TG TUTORIALS

TG TUTORIALS
TARUN GEHLOT

Saturday, 1 August 2015

CLASS 12 MATHS SAMPLE PAPER 2


Instructions
1. Questions 1 to 10 carry 1 mark each.
2. Questions 11 to 22 carry 4 marks each.
3. Questions 23 to 29 carry 6 marks each.
1. If f(x) = x + 7 and g(x) = x − 7, x ∈ R, find (fog) (7).
Sol. 
2. Evaluate
Sol.
Let log x = t ...(1)
Differentiating both sides:
3. If a, b, c, are in A.P., then find the determinant
Sol.

  (2b = a + c as a, b and c are in A.P.)
Applying R→ R1 – R2 and R3→ R3 - R2, we have :
Applying R→ R1 + R3, we have :
Here, all the elements of the first row (R1) are zero.
Hence, we have Δ = 0.
4. Find the value of  cot( tan-1a + cot-1a).
Sol. 
5. What positive values of x makes the following pair of determinant equal?
Sol.

As per the given information:

Thus, the determinants are equal for x = 4.
6. Find adjoint of  the matrix.
Sol.
7. For what value of λ are the vectors perpendicular to each other?
Sol.

If  and are perpendicular to each other, then must be 0.

Thus, the value of λ is 
8. Find the values of x and y so that the vectors  are equal.
Sol. The two vectors  will be equal if their correspondingcomponents are equal.
Hence, the required values of x and y are 2 and 3 respectively.
9. Evaluate: 
Sol.

It is known that 
10. Write down a unit vector in XY-plane, making an angle of 30° with the positive direction of x-axis.
Sol. If  is a unit vector in the XY-plane, then 
Here, θ is the angle made by the unit vector with the positive direction of the x-axis.
Therefore, for θ = 30°:

Hence, the required unit vector is 


11. From a bag containing 20 tickets, numbered from 1 to 20, two tickets are drawn at random. Find the probability that
(i) Both the tickets have prime numbers on them
(ii) On one there is prime number and on the other there is a multiple of 4.
Sol. It is given that from a bag containing 20 tickets numbered from 1 to 20, two tickets are drawn at random.
Out of 20 tickets, two tickets can be drawn in 20C2 ways.
Therefore, total number of elementary events = 20C2 = 190
(i) The prime numbers amongst the numbers from 1 to 20 are 2, 3, 5, 7, 11, 13, 17, 19
Out of these 8 prime numbers, two numbers can be selected in 8C2 ways.
Thus, number of outcomes favourable to the event “both the tickets have prime numbers on them = 8C2 = 28.
Therefore, required probability = 
(ii) The numbers which are multiples of 4 from amongst the numbers from 1 to 20 are 4, 8, 12, 16, 20
Out of these 5 numbers, one can be selected in 5C1 ways.
Out of 8 prime numbers from 1 to 20, one prime number can be selected in 8C1 ways.
Thus, the number of favourable outcomes for the event “on one there is prime number and on the other there is a multiple of 4” = 5C1 × 8C1 = 5 × 8 = 40
Therefore, required probability = 
12. Find the general solution of the given differential equation:
Sol. The given differential equation is:
Integrating both sides, we get:


Substituting the values of in equation (1), we get:
This is the required general solution of the given differential equation.
13. Prove: 
Sol.

Hence, the given result is proved.
14. Find the shortest distance between the lines 
Sol. The equations of the given lines are

It is known that the shortest distance between the lines,  and  given by,
    ...(1)
Comparing the given equations, we obtain

Substituting all the values in equation (1), we obtain

Therefore, the shortest distance between the two lines is units.
15. If , show that f o f(x) = x, for all  . What is the inverse of f?
Sol. It is given that 
Hence, the given function f is invertible and the inverse of f is f itself.
16. If are such that is perpendicular to then find the value of λ. 
Sol. The given vectors are 
Now,


