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# CBSE Set Qa1 Maths Sample Test Papers For Class 12th for students online

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Maths Class - XII  (CBSE)
You are on Set no 1 Answer 1 to 6

Q1) By using elementary row transformations, find the inverse of the matrix

 A = 1  2 2  -1

Ans1) Here

 A = 1  2 2  -1 and I = 1  0 0  1

Now A = IA

 = 1  2 2  -1 = 1  0 0  1 A

R2 R2 - 2R1

 1  2 0  -5 = 1  0 -2   1 A

= R2 (-1/5)R2

 1  2 0  1 = 1    0 2/5 -1/5 A

= R1 R1 - 2R2

 1   0 0   1 = 1/5  2/5 2/5 -1/5 A
 A-1 = 1/5    2/5 2/5   -1/5 = 1/5 1  2 2  -1

verify : AA-1 = I2

Q2) Express the matrix

 A = 3  -4 1  -1

as the sum of a symmetric and a skew symmetric matrix.
Ans2) For a given square matrix A. 1/2(A + A' ) is sym and 1/2(A - A' ) is skew-symmetric. Also
A = 1/2 (A + A' ) + 1/2 (A - A' )
Here

 A = 3   -4 1   -1 =A' = 3   1 -4  -1 ... 1/2 (A + A' )     = 3  -3/2 -3/2  -1 1/2(A - A' )     = 0  -5/2 5/2  0

Thus A =

 3  -3/2 -3/2  -1 + 0  -5/2 5/2  0

Q3) Show the vectors - 2 + 3 ,  -2 + 3 - 4 - 3 + 5 are coplanar.
Ans3)
For the vectors - 2 + 3 ,  -2 + 3 - 4 - 3 + 5  to be coplanar, we must have
[ , ,   ] = 0

 = 1      -2      3 -2      3      -4 1      -3       5 = 0

Apply  R2 R2 + 2R1,  R3 R3 - R on

 = 1   -2   3 0   -1   2 0   -1   2 = 1(-2 + 2) = 0

... The given vectors are coplanar

Q4) If x = x and x = x , prove that - is parallel to - , provided and .
Ans4) Given x = x   ...(i)
and x = x    ... (ii)
Now (i) - (ii) =
x - x x - x
= x () +  (  - ) x = 0
= x () -  (  - ) = 0
= ( - ) x () = 0
= ( - ) is parallel to ()
(as   and )

Q5) Verify Rolle's theorem for the function f(x) = x2 - x - 6 in the interval [-2,3]
Ans5) Here f(x) = x2 - x - 6 in [-2, 3]
Being a poly, f(x) is cont. on [-2, 3] and diff. on ]-2, 3[ Also f(-2) = 0 = f(3)
... All the conditions of Rolle's theorem are satisfied and so we must have
f '(c) = 0 for c ]-2, 3[
Now f '(x) = 2x - 1
... f '(x) = 0 = 2x - 1 = 0 = x = 1/2
... c = 1/2 ]-2, 3[
... Rolle's theorem is verified

Q6) Evaluate

 lim   x- (Sin x/(x - ))

Ans6)

 lim   x- (Sin x/(x - ))   putting x - = 0, as x - , - 0 = lim   -0 Sin (  + )/ = -lim  -0 Sin / = -1   (as Sin(  + ) = - Sin )

Q7) Differentiate tan-1 [cos x/(1 + sin x)] w.r.t. x
Ans7) let y = tan-1 [cos x/(1 + sin x)]
= tan-1 [sin (/2 - x)/(1 + cos (/2 - x)]
= tan-1 [2sin (/4 - x/2)cos(/4 - x/2) / 2cos2 (/4 - x/2)]
= tan-1 [tan (/4 - x/2)] = /4 - x/2
or y = /4 - x/2
= dy/dx = -1/2

Q8) Evaluate:-

 (log x)2 /x dx

Ans8) Let I =

 (log x)2 /x dx  putting log x = t, 1/x dx = dt (t)2 dt
= 1/3 t3 + c = 1/3 (log x)3 + c

Q9) Evaluate:-

 xex/(x + 1)2 dx

Ans9) Let I =

 xex/(1 + x)2 dx ex (x + 1 - 1)/(x + 1)2 dx ex [ 1/(x + 1) - 1/(x + 1)2] dx = ex/(x + 1) + c

