Cbse Sample Paper 1 For Mathematics

Posted on - 08-05-2017

CBSE Math

CBSE

SECTION – A (1 Mark)

q1.

The graph of the curve y = f(x) given below does not represent a function on R, give reasons?

Solution

f(x) is not a function as x Î R is not uniquely mapped i.e. for each x there exists 2 y values.

q2.

What is principal value of .

Solution

sin^-¹= - sin^-¹

= - sin^-¹= - sin^-¹

= - (since principal range of sin^-¹x is ).

q3.

If A = , B = and if A^T = B find x, y.

Solution

A^T

x + y = 6

2x + y = 1

⇒ x = - 5, y = 11

q4.

Find the point on the curve y = x² + 2x + 17, where the tangent is parallel to x-axis. .

Solution

= 2x + 2 = 0 ( since tangent is parallel to x-axis ⇒ slope = 0)

⇒ x = - 1, if x = - 1, y = 1 - 2 + 17 = 16

\point is (-1, 16).

q5.

If |a| = 3, |b| = 2 and find ||.

Solution

= |a| |b| cosq⇒ 3 = 3.2. cosq⇒cosq = ⇒q =

= |a| |b| sinq = = .

q6.

If are two unit vectors and ‘’ is the angle between them, then find the value
of .

Solution

1 + 1 + 2 cosq = 2 + 2 cosq = 2(1 + cosq) = 2. 2 cos²

⇒= .

SECTION B (4 Marks)

q7.

If f(x) = 2x + 3, g(x) = x² find fog (2), gof(2). .

Solution

fog(x) = f[g(x)] = f[x²] = 2x² + 3

⇒fog(x) = 2x² + 3

⇒fog(2) = 11

gof(x) = g[f(x)] = g[2x + 3] = (2x+3)²

⇒gof(x) = (2x + 3)²

gof(2) = 49

q8.

Prove that

Solve , |x| < 1.

Solution

Let

x = cos2q⇒ 2q = cos^-¹x

1 + cos2q = 2 cos²q

1 - cos2q = 2 sin²q

= = .

⇒

⇒ 2x²- 4 = - 3 ⇒ 2x² = 1 ⇒ x = ±.

q9.

Evaluate .

Solution

C₁ + C₂ + C₃

= C₁- C₂

= C₃- C₁

= C₂- C₃

= .

q10.

If y = (sinx)^x + x^sinx find

Solution

y = (sinx)^x + x^sinx,

Let y₁ = (sinx)^x⇒ x ln sinx = lny₁

⇒

Let y₂ = x^sinx ⇒ lny₂ = sinx .lnx.

q11.

A water tank has the shape of an inverted right circular cone with its axis vertical and vertex lower most. Its semi-vertical angle is tan^-¹. Water is poured into it at a constant rate of 5 cubic meter per minute. Find the rate at which the level of the water is rising at the instant when depth of water in the tank is 10 m.

Solution

Let at any instant radius be ‘r’ and ‘h’ be height of water and volume be V, given = 5 cu. mts

V =

5 =

m/ minutes.

q12.

If y = , z = , then find .

Solution

y = = 2 tan^-¹x,

z = cos^-¹= 2tan^-¹x ,

⇒

q13.

Evaluate dx as limit of sum.

Solution

Let us divide the interval [1, 3] in ‘n’ equal intervals

viz. [1 + h], [1 + 2h], [1 + 3h], … [ 1+ (n-1)h, 3].

\sum of area of all ‘n’ rectangle

= h{(2(1)² + 3) + (2(1 + h)² + 3) + (2(1 + 2h)² + 3) +...

+ (2(1 + (n-1)h)² + 3)]

= h[2(1² + 1² + 1² + … + n terms) + 4h (0 + 1 + 2 +

…+ n term) + 2h² (0 + 1² + 2² + 3² + … + n terms)

+ (3 + 3 + 3 + … n terms)

Where 3 - 1 = nh

⇒ h = and h→ 0 , n →¥.

