CBSE Eamination Paer (All India Set) (Detailed Solutions) Mathematics Class th z z. We have, z On aling R R R, we get z z z z (/) Taking common ( z) from R common from R, we get ( z)( ) z ( z)( ) [ R R are identical] (/). Given equation is d d d d On integrating both sides w.r.t, we get d d d d d c d d c (/) Clearl, the highest order derivative occurring in the differential equation is d/ d. So, its order is. Also, it is a olnomial equation in d/ d highest ower raised to d/ d is, so its degree is. Hence, the sum of the order degree of the above differential equation is. (/). The given differential equation is ( ) ( cot ) d d The above equation can be rewritten as (cot ) d d cot d d cot ( ) d d d cot d (/) Which is a linear differential equation of the form d P Q, where P d cot Q. Pd Now, integrating factor e e Put t d dt dt d IF e t log t e t [ ] (/). We have, a, b c are mutuall erendicular unit vectors, i.e. a b b c c a (i) a b c Now, a b c ( a b c) ( a b c) ( a a) ( a b ) ( a c) ( b a) [ a a a ] ( b b ) ( b c) ( c a) ( c b ) ( c c) [ dot roduct is distributive over addition] (/) ( a ) ( ) ( ) ( ) b ( ) ( ) ( ) c [from Eq. (i) a b b a] ( ) 6 a b c 6 (/) [ length cannot be negative]. (i) Firstl, determine erendicular vectors of a b, i.e. a b. (ii) Further, determine erendicular unit vector b using formula a b a b We have, a i j k b i j As we know that, the vectors a b is erendicular to both the vectors, so let us first evaluate a b. i j k Thus, a b i ( ) j ( ) k ( ) i j (/) Now, the unit vector erendicular to both a b is given b
a b a b i j i j i j ( ) ( ) (/) 6. Given equations of lines are 7 z (i) To convert the equation in stard form z z a b c Let us divide Eq. (i) b LCM (coefficient of, z), i.e. LCM (,, ) Now, the Eq. (i) become 7 z 7 z 7 z 6 (/) On comaring the above equation with Eq. (ii), we get 6,, are the direction ratios of the given line. Now, the direction cosines of given line are 6,, 6 ( ) 6 ( ) 6 ( ) 6 i.e. 7, 7, 7. [ 6 9 9 7 ] (/) 7. Let A be the matri that reresent the number of attemts made in three villages X, Y Z. And B be the matri that reresent the cost for each made er attemt. Thus, the matrices A B are given b X A Y Z House calls Letters Announcements 7 Cost (`) House calls B Letters Announcements Now, the cost incurred b the organisation for three villages is given b Cost (`) X AB Y 7 Z X Y 7 Z X 6 X Y Y Z 8 6 Z 9 Thus, the total cost incurred b the organisation for the three villages X, Y Z searatel are `, ` ` 9, r e s e c t i v e l. As societ is conservative about the health of women, therefore this will force the eole to make toilets, which will hel to maintain hgenic condition in the societ. 8. (i) Firstl, use the formula tan tan tan, then simlif it get the values of. (ii) Further, we have to verif the given equation b obtained values of. 8 We have, tan ( ) tan ( ) tan (i) ( ) ( ) tan tan ( )( ) 8 tan tan tan 8 tan tan ( ) 8 6 6 8 8 6 6 8 [dividing b ] 8 ( 8) ( 8) ( 8)( ) 8 or / But 8, make LHS of Eq. (i) negative as tan ( 8 ) tan ( 8 ) tan ( 7) tan ( 9) tan 7 tan 9 (tan 7 tan 9) RHS Hence, the solution of the given equation is / Note Sometimes, it ma be ossible that the value of does not satisf the given equation. So, lease be careful for some etra values of. To rove z z cot cot cot z z LHS cot (,, z ) z cot cot z z z z z tan tan tan z z cot tan