Q10) Evaluate

 8   |x - 5| dx 0

Ans10)

 8   |x - 5| dx 0 5   |x - 5| dx +  0 8   |x - 5| dx 5 -(x - 5) dx + 8                                  5                    8   (x - 5) dx = - [(x - 5)2/2]  +  [(x - 5)2/2] 5                                  0                    5

= -1/2 [0 - 25] + 1/2[9 - 0]
= 25/2 + 9/2 = 17

Q11) Evaluate

 dx        (2 - 4x + x2)

Ans11) Let

 I= dx         (x2 - 4x + 2) = dx        ((x2 - 4x + 4) - 2) = dx       (x - 2)2 - (2)2
= log [(x - 2) + (x2 - 4x + 2)] + c
 using [ dx        = (x2 - a2) log |x + (x2 - a2)| + c ]

Q12) Evaluate

 /2   sin x - cos x  dx    (1 + sin x cos x) 0

Ans12) Let

 I= /2   sin x - cos x  dx    (1 + sin x cos x) 0 = /2   sin [/2 - x] - cos [/2 - x]  dx    (1 + sin [/2 - x] cos [/2 - x]) 0 = /2     cos x - sin x    dx    (1 + cos x sin x) 0 = - /2     sin x - cos x    dx  = -1   (1 + sin x cos x) 0
= 21 = 0 = I = 0

Q13) Solve the differential equation. dy/dx + [(1 - y2)/(1 - x2)] = 0
Ans13) dy/dx + [(1 - y2)/(1 - x2)] = 0
= dy/dx = - (1 - y2)/(1 - x2)
= dy/(1 - y2) = - dx/(1 - x2)
on integrating we get sin-1 y + sin-1 x = c

Q14) Two unbiased dice are thrown. Find the probability that neither a doublet nor a total of 10 will appear.
Ans14)
Let S be the sample space so n(S) = 36
Let E be the event of getting a doublet
and F be the event of getting a total of 10
Then E = {(1, 1), (2, 2), (3, 3), (4, 4), (5, 5), (6, 6)}
and F = {(4, 6), (5, 5), (6, 4)}
= EF = {(5, 5)}
... P(E) = 6/36 = 1/6 , P(F) = 3/36 = 1/12 and P(EF) = 1/36
= P(EUF) = P(E) + P(F) - P(EF)
= 1/6 + 1/12 - 1/36 = 2/9
... P (neither a doublet nor a total of ten)
= 1 -P(EUF) = 1 - 2/9 = 7/9

Q15) Find the regression coefficient of y on x for the following data:
x = 24, y = 44, xy = 306, x2 = 164, y2 = 574, n = 4

Ans15) we know
byx = (1/n xy - )/(1/n x2 - 2) = [1/4 x 306 - (24/4)(44/4)]/[1/4(164) - (24/4)2]
= (153/2 - 6 x 11)/(41 - 36) = (153 - 22 x 6)/(2 x 5)
= (153 - 132)/10 = 21/10 = 2.1
bxy = (1/n xy - )/(1/n y2 - 2) = [1/4 x 306 - (24/4)(44/4)]/[1/4(574) - (44/4)2]
= (153/2 - 66)/(287/2 - 121) = (153 - 132)/(287 - 242)
= 21/45 = 7/15 = 0.467

Q16) Using the properties of determinants, Prove that

 a+b+c   -c       -b -c       a+b+c   -a -b         -a     a+b+c = 2 (a+b)(b+c)(c+a)

Ans16) Let

 D = a+b+c   -c       -b -c       a+b+c   -a -b         -a     a+b+c
applying  C1 C1 + C2 + C3
 D = a     -c       -b b   a+b+c   -a c     -a     a+b+c
 Apply: C1 C1 - C3, C2 C2 + C3 = = a+b      -c+b      -b b+a       b+c      -a -(a+b)    b+c    a+b+c = (a+b)(b+c) 1     -1     -b 1      1     -a -1     1   a+b+c Apply R1 R1 + R3, R2 R2 + R3
 = = (a+b)(b+c) 0    0   a+c 0   2    b+c -1  1  a+b+c

= (a+b)(b+c)(-1)(-2)(a+c)
= 2(a+b)(b+c)(c+a)

