2}\] \[=\hat{i}-\hat{j}+4\hat{k}\] \[\therefore \]\[|\overrightarrow{AD}|=\sqrt{{{1}^{2}}+{{(-1)}^{2}}+{{(4)}^{2}}}=\sqrt{18}\]

warning Report Error