Sample Model Papers For Maths CBSE Board for students online

Q 1 If A = - , B = 1 3 -6 , verify that (AB)/ = B/ A/ .


Ans.1 Given A = B = [1 3 -6 ]

therefore AB =    [ 1 3 -6]  = 

therefore (AB)^/ =

Q 2 Find Karl Pearson's coefficient of correlation between x and y for the following

Ans. 2

Prepare the table , here assumed me
Ans are a = 8 and b = 7

r = [ å(dX_I dy_I) - ådX_I ådy_I / n ] / Ö{å(dX_I)2 - å(dX_I)2/n}{å(dy_I)2 - å(dy_I)2 /n}

r = {10 - (-6) (-2)/5} / Ö {44 - 36/5}{20 - 4/5}

r = 0.28

Q 3 Find the magnitude of the vector a x b , if a = 2i + k and
b = i +j + k.

Therefore magnitude is given by Ö (-1)² + (-1)² + 2² = Ö 6

Q 4 Verify LaGrange's mean value theorem for the function
f(x) = x + 1/x in the interval [1,3]

Ans.4
The given function is f(x) = (x² + 1 ) / x .
It being a rational function such that x is not equal to 0, so continuous in [1,3].
Also f (x) = [1 - 1/x²] = [x²-1] / x²,which clearly exists for all values of x in ]1,3[.
So f(x) is differentiable in ]1,3[.
So there must exist some c such that
F/( c ) = f(3) - f(1) / 3 - 1 = 10/3 - 2 / 2 = 2/3
Therefore c2 - 1 / c2 = 2/3 which gives c = Ö3.

Q 5 Find the centre and radius of the sphere
5x² + 5y² +5z² +10x - 6y + 8z + 5 = 0

Ans. 5
5x² + 5y² +5z² +10x - 6y + 8z + 5 = 0
in standard form
x² + y² +z² +2x - 6/5y + 8/5z + 1 = 0
therefore u = 1, v = -3/5, w = 4/5 and d = 1
therefore centre is ( -u,-v,-w) = ( -1, 3/5, -4/5)
also radius is given by Ö u² + v² + w² - d = Ö1² + (-3/5)² +(4/5)² - = 1.

Q 6 Discuss the continuity of the function f(x) = { 3x - 2, when x £ 0
x + 1, when x 0 } at x = 0.

Ans. 6
Clearly f(0) = (3*0 - 2) = - 2
Lim f(x) = lim f(0+h) = lim f(h) = lim (h+1) = 1

since the limits are not equal therefore the function is discontinuous at x = 0.
The equation of tangent to the curve y = - 5x²+ 6x + 7 at the point
(1 / 2, 35 / 4 ) is given by
dy / dx = - 10x + 6
thetrefore dy / dx at ( 1 / 2, 35 / 4 ) = -10 *1/2 + 6 = 1
therefore eq. Is
y - 35/4 = 1 ( x - ½)
4y - 35 = 4x -2
4y - 4x = 33

Q 8 Evaluate 0^òp/4 log ( 1 + tanx ) dx

Ans. 8 oò^p/4 log ( 1 + tanx ) dx
let
I= oò^p/4 log(1+ tan(p/4-x)) dx
I= oò^p/4 log( 1+ (1-tanx)/(1+tanx)) dx
I=oò^p/4 log(2 / 1+tanx) dx
I = oò^p/4 log2 dx - 0ò ^p/2 log(1+tanx) dx
I= log2 * p/4 -I
I= p/8 log2

Q 9 Evaluate ò dx / (x+1)( x+2)(x+3) .

Ans.9 ò dx / (x+1)( x+2)(x+3) .
let
1 / (x+1)( x+2)(x+3) . = A / (x+1) +B / (x+2) + C / (x+3)
1 = A(x+2)(x+3) + B(x+1)(x+3) + C(x+1)(x+2)
which gives A = ½, B = - 1, C = ½
therefore the given integral can be written as
½ ò dx / x+1 - ò dx/ x+2 + ½ ò dx / x+3
= ½ log (x+1) - log(x+2) + 1/2log(x+3) + c

Q 10Evaluate ò ex[ (1 - sinx) / (1 - cosx)] dx.