(tan tan ) (tan tan z) (tan z tan ) (), z, z tan tan tan, if RHS Hence roved. a bc ac c 9. Let a ab b ac a b c ab b bc c On aling C a C, C b C C c C, we get a c a c abca b b a b b c c On aling C C ( C C ), we get a c abca b b b b b c b On aling C b C, we get a c ab ca b b b b c On aling C C bc C C bc, we get a c ab c aca c Aling C a C C c C, we get a b c On eing from R, we get a b c [ ( ) ( )] a b c Hence roved. Note When we multil an row (or column) b a constant k, then outside of the determinant divided b k.. We have, A Let A ij be the cofactor of the element a ij of A. Now, cofactors of A are A ( ) ( ) ( ) ( ) 6 ( ) ( ) 6 ( ) ( ) 6 ( ) ( ) A A A A ( ) ( ) 6 ( ) ( ) 6 ( ) ( ) 6 ( ) ( ) A A A A Now, the adjoint of the matri A is given b A A A 6 6 adj A A A A 6 6 A A A 6 6 Now, A ( ) ( ) ( ) ( ) ( 6) ( 6) 7 6 6 A( adj A) 6 6 6 6 6 6 6 6 6 6 6 6 6 6 6 6 7 7 7 7 I A I Hence roved. 7. We have, f( ), R The given function can be rewritten as ( ) ( ), f( ) ( ) ( ), ( ) ( ),,,, Differentiabilit at f h f L f ( ) lim ( ) ( ) h h ( h) lim h h
h lim lim ( ) h h h f h f R f ( ) lim ( ) ( ) h h lim lim h h h h L f ( ) R f ( ) (/) f is not differentiable at. (/) Differentiabilit at, f h f L f ( ) lim ( ) ( ) h h lim lim h h h h (/) f h f R f ( ) lim ( ) ( ) lim ( h ) h h h h h h lim lim h h h h (/) L f ( ) R f ( ) f is not differentiable at. Hence, f is not differentiable at. d. To show ( ) d m d d We have, e m sin (i) On differentiating both sides of Eq. (i) w.r.t., we get d msin d e d d m ( sin ) e msin m d m [ e m sin ] d Now, on squaring both sides, we get d ( ) m d On differentiating both sides of Eq. (ii) w.r.t, we get d d ( ) ( ) d m d d d d d d ( ) d m d d dividing on both sides b d d d ( ) d m Hence roved. d d, g( ) h( ) On differentiating above functions w.r.t., we get f ( ). We have, f( ) ( ) ( ) g( ) ( ) d v du u dv u d d d v v g( ) ( ) g( ) ( ) h( ) Now, f [ h{ g( )}] f h ( ) f () [ h( ) ]. (i) Firstl, use the method of integral f ( ) ( q) a b c d, i.e. ( q) A d ( a b c) B d Simlif get the values of A B. (ii) Simlif the integr use the formula a a d a sin Let I ( ) d C a (i) Put ( ) A d ( d ) B A( ) B ( A B) A On comaring the coefficient of constant term, we get A A A B B Eq. (ii) becomes ( ) ( ) Put ( ) in Eq. (i), we get I ( ) d d I I [sa] (iii) Now consider, I ( ) d Put t ( ) d dt I t dt ( t ) / C ( ) / C (iv)
I d ( ) d d 9 d d sin C a d a a C sin a ( ) 9 I sin C (v) From Eqs. (iii), (iv) (v), we get I / ( ) ( ) 9 sin C where, C C C Or (i) Firstl, use the artial fraction in the given integr, A B C i.e. ( )( ) Simlif it get the values of constants A, B C. (ii) Further, given integr becomes in simlest form then integrate it to get the result. Let I d ( )( ) Using artial fraction method. A B C Consider (i) ( )( ) A( ) ( B C)( ) ( A B) ( B C) ( A C) On comaring the coefficients of, constant terms both sides, we get A B B C (iii) A C (iv) On substituting the value of B from Eq. (ii) in Eq. (iii), we get ( A) C A C A C (v) Now, on solving Eqs. (iv) (v), we get C A Now, from Eq. (ii), we get B Thus, from Eq. (i), we have ( ) ( )( ) ( ) d ( ) I d d d I d log log tan C d tan C a a a f ( ) f d log ( ) f ( ) C /. Let I cos / d sin d cos ( sin cos ) / d / / cos cos sin / d 7 cos / sin / / sec 7 cos sin / [sin sin cos ] [on dividing numerator denominator b cos ] / sec / / cos sin d / Put tan d sec ( tan ) tan / d [ sec tan ] t sec d dt Lower limit, when, then t Uer limit, when, then t I t / t dt / / ( t t ) dt / / t t