Hence, the required value of λ is 8.
17. Write the function in the simplest form :
Sol.
18. Solve the following differential equation: x cos y dy = (xex log x + ex) dx.
Sol. The given differential equation is
x cos y dy = (xex log x + ex) dx
On integrating with respect to x, we have

   [Integrating by parts]
  where C is a constant
This is the required solution of the given differential equation.
19. Find the distance of the point P (6, 5, 9) from the plane determined by the points A (3, −1, 2), B (5, 2, 4) and C (−1, −1, 6).
Sol. It is known that equation of a plane passing through the points (x1, y1, z1), (x2, y2, z2) and (x3, y3, z3) is given by

Thus, equation of the plane passing through the points A (3, −1, 2), B (5, 2, 4) and C (−1, −1, 6) is given by
⇒ (x − 3) (12 − 0) − (y + 1) [8 − (−8)] + (z − 2) [0 − (−12)] = 0
⇒ 12 (x − 3) − 16 (y + 1) + 12 (z − 2) = 0
⇒ 3 (x − 3) − 4 (y + 1) + 3 (z − 2) = 0
⇒ 3x − 9 − 4y − 4 + 3z − 6 = 0
⇒ 3x − 9 − 4y − 4 + 3z − 6 = 0
⇒ 3x − 4y + 3z − 19 = 0
This is equation of plane through the points A, B, and C.
It is known that the distance (D) of a plane ax + by + cz + d = 0 from a point (x1, y1, z1) is given by 
Thus, distance (D) of the plane 3x − 4y + 3z − 19 = 0 from the point (6, 5, 9) is
Thus, the distance of the point P (6, 5, 9) form the plane determined by the points A (3, −1, 2), B
(5, 2, 4) and C (−1, −1, 6) is 
20. For what value of k is the following function continuous at x = 2?
Sol. The given function f (x) will be continuous at x = 2, if:

Left hand limit of f (x) at x = 2, is 

Right hand limit of f (x) at x = 2, 

Thus, for k = 5, the given function is continuous at x = 2.
21. If , for, −1 < x <1, prove that 
Sol. It is given that,


Differentiating both sides with respect to x, we obtain
Hence, proved.
22. Find the equations of the normals to the curve y = x3 + 2x + 6 which are parallel to the line x + 14y + 4 = 0.
Sol. The equation of the given curve is y = x3 + 2x + 6.
The slope of the tangent to the given curve at any point (x, y) is given by .
∴ Slope of the normal to a curve at any point (x, y) 
∴ Slope of normal of given curve 
It is given that normals to the given curve are parallel to the line x + 14y + 4 = 0.
x + 14y + 4 = 0
⇒ 
∴ Slope of the given line = 
When x = 2, y = 8 + 4 + 6 = 18.
When x = −2, y = −8 − 4 + 6 = −6.
Therefore, there are two normals to the given curve with slope and passing through the points (2, 18) and (−2, −6).
Thus, the equation of the normal through (2, 18) is given by:
The equation of the normal through (−2, −6) is given by:

Thus, the equations of the normals to the given curve are x + 14y – 254 = 0 and x + 14y + 86 = 0.
23. Find the equation of the plane passing through the point (−1, 3, 2) and perpendicular to each of the planes x + 2y + 3z = 5 and 3x + 3y + z = 0. 
Sol. The equation of the plane passing through the point (−1, 3, 2) is
a (x + 1) + b (y − 3) + c (z − 2) = 0 … (1)
where, a, b, c are the direction ratios of normal to the plane.
It is known that two planes, and are perpendicular, if 
Plane (1) is perpendicular to the plane, x + 2y + 3z = 5

Also, plane (1) is perpendicular to the plane, 3x + 3y + z = 0

From equations (2) and (3), we obtain
Substituting the values of a, b, and c in equation (1), we obtain
This is the required equation of the plane.
24. Find the values of x, y, z if the matrix satisfy the equation 
Sol.