\ A = .

= = .

q14.

Evaluate

Solution

Let x- b = t ⇒ dx = dt , x - a = t + b - a

= cos(b - a) + sin(b - a)

= t cos(b - a) + sin(b - a) . log|sint| + c .

= (x - b) cos(b - a) + sin(b - a) log|sin(x - b)| + c.

q15.

Solve the differential equation

, where y(0) = 1.

Solution

IF =

Solution is tanx = t, sec²x dx = dt

y.⇒ y e^tanx = [t e^t-e^t ] + c

ye^tanx = [tanx- 1] e^tanx + c

⇒ y = (tanx- 1) + c e^-^tanx .

q16.

If are 3 mutually perpendicular vectors with magnitudes |a| = 2, |b| = 3, |c| = 4 find .

Solution

= 4.4 + 9.9 + 16.16 since

= 16 + 81 + 256 = 353

= .

q17.

Find the equation of the plane passing through (1, 2, 3); (-1, 4, 7), (0, 0, 2). .

Find the vector, Cartesian equation of line passing though (1, 2, -7); (0, 3, 8). .

Solution

Equation of plane passing (a₁, b₁, c₁), (a₂, b₂, c₂), (a₃, b₃, c₃)

i.e.

and also Cartesian form

q18.

A person speaks truth in 60% of cases and his wife speaks false in 80% of cases if both speak on same issue find the probability they agree upon each other. .

Solution

P(H) = , P() =

, P(w) =

= 44%

Question 19.

Evaluate

Solution

⇒

SECTION C

q20.

Show that the volume of largest cone that can be inscribed in a sphere is of the volume of sphere.

Solution

r =

h = R + x

=> volume of cone = V_max =

= (let)

f’(x) =

x = (point of maxima) (Check using double derivative test)

V_max = =

Also volume of sphere =

V_max = V

Hence Proved.

q21.

Find the Area common to y² = 4ax, x² = 4ay (a > 0). .

Find the area of the region {(x, y) | 0 ≤ y ≥ x² + 1, 0< y < x + 1, 0 < x < 2}

Solution

Required Area =

= = sq. units.

(y - 1) = x² and

y = x + 1

x -y = - 1;

Solve

y = x²+ 1

y = x + 1

to get point of intersections x² + 1 = x + 1

= x(x - 1) = 0 , x = 0, 1

Required area =

= sq. units.

q22.

A variance plane is at constant distance P from the origin and meets the co-ordinates axes at A, B and C. Show that the locus of centroid of triangle ABC .

Find the vector equation of the plane passing through the point of intersection of planes

, and passing through (1, 1, 1).

Solution

Let the plane meets area at

A(a, 0, 0), B(0, b, 0),

C(0, 0, c). Equation of plane .

= (x₁, y₁, z₁)

perpendicular distance from (0, 0, 0)

= ==

=>= .

q23.

Evaluate .

Solution

= tan²x = 1 ⇒ 2 tanx sec²x dx = dt

I = .

q24.

A fair die is tossed 3 times

(i) Find the probability of getting a number greater than 4 at last once. .

(ii) Find the probability getting an even number of odd number of times. .

Solution

(i) Let getting number > 4 is success

P(success) = , P(failure) = q = , n = 3

P(number greater than 4 at Least once) = 1 - P(not even once)

= 1 -³C₀ a³p⁰ = 1 -³C₀=

(ii) Let getting even is success p = , q =

odd no. of time P(x = 1) or P(x = 3) .

P(x = 1) + P(x - 3) = ³C₁ a² p¹ + ³C₃ p³ = 3 =

q25.

A manufacturer produces nuts and bolts. It takes 1 hour of work on machine A and 3 hours on machine B to produce a package of nuts while it takes 3 hours on machine A and 1 hour an machine to produce a package of bolts. He earns a profit of Rs 17.50 per package on nuts and 7.00 per package on bolts. How many packages of each should be produced each day so as to maximize his profit? If he operates his machines for a most 12 hours a day.