/ / ( ) ( ) 6. Let I log d ( ) log d ( ) I 6 On aling integration b arts, we get d d d I log (log ) ( ) d ( ) II d d u v d u v d d u v d d I II ( ) ( ) log d ( ) d ( ) ( ) log I [sa] (i) d Where, I ( ) Now, using artial fraction method consider, A B ( ) A( ) B On utting, we get A on utting, we get B I d d d log log C Now, from Eqs. (i) (ii), we get I log log log C log log C m log m log n log n 7. We have, a i j k, b i j c i j k Now, a b ( i j k ) ( i j ) i j k c b ( i j k ) ( i j ) i j k Now, a vector erendicular to ( a b) ( c b) i j k is given b ( a b) ( c b) i ( ) j ( ) k ( ) i ( ) j( ) k ( ) j k unit vector along ( a b) ( c b) is given b j k j k j k j k ( ) j k j k 8. Let the equation of line assing through (,, ) is r ( i j k ) ( b i b j b k ) (i) Since, the line (i) is erendicular to the given lines r ( 8 i 9 j k ) ( i 6 j 7 k ) r ( i 9 j k ) ( i 8 j k ) Therefore, we have b 6b 7b b 8b b (iii) If two lines r a b r a b are erendicular, then b b. Now, on solving Eqs. (ii) (iii), we get b b b 8 6 8 b b b 6 7 b b b 6 7 b b b 6 [multiling b ] Hence, the required equation of line is r ( i j k ) ( i j k 6 ) Or Let the equation of lane assing through the oint (,, ) is A( ) B( ) C( z ), (i) where, A, B C are direction ratios of normal to the lane. (/) Since, the lane asses through (,, ), therefore we have A( ) B( ) c( ) A B C (/) Also, the lane arallel to the line z z or Therefore, we have A B C (iii) [ the normal to the lane will be erendicular to the line, i.e. a a b b cc ] On solving Eqs. (ii) (iii), we get A B C
A B C A B C (sa) On substituting the values of A, B C in Eq. (i), we get the required equation of lane is ( ) ( ) z z z 9. Let X be a rom variable that denote the number of sades in a draw of three cards. So, X can takes values,,,. Now, P( X ) P (no sade) 9 9 9 7 6 [ cards are drawn with relacement] P( X ) P (one sade two non-sade) P (I st card is sade, IInd IIIrd cards are non-sade) P (Ist card is non-sade, IInd card is sade III is non-sade) P (Ist IInd cards are non-sade IIIrd card is sade) 9 9 9 9 9 9 7 6 P ( X ) P (two sade one non-sade) 9 9 9 9 6 P( X ) P(all are sade cards) 6 Thus, the required robabilit distribution is X P( X) 7 6 7 6 9 6 6 Now, mean of above distribution is given b E( X) P( X ) P( X ) P( X ) P( X ) P( X ) 7 7 9 6 6 6 6 7 8 8 6 6 Or Let Probabilit of success, q Probabilit of failure X be a rom variable that denote the number of success in 6 trials. As, X follows binomial distribution with arameters, n 6. 6 6 P( X ) C q where,,,,,, 6 q Also, given, 9P( X ) P( X ) 6 6 9 ( C q ) C q 9 q [ 6 6 C C ] 9 ( ) [ q ] 9 8 8 ( ) ( ) ( )( ) or Thus, [robabilit cannot be negative]. (i) Firstl, rove that f is one-one onto. (ii) Further, consider given function as convert the given function in terms of, i.e. f ( ) is evaluated. We have a maing f : R [ 9, ) given b f( ) 6 9 To rove f is invertible with f ( ). For f is one-one Let, R be an arbitrar elements such that f( ) f( ) 6 9 6 9 ( ) 6( ) ( )( ) 6 ( ) ( )[ ( ) 6] or 6 [ 6 as, R ],, R f is one-one For f is onto Let [ 9, ) be an arbitrar element f( ) Then, 6 9 6 9 6 6 ( 9 ) b b ac a 6 6 6
