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

Ans1) Here

Now A = IA

R₂ R₂ - 2R₁

= R₂ (-1/5)R₂

= R₁ R₁ - 2R₂

verify : AA^-1 = I₂

Q2) Express the matrix 

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

Thus A =

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

Apply  R₂ R₂ + 2R_1,  R₃ R₃ - R₁  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

Ans6)

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

Ans11) Let 

 = log [(x - 2) + (x² - 4x + 2)] + c

Q12) Evaluate

Ans12) Let 

 = 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

Ans16) Let

          applying  C₁ C₁ + C₂ + C₃

= (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

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

Q24) Evaluate

Ans24) Let 

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(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|

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 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' x_o 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 = x_o
.^.. log x_o = c
hence from (i) 
log x = -kt + log x_o
= log(x/x_o) = - kt   -(ii)
Also it is given that 
at t = 1 hr, x = x_o/2
.^.. log(1/2) = -k
or - log 2 = -k
.^.. k = log 2
.^.. from (ii) 
log (x/x_o) = (-log 2)t   -(iii)
when it loses 95% of moisture the moisture content present = 5% of initial
i,e, x = x_o/20 = 5/100 x_o = x_o/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: