# CBSE Sample Papers For Class 12 Maths Mock Paper 1

Q1. If A = diag(1, –2, 4) and B = diag(6, 2, -1), then the value of 3A + 5B is

Opt:

diag(7, 0, 3).

diag(7, 4, 5).

diag(27, 16, –17).

diag(33, 4, 7).

Ans:

diag(33, 4, 7).

Q2. If A is a skew-symmetric matrix of odd order, then |A| is equal to

Opt:

A’.

1.

0.

-A.

Ans:

0.

Q3.

$The value of sincot−1tan cos−110 is$

Opt:

-10.

–102.

10.

102.

Ans:

10.

Q4.

$Let f : R → R be​ a function given by fx = x2 + 1, then thevalue of f−1 82 is$

Opt:

{-3, 3}.

{3, 9}.

{-3, -9}.

{-9, 9}.

Ans:

{-9, 9}.

Q5.

Opt:

Ans:

Q6.

$Let f(x)=x+1cotx be continuous at x=0, then f(0) will beequal to$

Opt:

1/e.

e.

e2.

e3.

Ans:

e.

Q7.

Opt:

Ans:

Q8. A fair die is thrown twenty times. The probability that on
the tenth throw the fourth six appears is”
Opt:

Ans:

Q9. The value of objective function is minimum under linear constraints at
Opt: any point outside the feasible region. the centre of the feasible region. any vertex of the feasible region. the vertex which is minimum distance from (0, 0).
Ans: any vertex of the feasible region.

Q10.

$The value of sin−12x1+x2−2cos−11−x21+x2+3tan−12x1−x2 is equal to$

Opt:

$1$ $13$ $13$ $123$

Ans:

$13$

Q11.

Three persons A, B and C are to speak at a function
along with five others. If they all speak in random order,
the probability that A speaks before B and B speaks before C is
Opt:

Ans:

Q12.

$The value of x if3sin−12x1+x2−4cos−11−x21+x2+2tan−12x1−x2 = π3$

Opt:

$1$ $13$ $13$ $123$

Ans:

$13$

Q13.

$If fx = 12x3xx−1xxx−1xx−1x−2x+1xx+1xx2−1, then f100is equal to$

Opt:

100.

1.

0.

–100.

Ans:

0.

Q14. If f(x) = |7x + 2| and g(x) = f(f(x)), then for x > 20, g’(x) equals

Opt:

-1.

0.

1.

7.

Ans:

7.

Q15.

$If the function fx=x3+x2−16x+20(x−2)2, if x≠2k , if x=2is continuous for all x, then k is equal to$

Opt:

0.

5.

7.

9.

Ans:

7.

Q16.

Given the relation R = {(1, 2), (2, 3)} on the set A = {1, 2, 3}, the minimum number of ordered-pairs, which, when added to R make it an equivalence relation, is

Opt:

5.

6.

7.

8.

Ans:

7.

Q17.

Opt:

Ans:

Q18.

$Show that sin−12x1−x2 = 2sin−1x$

Opt:

$sin−12x1−x2= 2sin−1xlet x = sinθsin−12x1−x2= sin−12 sinθ 1−sin2θ = sin−12 sinθ cosθ = sin−1 sin2θ = 2θ = 2sin−1x$

Ans:

$sin−12x1−x2= 2sin−1xlet x = sinθsin−12x1−x2= sin−12 sinθ 1−sin2θ = sin−12 sinθ cosθ = sin−1 sin2θ = 2θ = 2sin−1x$

Q19.

$Show that the functionf: →R given by f(x) = 1 if x > 00 if x = 0−1 if x < 0is neither one−one nor onto.$

Ans:

(a) f(1) = f(2) = 1

1 and 2 have the same images.

⇒ f(x1) = f(x2) = 1 for x > 0
But x1x2

Similarly -1 and -2 have the same images.

Hence f is not one-one.

(b) Except -1, 0, 1 no other element of co-domain of f has any pre-image in its domain.

f is not onto.