or R, therefore Now, R 9 Minimum value of is. Min 9 Thus, for each [ 9, ) their eist R such that f( ) f is onto f is invertible inverse of f is given b f :[ 9, ) R defined as f ( ) Or We have, * is a binar oeration defined on the set X R { } given b *,, X (i) Commutative Let, X be an arbitrar elements. To verif, * * Consider, * [ usuall addition multilication is commutative *, X are arbitrar. on the set of real numbers] * is commutative (ii) Associative Let,, X be an arbitrar elements Consider, * ( * ) * ( ) ( ) [ usuall multilication is distributive over usual addition on set of real numbers] Now consider, ( * )* ( )* ( ) [ usuall multilication is distributive over addition on the set of real numbers] [ usuall addition multilication is commutative on the set of real numbers] From Eqs.(i) (ii), we get * ( * ) ( * )*, ad are arbitrar, * is associative. Eistence of identit element Now, let e Xbe an identit element of X, i.e. * e e *, X Consider, * e e e e e e( ) e [ ] Similarl, e * e Thus, e X is an identit element for X. Eistence of inverse Let X be an arbitrar element. Now, if ossible, let X be the inverse of, i.e. * * Now, * ( ) X not true Thus, for each X, X such that * * Hence, inverse of X is.. Given equations of curves are 9 ( 9 ) (i) ( ) As these curves cut each other at right angles, therefore their tangent at oint of intersection are erendicular to each other. So, let us first find the oint of intersection sloe of tangents to the curves. From Eqs. (i) (ii), we get 9 ( 9 ) ( ) 9 ( 9 ) [, as if, then curves becomes straight, which will be arallel] 8 9 8 8 On substituting the value of in Eq.(i),we get 9 Thus, the oint of intersection are (, 8) (, 8). Now, consider Eq.(i),we get
9 9 9 9 On differentiating w.r.t, we get d d 9 From Eq. (ii), we get On differentiating w.r.t, we get d d (iii) (iv) Now, for intersection oint (, 8), we have sloe of tangent to first curve 6 9 [using Eq. (iii)] sloe of tangent to second curve 6 [using Eq. (iv)] Tangents are erendicular to each other.then, 6 6 [ m m ] 9 for intersection oint (, 8), we have sloe of tangent to first curve 6 [using Eq. (iii)] 9 sloe of tangent to second curve 6 [using Eq. (iv)] Tangents are erendicular to each other. Then, 6 6 [ m m ] 9 Hence, the value of is.. Given equations of curves the lines are ; ; (i) Let us first find the oint of intersection of curves (i) (ii), we get 6 6 6 ( 6) or [consider onl real values] Now, if, then from Eq.(ii), if, then from Eq. (ii), Thus, the oint of intersection of curves are (, ) (, ). Now, these curves lines can be reresent grahicall as. X Now, firstl find area of region (II) ( ) d d () d [ ] 6 8 6 6 sq units (i) Area of region (I) Area of region OABO Area of region (II) 6 6 d 8 6 6 sq units area of region (III) Area of square OABC Area of region (I) Area of region (II) 6 6 ( ) 8 6 sq units (iii) From Eqs. (i), (ii) (iii), we get Area of region (I) Area of region (II) Area of region (III). Hence roved. (i) Firstl, consider d as equal to F(, ). Then, d relace b b on both sides, we get F(, ) F(, ) (ii) Put v convert the given equation in terms of v. Y = C. B = (, ) III II I A... X O Y = = (iii) Searate the variables integrate it. Further ut v simlif it to get the required result. Given differential equation is d (i) d
Let F(, ) Now, on relacing b b, we get F(, ) F(, ) ( ) Thus, F(, ) is a homogeneous function of degree zero hence, the given differential equation is a homogeneous differential equation. Now, to solve it, ut v d v dv d d From Eq. (i), we get v dv v v d v v dv v v v v v d v v dv v d v v d dv v On integrating both sides, we get v dv d v log v log C log log C ut v log log log C m log log m log n n log C which is the required solution. Or (i) Firstl, convert the given differential equation in the form of d P Q d (ii) Determine integrating factor b using formula IF e Pd (iii) Further, find the solution b using formula IF IF Q d