Ans. 10 ò ex[ (1 - sinx) / (1 - cosx)] dx.
ò e^x {1 / 1- cosx - sinx / 1- cosx } dx
ò e^x {1 / 2sin²x/2 - 2sinx/2 cosx/2 / 2sin²x/2} dx
ò e^x {(1/2)cosec²x/2 - cotx/2} dx = - ò e^x cotx/2 dx + 1/2ò e^x cosec²x/2 dx
= - e^x cotx/2 - ½ ò e^x cosec²x/2 dx + C + ½ ò e^x cosec²x/2 dx
= - e^x cotx/2 + C.

Q 11 ò [(x² - 1) / (1 + x⁴)] dx.

Ans. 11 ò [(x² - 1) / (1 + x⁴)] dx.
= divide Nr. And Dr. by x² we have
= . ò [(1 - 1/ x² ) / (1/x2 + x²)] dx
put x + 1/x = t
( 1 -1/x²)dx = dt
also x² + 1/x² = t² -2
therefore . ò dt / t² - 2
= 1/ 2Ö2 log (t - Ö2)/ (t+Ö2).
= 1/2Ö2 log [( x+1/x -Ö2) / (x+1/x + Ö2)] + C

Q 12 One card is drawn from a set of 17 cards numbered 1 to 17. Find the probability that the number drawn is divisible by 3 or 7.

Ans.12 single card can be drawn in 17 ways
no. divisible by 3 are 3,6,9,12,15.
No divisible by 7 are 7 and 14
Therefore total no. of favourable events are 7
Therefore the required probability is 7 / 17.

Q 13 Out of 9 outstanding students in a college, there are 4 boys and 5 girls. A team of 4 Students are to be selected for a quiz programme. Find the probability that two are boys and two girls.

Ans. 13 Students out of nine can be selected in 9C4 ways i.e. 126
also 2 out of 4 can be selected in 4C2 ways i.e. 6 and 2 out of 5 can be selected in 5C2 ways i.e. 10.
Therefore the required probability is 10 * 6 / 126 = 10 / 21 .

Q 14 A particle follows the path S = 3t2 + 2t - 4. Find the velocity and acceleration of the Particle at t = 4sec. Assume S in meters.

Ans. 14 Given S = 3t² + 2t - 4, we know that velocity is given by ds/ dt and acceleration is given by d²s / dt² therefore we have
V = ds / dt = 6t + 2 therefore at t= 4 we have v = 6*4 + 2 = 26 m/s
Also A = d²s/dt² = 6m/s².

Q 15 Use differential to approximate the value of cube root of 0.009.

Ans. 15
let y = x^1/3 and take x = 0.008, dx = + .001
so that x + dx = 0.009
now Dy = (x + Dx)^1/3 - x^1/3
= (0.008 + .001)^1/3 - 0.008^1/3
therefore 0.009^1/3 = Dy + 0.2
here Dy is approximately equal to dy and
dy = (dy/dx) dx = (1/3 x^-2/3) dx
= 1/3 (0.008)^-2/3 (0.001) at x = 0.008
= (1/3 )(1/0.04)(0.001) = 0.00833
therefore 0.009^1/3 = 0.00833 + 0.2 = 0.2083

Q 16 Solve the differential equation dy / dx = sin (x + y).

Ans. 16 dy / dx = sin(x + y)
put x + y = z
1 + dy/dx = dz /dx
dy/dx = dz/dx - 1
substituting we get
dz/dx - 1 = sin z
dz/sinz +1 = dx
òdz / sinz +1 = ò dx
x = ò (1/ 1 + sinz)* (1 - sinz)/(1 - sinz) dz
= ò (1 - sinz)/cos²z dz
x = òsec²z dz - òsecz tanz dz
x = tanz - secz +C = tan(x+y) - sec(x+y) +C

Q 17 Solve (1 + x²) dy/dx + 2xy - 4x² = 0.

Ans. 17 (1 + x²) dy/dx + 2xy - 4x² = 0.
The above equation is in the form
dy/dx + Py = Q.
It can be rewritten as
dy/dx + {2x/(1+x2)} y = 4x².

Q 19 A ladder 13m long is leaning against a wall. The bottom of the ladder is pulled along the ground, away from the wall, at the rate of 2m/s.How fast is its height on the wall decreasing when the foot of the ladder is 5m away from the wall?

Ans. 19 Let the ladder is kept vertical initially.
When the horizontal distance is 5 m,
the vertical distance would be Ö(13² - 5²) =12 m.
Speed when the ladder is pulled horizontally = 2m/s

Time taken for covering 5m = distance/ velocity = 5/2 =2.5 s
In this time the ladder is pulled 1 m vertically.
Therefore, vertical speed = 1/(2.5) = 0.4 m/s

Ans = 0.4 m/s

Q 20 Evaluate dx as a limit of sums.

Ans. 20) Evaluate ₀ò 2(1 + x²) dx
_aò ^b f(x)dx = (b-a)lim 1/n[f(a) + f(a+h) +……………..+f(a + (n-1)h)]
n-µ............................................................................. where h = (b-a)/n

Here a=0, b=2, f(x) = 1 + x², h= 2/n,
₀ò ²(1 + x2) dx = 2 lim 1/n [f (0) + f(2/n) +f(4/n)……………..+f((2n-2)/n)]
= 2 lim 1/n [1 + ((2²/n²)+1) +((4²/n²)+1)……………..+((2n-2)²/n² +1 )]
= 2 lim 1/n[(1+1+….n times)+1/n²(2²+4²+…………..+(2n-2)²]
= 2 lim 1/n[n + 2²/n²(1²+2²+…………..+(n-1)²)]
= 2 lim 1/n[n + 4/n².{(n-1)n(2n-1)}/6]
= 2 lim 1/n[n + 2/3.{(n-1)(2n-1)}/n]
= 2 lim [1 + 2/3{(n-1)(2n-1)}/n² ]
= 2 lim [1 + 2/3(1-1/n)(2-1/n)]
= 2 [1 +4/3]
= 14/3.

Ans. = 14/3

Q 21 Evaluate

Ans. 21

Expanding the determinant, we get
D= y[(2x+y)(x+y)-2x²]
= y[ 3xy + y²]
= y²[3xy + y]

Ans = y²[3xy + y]

Q 22 Find the correlation coefficient between x and y, when the lines of regression are: 2x - 9y + 6 = 0 and x - 2y + 1 = 0.

Ans.22) The regression line of y on x is
y = (2x+6)/9
thus b_yx =2/9
The regression line of x on y is
x = 2y -1
b_xy = 2
Correlation coefficient,
r = Ö (b_yx b_xy)
= Ö (4/9)
=

Q 23 Using matrices solve the following
x + y - 2z = 0;
2x + y - 3z = 0;
5x + 4y - 9z = 0

Ans. 23
Writing in the matrix form

lAl = 1(-9+12) -1(-18+15)-2(8-5)
= 3 + 3 -6 = 0


Q 24 Evaluate ^{p/ 2} (log sinx) dx

Ans. 24 p/2
Let I = 0ò (log sinx) dx ………………(1)
p/2
Then , I = oò log [ sin(p/2 - x ) ]dx
[since oò^a f(x) dx = oò^af(a- x)dx
Or
I = oò^p/2 log(cos x ) dx ……………………………….(2)
Adding equation (1) and (2),
2I = oò^p/2 [log (sin x) +log (cos x)] dx
= oò^p/2 log( sin 2x/2) dx
= oò^p/2 log(sin2x)dx - oò^p/2 log(2) dx
= ½ oò^p/2 log sin t dt - (log 2)* oò^p/2 dx { putting 2x = t in the 1st integral}
= ½*2 oò^p/2 logsinx dx - p/2(log 2)
2I = I - p/2 (log 2)
or
I = - p/2 (log2)
Therefore
oò^p/2 log sin x = - p/2 (log 2)

Q 25 Find the magnitude and direction of the resultant of two forces 14N and 22N inclined at an angle of 30^o to each other.

Ans. 25
Let the resultant force 'R', the magnitude of the resultant force can be determined by
Given force1 A = 14 N, force B = 22 N,
R = Ö A² + B²+ 2A*B cos q
Here q = 30^o
Therefore
R = Ö 14²+ 22² + 14*22 cos 30^o
R = 32.181 N

Direction can be calculated by
tana = ( B sinq)/A + B cosq
= 22 sin 30^o/ 14 + 22 cos 30^o
tana = 0.4119
a = tan^-1 0.4119

Q 26 Prove that the points with position vectors ( - j - k ), ( 4i + 5j + k ), ( 3i + 9j + 4k ) and ( - 4i + 4j + 4k ) are coplanar.