Q17) A variable plane passes through a fixed point (1, 2, 3). Show that the locus of the foot of the perpendicular drawn from origin to this plane is the sphere given by the equation:
x2 + y2 + z2 - x - 2y - 3z = 0
Ans17)

Let P(x, y, z) be the foot of  drawn from origin on the variable plane (there can be infinite planes through a fixed pt.) passing through the pt A(1, 2, 3)
dr's of OP are
(x - 0) : (y - 2) : (z - 3)
whereas dr's of PA are
(x - 1) : (y - 2) : (z - 3)
... OP is  AP
... the sum of product of respective dr's = 0
= x(x - 1) + y(y - 2) + z(z - 3) = 0
or x2 + y2 + z2 - x - 2y - 3z = 0
is the required equation of locus of pt P. It is a sphere whose centre is mid pt of OA, with OA being the diameter.

Q18)  If a, b, c are the lengths of the sides opposite respectively to the angles A, B, C of a ABC, using vectors prove that cos c = (a2 + b2 - c2)/2ab
Ans18)

Let us consider ABC. Let the sides BC, CA and AB be represented by , ,
Let || = a, || = b, || = c
Now , In ABC,
+   + = 0
= + = -
= ( + ) . ( + ) = - . -
= ||2 + ||2 + 2. = ||2
= 2. = ||2 - ||2 - ||2
= 2|||| cos (  - c) = ||2 - ||2 - ||2
( makes angle - c with )
= - 2ab cos c = c2 - a2 - b2
= cos c = a2 + b2 + c2/2ab

Q19) Two balls are drawn at random from a bag containing white, 3 red, 5 green and 4 black balls, one by one without replacement. Find the probability that both the balls are of different colours.
Ans19)
3   5  4
R  G  B
If two balls of different colours are to be drawn from the bag the following cases are possible:
RG or GR or GB or BG or RB or BR
also the balls are drawn without replacement
... req probability
= 3/12 x 5/11 + 5/12 x 3/11 + 5/12 x 4/11 + 4/12 x 5/11 + 3/12 x 4/11 + 4/12 x 3/11
= (15 + 15 + 20 + 20 + 12 + 12)/132
= 94/132 = 47/66

Q20)  Find the probability distribution of the number of heads in three tosses of a coin
Ans20)
If a given coin is tossed three times the sample space is given by
S = { HHH, HTT, HHT, HTH, TTT, THH, TTH, THT}
Considering getting a head to be a success, the prob. distribution of number of heads is given by

 X 0 1 2 3 P(X) 1/8 3/8 3/8 1/8

is the required prob distribution for no. of heads (pls Note P(X) = 1)

Q21) For the function f(x) = -2x3 - 9x2 - 12x + 1, find the interval (s):
(i) in which f(x) is increasing.
(ii) in which f(x) is decreasing.
Ans21)
f '(x) = -6x2 - 18x - 12
= - 6(x2 + 3x + 2)
= - 6(x + 1) (x + 2)
putting f '(x) = 0
= x = -2, -1, which divide the no. line in the following intervals (-, -2), (-2, -1), (-1, ). Let us look at the sign of f '(x) in the intervals.

 Interval Sign of (x + 1) Sign of (x = 2) Sign of f '(x)= -6(x + 1)(x + 2) Nature of f(x) (- , -2) (-2, -1) (-1, ) - - + - + + - + -

... f(x) is decreasing on (-, -2) U (-1, ) and is increasing on (-2, -1)

Q22) Find from the first principle derivative of cosec x w.r.t. x.
Ans22) Let y = cosec x
= y + y = cosec (x + x)
... y = cosec (x + x) - cosec x
= [sin x - sin (x + x)] / [sin x . sin (x + x)]
= [sin x - sin (x + x)]/ [sin x . sin (x + x) (sin x + sin (x + x))]
= (-2cos ((2x + x)/2) . sin x/2)/ [sin x . sin (x + x) (sin x + sin (x + x))]
divide by x and proceed to limits as x  0, to get

 dy  =  lim    y  = -2cos x . 1/2 . 1 dx   x0  x     sin x. 2 sin x

= -1/2 cosec x . cot x

Q23) Find the value of

 2 0 (3x2 - 2)dx  as limit of sum.

Ans23) Here  f(x) = 3x2 - 2, a = 0 , b = 2
... nh = b - a = 2 - 0 = 2

 2 0 (3x2 - 2)dx
 = h f(0) + f(0+h) +....+ f(0+(n-1)h) = h -2 + (3h2 - 2) +...+{3(n-1)2h2 -2} = h -2n + 3h2{ 12 + 22 + 32 +...+ (n-1)2} = -2nh + h3 {(n-1)(n)(2n-1)}/6 = -2 x 2 + 1/2{(2-h)(2)(2 x 2 -h)} = -4 + 1/2 x 2 x 2 x 4 = 4

Q24) Evaluate

 3           dx  (1 - x)(1 + x2)

Ans24) Let

 I= 3           dx  (1 - x)(1 + x2)

Let          3             A    Bx + C
(1 - x)(1 + x2)    (1 - x)     (1 + x2)
A(1 + x2) + (1 - x)(Bx + C)
(1 - x)(1 + x2)
= 3 = A(1+ x2) + (1 - x)(Bx + C)
putting x = 1
3 = 2A = A = 3/2
putting x = 0
3 = A + C = C = 3 - 3/2 = 3/2
putting x = -1
3 = 2A + 2(-B + C)
= 3 = 2A - 2B + 2C = B = A + C - 3/2
or B = 3/2 + 3/2 - 3/2 = 3/2

 I = 3/2    +  3/2x + 3/2 dx (1 - x)        (1 + x2)
 = 3/2 1/(1 - x) dx + 3/4 2 . x/(1 + x2) dx + 3/2 1/(1 + x2) dx

= -3/2 log |1 - x| + 3/4 log|1 + x2| + 3/2 tan-1 x + c

Q25) Draw a rough sketch and find the area of the region bounded by the two parabolas y2 = 4x and x2 = 4y by using methods of integration.
Ans25) The given parabolas are
y2 = 4x   ...(i)
x2 = 4y   ...(ii)
solving (i) and (ii)
x4/16 = 4x = x(x3 - 64) = 0
= x = 0, 4
... y = 0, 4
(i) and (ii) meet at (0, 0) and (4, 4)

the shaded region is the required bounded region whose area we have to find
Area of the reqd region

 = 4   (y1 - y2) dx 0 = 4   [4x - x2/4] dx 0 = 2. (x3/2)/(3/2) - 1/4 . x3/3 4 0

= 4/3 . 8 - 1/12 . 64 = 16/3 sq. Units

Q26) If A =

 4  -5  -11 1   -3    1 2   3   -7 , find A-1

Using A-1, solve the following system of equations:
4x - 5y - 11z = 12
x - 3y + z = 1
2x + 3y - 7z = 2

Ans26)
Given

 A = 4  -5  -11 1  -3   1 2   3   7

= |A| = 4(21 - 3) + 5(-7 - 2) - 11(3 + 6)
= 72 - 45 - 99 = -72 0
= A-1 exists
calculating co-factors, we get
A-1 = adj A / |A|

 A-1 = -1/72 18  -68   -38 9    -6    -15 9    -22   -7

The given system of linear eqs is
4x - 5y - 11z = 12
x - 3y + z = 1
2x + 3y - 7z = 2
= AX = B

 where A = 4  -5  -11 1  -3   1 2   3   7 ; X = x y z ; B = 12 1 2
since A-1 exist, AX = B
A-1(AX) = A-1B = (A-1A)X = A-1B
= IX = A-1B = X = A-1B
=
 x y z = -1/72 18  -68   -38 9    -6    -15 9    -22   -7 12 1 2 = -1/72 72 72 72 = -1 -1 -1

x = y = z = -1 Ans.

Q27) Define the line of shortest distance between two skew lines. Find the shortest distance and the vector equation of the line of shortest distance between the lines given by
= 2 - 3 + (2 - ) and
= 4 + 3 + (3 + +)

Ans27) The given lines are
= 2 - 3 + (2 - ) and
= 4 + 3 + (3 + +)
or = 2 + (2 - ) - 3
and = (4 + 3) + +(3 + )

The general pts P and Q on L1 and L2 are respectively P(2, 2 - , -3) and Q(4 +3, , 3 + )
Let || be the shortest distance for some value of and
Now dr's of are
3 - 2 + 4 : + - 2, 3 + + 3
...   L1
= 2(4 + 3 - 2) - 1( + - 2) + 0(6 + ) = 0
(dr's of L1 are 2, -1, 0)
Also   L2 =
3(4 + 3 - 2) + 1( - 2 + ) + 1(6 + ) = 0
(dr's of L2 are 3, 1, 1)
... we have
-   = 2
and 5 - 11 = 16
solving for and
= 1, = 1
... co-ordinates of P and Q are
(2, 1, -3) and (1, -1, 2)
hence shortest distance = ((2 - 1)2 + (1 + 1) 2 + (-5)2)
= (1 + 4 + 25) = 30
and the vector equation of line PQ is
= 2 + - 3 + ()
= 2 +   - 3(- - 2  + 5)

Q28)  A wet porous substance in open air loses it's moisture at the rate proportional to the moisture content. If a sheet hung in wind loses half of it's moisture during the first hour, when will it have lost 95% moisture, weather conditions remaining the same
Ans28) Let x be the moisture of the porous substance present any time 't' xo be initial moisture
... from the given statement
dx/dt x
= dx/dt = -k x
where k is proportionality constant

 dx/x = - k dt

or log x = -kt + c   - (i)
at t = 0 , x = xo
... log xo = c
hence from (i)
log x = -kt + log xo
= log(x/xo) = - kt   -(ii)
Also it is given that
at t = 1 hr, x = xo/2
... log(1/2) = -k
or - log 2 = -k
... k = log 2
... from (ii)
log (x/xo) = (-log 2)t   -(iii)
when it loses 95% of moisture the moisture content present = 5% of initial
i,e, x = xo/20 = 5/100 xo = xo/20
... from (iii)
log (1/20) = (-log 2)t
or -log 20 = (-log 2)t
... t = log 20/log 2 hours

Q29) A rectangle is inscribed in a semi circle of radius 'r' with one of it sides on the diameter of the semi circle. Find the dimensions of the rectangle so that it's area is maximum. Find also this area.
Ans29)

Let us consider a rectangle ABCD with side AB along the diameter of the semi circle. let O be the centre of the circle
Now  OBC ~ OAD (by RHS rule)
... OA = OB = x (say)
Let S be the area of rectangle
then S = 2xy
or S = 2x(r2 - x2)
squaring both sides
S2 = 4x2 (r2 - x2) = (4r2x2 - 4x4)
for S to be maximum , S2 has to be maximum
... dS2/dx = 0
Now dS2/dx = 8r2x - 16x3 = 8x[r2 - 2x2]
putting dS2/dx = 0
= x = 0  or x = r/2
not possible
Also d2S2/dx2 = 8r2 - 48x2
... (d2S2/dx2)at x =r/2 = -16r2
= -ve
... x = r/2 is the point of maxima and hence the max area is given by:-
S2 = 4r2/2 (r2 - r2/2)
S2 = r2 . r2
... Max area S = r2

Q30) Calculate Karl Pearson's coeff of correlation between x and y for the following data:

 x: 16 18 21 20 22 26 27 15 y: 22 25 24 26 25 30 33 14
Also, give the interpretation of coeff of correlation thus obtained.
Ans30) Let a = 21, b = 25
 x y u = x - 21 v = y - 25 uv u2 v2 16 18 21 20 22 26 27 15 22 25 24 26 25 30 33 14 -5 -3 0 -1 1 5 6 -6 -3 0 -1 1 0 5 8 -11 15 0 0 -1 0 25 48 66 25 9 0 1 1 25 36 36 9 0 1 1 0 25 64 121 u = -3 v = -1 uv = 153 u2 = 133 v2 = 221
... r = [1/nuv - ] / [1/nu2 - 2 ] . [1/nv2 - )2]
= [1/8 x 153 - (-3/8)(-1/8)] / [(1/8 x 133 - (-3/8)2) (1/8 x 221 - (-1/8)2)]
= (153 x 8 - 3) / (133 x 8 - 9) . (221 x 8 - 1) = (1224 - 3)/(1064 - 9) . (1768 - 1)
= (1221)/ [(1055) . (1767)]
Taking log on both sides
log r = log 1221 - 1/2 (log 1055 + log 1767)
log r = 3.0867 - 1/2[3.0232 + 3.2472]
= 3.0867 - 1/2[6.2704]
= 3.0867 - 3.1352
= - 0.0485 = - 1 + 1 - 0.8485
log r = 1.9515
... r = antilog of 1.9515
= 0.8943
It is a high degree positive correlation between x and y