Q20. Find two numbers whose sum is 24 and whose product is as large as possible.

Ans:

Let one number be x. Then, the other number is (24 − x).

Let P(x) denote the product of the two numbers. Thus,

P(x) = x(24 – x) = 24 – x2

∴ P'(x) = 24 – 2x

P”(x) = -2

Now, P'(x) = 0 ⇒ x = 12

Also, P”(12)= -2 < 0

∴ By second derivative test, x = 12 is the point of local maxima of P. Hence, the product of the numbers is the maximum when the numbers are 12 and 24 − 12 = 12.

Q21.

$Find a→ and b→, if a→ − b→. a→ + b→=27 and a→= 2b→.$

Ans:

$a→ − b→. a→ + b→=27⇒ a→2− b→2 = 27⇒ 4b→2− b→2 = 27⇒ 3 b→2 = 27⇒ b→2 = 9⇒ b→ = 3Now, a→ = 2b→⇒ a→ = ​6Thus, a→ = ​6 and b→ = 3$

Q22.

$Find a vector in the direction of vector 2 i^ − j^ +3 k^which has magnitude 7 units.$

Ans:

$The unit vector of 2i^ − j^ +3k^ = 2i^ − j^ +3k^2i^ − j^ +3k^ = 2i^ − j^ +3k^22 − −12 +32 = 2i^ − j^ +3k^4 + 1 +9 = 2i^ − j^ +3k^14Thus, the vector in the direction of 2i^ − j^ +3k^ which hasmagnitude 7 units is = 7 × 2i^ − j^ +3k^14 = 14i^ − 7j^ + 21k^14 = 1414i^ − 714j^ + 2114k^$

Q23. Form the differential equation of the family of parabolas having vertex at originand axis along positive x-axis.

Ans:

The equation of parabolas having vertex at origin
and axisalong positive y-axis is as follows
y2 = 4ax …(i)
Differentiating w.r.t. x, we get
2yy’ = 4a

Putting value of 4a in equation (i), we get
y2 = 2yy’ x

y = 2y’ x

or2y’x – y =0

Thus, the required differential equation is
2y’x – y =0.

Q24. Given that E and F are events such that P(E) = 0.5, P(F) = 0.4 and P(E ∩ F) = 0.1, find P (E | F) and f (F | E).

Ans:

$Since, P(E | F)=P(E ∩ F)P(F)∴ P(E | F)=0.10.4 =14And P(F | E)=P(E ∩ F)P(E) =0.10.5 =15$

Q25. Find the rate of change of the area of a circle with respect to its radius r when r = 5cm.

Ans:

The area A of a circle with radius r is given by A =

$π r2$

.

Therefore, the rate of change of the area A with respect to its radius r is given by

dA/dr = 2

$π$

r = 10

$π$

(Since, r = 5 cm given)

Thus, the area of the circle is changing at the rate of 10

$π$

cm2/s.

Q26.

$If the line r→ = i^ − 2 j^ + k^ + λ 2i^ + j^ + 2 k^ and the plane r→. 3 i^ −2 j^ +m k^ = 14.Find the value of m.$

Ans:

$The given line is parallel to the vector b→ = 2i^ + j^ + 2 k^and the given vector is normal to the vector n→ = 3 i^ −2 j^ +m k^If the line is parallel to th the plane then normal is perpendicular to the line.∴b→⊥n→⇒b→ . n→ =0⇒2i^ + j^ + 2 k^ .3 i^ −2 j^ +m k^=0⇒ 6−2+2m=0⇒ m=−2$

Q27. Show that for any square matrix A, A + A’ is a symmetric matrix and A – A’ is a skew symmetric matrix.

Ans:

Let B = A + A’
B’ = (A + A’)’
= A’ + (A’)’ [as (A + B)’ = A’ + B’]
= A’ + A [ ∵ (A’)’ = A]
= B
∴ B = A + A’ is a symmetric matrix.

Now let C = A – A’
C’ = (A – A’)’
= A’ – (A’)’
= A’ – A
= -(A – A’)
∴ C = A – A’ is a skew symmetric matrix.

Q28.

$If A = 1 0−1 7 and I = 1 00 1, then find k so that A2=8A+kI.$

Ans:

$A2= AA = 1 0−1 7 1 0−1 7 = 1+0 0 + 0−1−7 0+49 = 1 0−8 498A+ KI = 8 1 0−1 7 +K1 00 1 = 8+k 0−8 56+k Given A2=8A+kI⇒ 1 0−8 49 = 8+k 0−8 56+k⇒ 8+k=1 and 56+k = 49⇒ k=−7$

Q29. A firm deals with two kinds of fruit juices – pineapple and orange juice. These are mixed and two mixtures are sold as soft drinks A and B. One tin of A requires 4 litres of pineapple and 1 litre of orange juice. One tin of B requires 2 litres of pineapple and 3 litres of orange juice. The firm has only 46 litres of pineapple juice and 24 litres of orange juice. Each tin of A and B is sold at a profit of Rs. 4 and Rs. 3 respectively. How many tins of each type should the firm produce to maximise the profit? Solve the problem graphically.

Ans:

Let the number of tins of A be x and that of B be y.
According to the given conditions, the L.P.P. is
x ≥ 0, y ≥ 0 … (1)
4x + 2y ≤ 46 ⇒ 2x + y ≤ 23 … (2)
x + 3y ≤ 24 … (3)
and to maximise Z = 4x + 3y.
Plotting the inequations (i), (ii) and (iii) on the graph,

We get the shaded portion as the optimum solution. Possible points for maximum profit are A(23/2,0), B(9,5) and C(0,8)

$PointsZ=4x+3yValueA232, 046+046B(9, 5)36+1551 [maximum]C(0, 8)0+2424$

We notice profit is maximum at B (9,5), i.e., x = 9, y = 5.
Hence, 9 tins of A and 5 tins of B should be produced.

Q30.

$If a, b, care in A.P then without expanding show thatx+2 x+3 x+2ax+3 x+4 x+2bx+4 x+5 x+2c = 0$

Ans:

$Here, Δ = x+2 x+3 x+2ax+3 x+4 x+2bx+4 x+5 x+2c = x+2 x+3 x+2ax+3 x+4 x+2b 1 1 2c−b by applying R3→R3−R2 = x+2 x+3 x+2a 1 1 2 b−a 1 1 2c−b by applying R2→R2−R1 Now because a, b, c are in A.P., we haveb−a = c −bHence in Δ R2= R3If any two rows or columns of a determinant are identicalthen the value of determinant is zero.⇒ Δ = 0$

Q31.

$Using properties of determinants, show that 1+a 1 11 1+b 11 1 1+c = abc 1+1a+1b+1c = abc +bc + ca + ab.$

Ans:

$Let Δ = 1+a 1 11 1+b 11 1 1+cBy applying c1→c2 and c3→c3−c2, we geta 1 0−b 1+b −b0 1 cExpanding along R1, we obtain Δ = a 1+b −b1 c−1 −b −b0 c+ 0 = a c+bc+b−−bc = ac+abc+ab+bc = abc+bc+ca+ab = abc a+1a+1b+1c$

Q32. Determine whether the following relation is reflexive, symmetric and transitive: Relation R on the set N of all natural numbers is defined as
R = {(x, y): y = x + 5 and x < 4}

Ans:

R = {(x, y): y = x + 5 and x < 4}
R = {(1, 6), (2, 7), (3, 8)}

Reflexivity: (1, 1), (2, 2) etc. are not in R. So, R is not reflexive.

Symmetry: (1, 6) ∈ R but (6, 1) ∉ R. So, R is not symmetric.

Transitivity: Since (1, 6) ∈ R and there is no ordered pair in R which has 6 as the first element. Same is the case for (2, 7) and (3, 8). So, R is not transitive.

Q33. Show that the Relation R on the set A = {x ∈ Z : 0 ≤ x ≤ 12}, given by R = {(a, b) : |a – b| is a multiple of 4} is an equivalence relation.