Ans. 26 Given pts. are ( - j - k ), ( 4i + 5j + k ), ( 3i + 9j + 4k ) and
( - 4i + 4j + 4k )
let the pts. Are A,B,C and D respectively, so
AB = ( 4i + 5j + k ) - ( - j - k ) = 4i + 6j + 2k
BC = ( 3i + 9j + 4k ) - ( 4i + 5j + k ) = -i + 4j + 3k
CA = ( - j - k ) - ( - 4i + 4j + 4k ) = 4i - 5j - 5k
Now if they are coplaner then

4 ( -20 + 15) - 6 ( 5 - 12 ) + 2 ( 5 - 16 ) which is equal to 0. Hence the pts. Are coplanar
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Q 27 Find the equation of plane passing through the origin and perpendicular to each of the planes, x + 2y - z = 1 and 3x - 4y + z = 5.
Ans. 27
Given equation
x + 2y - z = 1
3x- 4y +z = 5
Any plane through (0,0,0) is
a( x- 0) + b(y - 0) + c(z - 0) = 0,
i.e. ax + by + cz = 0 ……………………….(1)
If (1) is perpendicular to each one of the given planes, we have:
A +2b -c = 0 ……………………………..(2)
3a -4b + c = 0 …………………………….(3)
on solving (2) and (3), we get
a/-2 = b/-4 = c/-10 or a/1 = b/2 = c/5 = k.
putting a= k, b= 2k and c = 5k in eq. (1), we obtain the required equation of the plane as
kx + 2ky + 5kz =0
or
x +2y +5z = 0. (
Ans.)

Q 28 If a car goes from point A to B with a velocity of 40Km/Hr and returns back with a speed of 60Km/Hr.Find the average speed of the car.

Ans. 28 Let the distance between the two pts. be S Km, then time taken in first ride will be
T₁ = S / 40 also time taken in second ride will be
T₂ = S / 60
Therefore total time will be T = S / 40 + S / 60
Avg. Velocity = total distance travelled / total time taken
= 2S / (S / 40 + S / 60)
= 48 Km / Hr.

Q 29 The ceiling of a long hall is 25m high. What is the maximum horizontal distance that a ball thrown with a speed of 40m/s can go without hitting the ceiling of the hall?

Ans. 29 we know that h = u² sin²q / 2g
therefore 25 = 40* 40 sin²q / 2 * 10 ……………………………taking
g = 10 m/s²
sinq = 51/2 / 4
therefore R = u²sin2q / g
= 40 * 40 * 2 sinq cosq / 10
= 40 * 40 * 2* 5^1/2 * 11^1/2 / 4*4*10
= 148.32 m

Q 30 Solve the differential equation
(x³ - 3xy²) dx = (y³ - 3x²y) dy.

Ans. 30
(x³ - 3xy²)dx = (y³- 3x²y)dy
the given equation can be written as,
dy/dx = x³ - 3xy²/ y³- 3x²y which is clearly homogeneous.
Putting y= vx and dy/dx = v + x dy/dx in it, we get:
v+ x dv/dx = x³ - 3x3v2/v³x³- 3vx³ = 1 - 3v²/ v³-3v
x dv/dx = [1 - 3v²/ v³-3v] - v = 1 - v⁴/v³ - 3v
taking integral both sides,
ò dx/x = ò v³ - 3v/1 - v⁴ dv
by partial fraction
= ò[ 1/ 2(v +1) + ½(v - 1) - 2v/v²+1]dv
log | x | + log C =1/2 log | v+1| + ½ log |v - 1| - log | v²+1|
log|Cx| = log | (Ö v+1) (Öv - 1)/ (v²+1)|
therefore
Cx = Ö v² - 1/( v²+1)
C²x² = (v² - 1)/( v²+1)
Putting v = y/x
C²(y²+ x²)2 = ( y² - x²)
Therefore the solution of the differential equation is
C²(y²+ x²)² = ( y² - x²) (
Ans.)