Now, 


  
On comparing the corresponding elements, we have:
25. A card from a pack of 52 cards is lost. From the remaining cards of the pack, two cards are drawn and are found to be both diamonds. Find the probability of the lost card being a diamond.
Sol. Let E1 and Ebe the respective events of choosing a diamond card and a card which is not diamond.
Let A denote the lost card.
Out of 52 cards, 13 cards are diamond and 39 cards are not diamond.
When one diamond card is lost, there are 12 diamond cards out of 51 cards.
Two cards can be drawn out of 12 diamond cards in 12C2 ways.
Similarly, 2 diamond cards can be drawn out of 51 cards in 51C2 ways. The probability of getting two cards, when one diamond card is lost, is given by P (A|E1).
When the lost card is not a diamond, there are 13 diamond cards out of 51 cards.
Two cards can be drawn out of 13 diamond cards in 13C2 ways whereas 2 cards can be drawn out of 51 cards in 51C2 ways.
The probability of getting two cards, when one card is lost which is not diamond, is given by   P (A|E2).
The probability that the lost card is diamond is given by P (E1|A).
By using Bayes’ theorem, we obtain
            
26. Find the area bounded by the circle x2 + y2 = 16 and the line y = x in the first quadrant.
Sol. Equation of the circle is x2 + y2 = 16 and the line y = x. Solving these two equations, we obtain x = y = 
Hence, the line intersects the circle at the point  in the first quadrant.
The area enclosed between the circle x2 + y2 = 16 and the line y = x in the first quadrant is as shown by the shaded area OABCO.
ar (OABCO) = ar (OACO) + ar (ABCA)


                                                      
27. Evaluate: 
Sol.

where which is the integral of even function
and which is the integral of odd function, and so I2 = 0
Now,
28. A dietician wishes to mix two types of foods in such a way that vitamin contents of the mixture contain atleast 8 units of vitamin A and 10 units of vitamin C. Food ‘I’ contains 2 units/kg of vitamin A and 1 unit/kg of vitamin C. Food ‘II’ contains 1 unit/kg of vitamin A and 2 units/kg of vitamin C. It costs Rs 50 per kg to purchase Food ‘I’ and Rs 70 per kg to purchase Food ‘II’. Formulate and solve this problem as a linear programming problem to minimise the cost of such a mixture.
Sol.
Resources
Food
Diet requirement
I
(x)
II
(y)
Vitamin A (units/kg)
2
1
8
Vitamin C (units/kg)
1
2
10




Cost(Rs/kg)
50
70


Minimise cost: Z = 50x + 70y   (1)
Equation of vitamin A
2x + y ≥ 8                               (2)
Equation of vitamin C
x + 2y ≥ 10                             (3)
x, y ≥ 0                                   (4)
Graph the inequalities (2) to (4). The feasible region determined by the system is shown below:
Corner Point
Z = 50x + 70y
A(0, 8)
560
B(2, 4)
380 Maximum
C(10, 0)
500
Hence, the optimal mixing strategy for the dietician would be to mix 2 kg of Food ‘I’ and 4 kg of Food ‘II’, and with this strategy, the minimum cost of the mixture will be Rs 380.
29. Find the equation of the plane which is perpendicular to the plane 5x + 3y + 6z + 8 = 0 and which contains the line of intersection of the planes x + 2y+ 3z − 4 = 0 and 2x + y − z + 5 = 0.
Sol. The equation of the plane through the line of intersection of the planes x + 2y + 3z − 4 = 0 and 2x + y − z + 5 = 0 is
(x + 2y + 3z − 4) + λ (2x + y − z + 5) = 0, where λ is a constant
⇒ x (1 + 2λ) + y (2 + λ) + z (3 − λ) − 4 + 5λ= 0 … (1)
This is perpendicular to the plane 5x + 3y + 6z + 8 = 0. Therefore, we have
5 (1 + 2λ) + 3 (2 + λ) + 6 (3 − λ) = 0
⇒ 7λ + 29 = 0
⇒ 
On putting in equation (1), we obtain

This is the equation of the required plane.