Ans:

R = {(a, b) : |a – b| is a multiple of 4}, where a, b ∈ A ={x Z : 0 ≤ x ≤ 12} = {0, 1, 2, …, 12}.

Reflexivity: For any a ∈ A, |a – a| = 0, which is a multiple of 4.

⇒ (a, a) ∈ R, for all a ∈ A.

So, R is reflexive.

Symmetry: Let (a, b) ∈ R. Then,

(a, b) ∈ R

⇒ |a – b| is a multiple of 4
⇒ |a – b| = 4k for some k ∈ N
⇒ |b – a| = 4k for some k ∈ N
⇒ (b, a) ∈ R

So, R is symmetric.

Transitive: Let (a, b) ∈ R and (b, c) ∈ R. Then,
⇒ |a – b| is a multiple of 4 and |b – c| is a multiple of 4
⇒ |a – b| = 4k and |b – c| = 4m for some k, m ∈ N
⇒ a – b = ± 4k and b – c = ± 4m
⇒ a – c = ± 4k ± 4m
⇒ a – c is a multiple of 4
⇒ |a – c| is a multiple of 4
⇒ (a, c) ∈ R

So, R is transitive.

Hence, R is an equivalence relation.

Q34.

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Ans:

The area of the smaller region bounded by the hyperbola and line is represented by shaded region ABCA.