C. Simlif it get the required solution. Given, differential equation is (tan ) d ( ) d This equation can be rewritten as d d d d tan tan Which is a linear differential equation of the form (/) where, d d P Q P Now, Integrating factor tan Q Pd e e d tan e the solution of given differential equation is given b tan tan tan e e d C I C (i) (/) tan tan Now, consider I e d Put tan t d dt I e t dt t t t e II I d u v d u vd d u v d d I II [ ] t t e t t e e C e t ( t ) C tan e (tan ) C [ut t tan ] From Eq. (i), we get tan tan e e (tan ) C C tan tan e e (tan ) C It is given that, when. Therefore, we have e e ( ) C [ut C C C ] C C Hence, the required solution is tan tan e e (tan ). The direction ratios of line joining A(,, ) B(,, ) are [( ),( ), ( )], i.e. (,, 6). Now, the equation of line assing through (,, ) having DR (,, 6) is given b z z z 6 a b c Now, let z 6 (sa), z 6 Thus, the general oint on the line is given b (,, 6 ). Since, line intersect the lane z 7. So, general oint on the line (,, 6 ) satisf the equation of lane. ( ) 6 7
6 6 7 So, the oint of intersection of line lane is (,, 6 ), i.e.(,, 7). Now, distance between (,, ) (,, 7) is given b ( ) ( ) ( 7) 6 9 9 7 units. Let factor (I) run das factor (II) run das. Then, Total cost (in `) Then, Minimize Z Subject to the constraints 6 or 6 (i) or 8 or 8 (iii) (iv) The table for line 6 is X 8 6 8 6 O Y Y D(, ) C(, ) 6 8 6 8 + = B(8, 6) + =6 A(6, ) X + =8 + <86 8 6 So, the line asses through (8, ) (, 6) On utting (, ) in the inequalit 6, we get 6 (which is false) So, the half lane is awa from the origin. The table for line is 8 So, the line ases through (8, ) (, ). On utting (, ) in the inequalit, we get (which is false) So, the half lane is awa from the origin. The table for line 8 is 6 On utting (, ) is the inequalit 8, we get 8 (which is false) So, the half lane is awa from the origin. Since,, so the feasible region lies in the first quadrant. The oint of intersection of lines (i) (iii) is B (8, 6) lines (i) (ii) is (, ) The above Eqs. (i), (ii), (iii) (iv) can be reresented grahicall as Clearl, fessible region is ABCD (shaded region), where corner oints are A( 6, ), B ( 8, 6), (, ) D (, ). The value of Z at corner oints are Corner Points Z A( 6, ) Z 6 9 B( 8, 6) Z 8 6 86 Minimum value C(, ) Z 8 D(, ) Z In the table, we find that minimum value of Z is 86 occur at the oint (8, 6). But we can t sa that it is a minimum value of Z as region is unbounded. Therefore, we have to draw the grah of the inequalit 86 or 86 i.e. To check whether the resulting oen half lane has an oint common with feasible region. From figure, we see that it has no oint in common. Thus, the minimum value of Z is ` 86 attained at the oint (8, 6). Hence, factor (I) should run for 8 das factor (II) should run for 6 das. 6. Let E be the event that the selected bolt is manufactured b machine A, E be the event that the selected bolt is manufactured b machine B,
E be the event that the selected bolt is manufactured b machine C, E be the event that the selected bolt is defective. Then, we have P( E) % P( E) % P( E) % Also, given that %, % % bolts manufactured b machines A, B C resectivel are defective. So, P E % E P E % E P E % E Now, firstl find the robabilit that selected bolt which is defective is manufactured b machine B is So, P E P E P E ( ) E E P E P E ( ) P( E ) P E P E P E ( ) E E E 9 The robabilit that selected bolt which is defective is not manufactured b machine B is P E P E A E