$∴Area( ABCA )=Area( OBCAO )−Area( OBAO ) = ∫ 0 3 2 1− x 2 9 dx − ∫ 0 3 2( 1− x 3 )dx = 2 3 ∫ 0 3 9− x 2 dx − 2 3 ∫ 0 3 ( 3−x )dx = 2 3 [ x 2 9− x 2 + 9 2 sin −1 ( x 3 ) ] 0 3 − 2 3 [ 3x− x 2 2 ] 0 3 = 2 3 { 3 2 9− 3 2 + 9 2 sin −1 ( 3 3 )− 0 2 9− 0 2 − 9 2 sin −1 ( 0 3 ) } − 2 3 { 3( 3 )− 3 2 2 −3( 0 )+ 0 2 2 } = 2 3 { 9 2 × π 2 }− 2 3 { 9− 9 2 } = 3π 2 −3= 3 2 ( π−2 ) square units MathType@MTEF@5@5@+= feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbwvMCKf MBHbqedmvETj2BSbqefm0B1jxALjhiov2DaerbuLwBLnhiov2DGi1B TfMBaebbnrfifHhDYfgasaacH8Mrpy0xbbf9q8WrFfeuY=Hhbbf9v8 qqaqFr0xc9pk0xbba9q8WqFfea0=yr0RYxir=Jbba9q8aq0=yq=He9 q8qqQ8frFve9Fve9Ff0dmeaabaqaaiaacaGaaeqabaWaaqaafaaakq aabeqaaiabgsJiCjacyc4GbbGaiGjGdkhacGaMaoyzaiacyc4GHbWa iGjGbmaabGaMakacyc4GbbGaiGjGdkeacGaMao4qaiacyc4GbbaacG aMaAjkaiacycOLPaaacqGH9aqpcaWGbbGaamOCaiaadwgacaWGHbWa aeWaaeaacaWGpbGaamOqaiaadoeacaWGbbGaam4taaGaayjkaiaawM caaiabgkHiTiaadgeacaWGYbGaamyzaiaadggadaqadaqaaiaad+ea caWGcbGaamyqaiaad+eaaiaawIcacaGLPaaaaeaacaWLjaGaaCzcai aaxMaacaaMc8UaaGPaVlabg2da9maapedabaGaaGOmamaakaaabaGa aGymaiabgkHiTmaalaaabaGaamiEamaaCaaaleqabaGaaGOmaaaaaO qaaiaaiMdaaaaaleqaaOGaamizaiaadIhaaSqaaiaaicdaaeaacaaI ZaaaniabgUIiYdGccqGHsisldaWdXaqaaiaaikdadaqadaqaaiaaig dacqGHsisldaWcaaqaaiaadIhaaeaacaaIZaaaaaGaayjkaiaawMca aiaadsgacaWG4baaleaacaaIWaaabaGaaG4maaqdcqGHRiI8aaGcba GaaCzcaiaaxMaacaWLjaGaaGPaVlaaykW7cqGH9aqpdaWcaaqaaiaa ikdaaeaacaaIZaaaamaapedabaWaaOaaaeaacaaI5aGaeyOeI0Iaam iEamaaCaaaleqabaGaaGOmaaaaaeqaaOGaamizaiaadIhaaSqaaiaa icdaaeaacaaIZaaaniabgUIiYdGccqGHsisldaWcaaqaaiaaikdaae aacaaIZaaaamaapedabaWaaeWaaeaacaaIZaGaeyOeI0IaamiEaaGa ayjkaiaawMcaaiaadsgacaWG4baaleaacaaIWaaabaGaaG4maaqdcq GHRiI8aaGcbaGaaCzcaiaaxMaacaWLjaGaaGPaVlaaykW7cqGH9aqp daWcaaqaaiaaikdaaeaacaaIZaaaamaadmaabaWaaSaaaeaacaWG4b aabaGaaGOmaaaadaGcaaqaaiaaiMdacqGHsislcaWG4bWaaWbaaSqa beaacaaIYaaaaaqabaGccqGHRaWkdaWcaaqaaiaaiMdaaeaacaaIYa aaaiGacohacaGGPbGaaiOBamaaCaaaleqabaGaeyOeI0IaaGymaaaa kmaabmaabaWaaSaaaeaacaWG4baabaGaaG4maaaaaiaawIcacaGLPa aaaiaawUfacaGLDbaadaqhaaWcbaGaaGimaaqaaiaaiodaaaGccqGH sisldaWcaaqaaiaaikdaaeaacaaIZaaaamaadmaabaGaaG4maiaadI hacqGHsisldaWcaaqaaiaadIhadaahaaWcbeqaaiaaikdaaaaakeaa caaIYaaaaaGaay5waiaaw2faamaaDaaaleaacaaIWaaabaGaaG4maa aaaOqaaiaaxMaacaWLjaGaaCzcaiaaykW7caaMc8Uaeyypa0ZaaSaa aeaacaaIYaaabaGaaG4maaaadaGadaqaamaalaaabaGaaG4maaqaai aaikdaaaWaaOaaaeaacaaI5aGaeyOeI0IaaG4mamaaCaaaleqabaGa aGOmaaaaaeqaaOGaey4kaSYaaSaaaeaacaaI5aaabaGaaGOmaaaaci GGZbGaaiyAaiaac6gadaahaaWcbeqaaiabgkHiTiaaigdaaaGcdaqa daqaamaalaaabaGaaG4maaqaaiaaiodaaaaacaGLOaGaayzkaaGaey OeI0YaaSaaaeaacaaIWaaabaGaaGOmaaaadaGcaaqaaiaaiMdacqGH sislcaaIWaWaaWbaaSqabeaacaaIYaaaaaqabaGccqGHsisldaWcaa qaaiaaiMdaaeaacaaIYaaaaiGacohacaGGPbGaaiOBamaaCaaaleqa baGaeyOeI0IaaGymaaaakmaabmaabaWaaSaaaeaacaaIWaaabaGaaG 4maaaaaiaawIcacaGLPaaaaiaawUhacaGL9baaaeaacaWLjaGaaCzc aiaaxMaacaWLjaGaeyOeI0YaaSaaaeaacaaIYaaabaGaaG4maaaada GadaqaaiaaiodadaqadaqaaiaaiodaaiaawIcacaGLPaaacqGHsisl daWcaaqaaiaaiodadaahaaWcbeqaaiaaikdaaaaakeaacaaIYaaaai abgkHiTiaaiodadaqadaqaaiaaicdaaiaawIcacaGLPaaacqGHRaWk daWcaaqaaiaaicdadaahaaWcbeqaaiaaikdaaaaakeaacaaIYaaaaa Gaay5Eaiaaw2haaaqaaiaaxMaacaWLjaGaaCzcaiaaykW7caaMc8Ua eyypa0ZaaSaaaeaacaaIYaaabaGaaG4maaaadaGadaqaamaalaaaba GaaGyoaaqaaiaaikdaaaGaey41aq7aaSaaaeaaiiaacqWFapaCaeaa caaIYaaaaaGaay5Eaiaaw2haaiabgkHiTmaalaaabaGaaGOmaaqaai aaiodaaaWaaiWaaeaacaaI5aGaeyOeI0YaaSaaaeaacaaI5aaabaGa aGOmaaaaaiaawUhacaGL9baaaeaacaWLjaGaaCzcaiaaxMaacaaMc8 UaaGPaVlabg2da9maalaaabaGaaG4maiab=b8aWbqaaiaaikdaaaGa eyOeI0IaaG4maiabg2da9maalaaabaGaaG4maaqaaiaaikdaaaWaae WaaeaacqWFapaCcqGHsislcaaIYaaacaGLOaGaayzkaaGaaGPaVlaa dohacaWGXbGaamyDaiaadggacaWGYbGaamyzaiaabccacaqG1bGaae OBaiaabMgacaqG0bGaae4Caaaaaa@3889@$

Q35.

$For the differential equation xydydx=x+2y+2, find the solution curve passing through the point 1, -1.$

Ans:

$The given differential equation is : xydydx=(x+2)(y+2) ydy(y+2)= (x+2)xdx (y+2−2)dy(y+2)= (x+2)xdx 1dy−2(y+2)dy=(1+2x)dxIntegrating both sides, we get∫1 dy−∫2(y+2) dy=∫1 dx+2∫1x dx y−2log(y+2)=x+2logx+C y−x−C=logx2+log(y+2)2 y−x−C=logx2(y+2)2 ...(ii)Since, the curve passes through (1,−1). −1−1−C=log(1)2(−1+2)2 −2−C=log(1)2⇒ C=−2Putting the value of C in equation(ii), we get y−x+2=logx2(y+2)2This is the required solution of the curve.$

Q36.

$Show that height of the cylinder of greatest volume which can be inscribed in a right circular cone of height h and semi vertical angle θ is one–third that of the cone and the greatest volume of cylinder is 427πh3 tan2α.$ 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Ans:

$In given right circular cone of fixed height is h and semi−vertical angle be α.Then a cylinder of radius R and height H is inscribed in the cone. Then ∠GAO=α, OG=r, OA=h, OE=R​ and CE=H.We have, r=htanαNow, since ΔAOG is similar to ΔCEG, we have:AOOG=CEEG⇒hr=Hr−R[EG=OG−OE$ $⇒ H= h r ( r−R ) = h htanα ( htanα−R ) = 1 tanα ( htanα−R ) Now, the volume (V) of the cylinder V=π R 2 h = π R 2 tanα ( htanα−R ) =π R 2 h− π R 3 tanα Differentiating w.r.t. R, we get dV dR =2πRh− 3π R 2 tanα and d 2 V d R 2 =2πh− 6πR tanα For maxima or minima, we have dV dr =0⇒2πRh− 3π R 2 tanα =0 ⇒πR( 2h− 3R tanα )=0⇒R= 2htanα 3 So,​ d 2 V d R 2 <0 for R= 2htanα 3 ∴By second derivative test, the volume of the cylinder is the greatest when R= 2htanα 3 . Then, H= 1 tanα ( htanα−R ) = 1 tanα ( htanα− 2htanα 3 ) = h 3 Thus, the height of the cylinder is one-third the height of the cone when the volume of the cylinder is the greatest. Now, the maximum volume of the cylinder V=π R 2 H =π ( 2htanα 3 ) 2 ( h 3 ) ∴ V= 4 27 π h 3 tan 2 α 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