Please refer to Class 12 Mathematics Sample Paper With Solutions Set K provided below. The Sample Papers for Class 12 Mathematics have been prepared based on the latest pattern issued by CBSE. Students should practice these guess papers for class 12 Mathematics to gain more practice and get better marks in examinations. The Sample Papers for Mathematics Standard 12 will help you to understand the type of questions which can be asked in upcoming examinations.

**SECTION A**

**1. What is the principle value of tan−1 (tan (2π/3) ?****Sol. **tan−1 (tan (2π/3)) = tan^{−1} (− tan (π/3)) = −π/3.

**2. A and B are square matrices of order 3 each, |A| = 2 and |B| = 3. Find |3AB| .****Sol. |**3AB| = 3^{3} | A|| B| = 27×2×3 =162 .

**3. What is the distance of the point (p, q, r) from the x-axis?****Sol.** Distance of the point (p, q, r) from the x-axis is same as the distance of (p, q, r) from (p, 0, 0).

So, the required distance is

**4. Let f :R → R and g :R → R be defined by f (x) = 3x ^{2} − 5 and g(x) = x /x^{2}+1 .Find gof .**

**Sol.**

**SECTION B**

**5. How many equivalence relations on the set {1, 2, 3} containing (1, 2) and (2, 1) are there in ****all? Justify your answer.****Sol. **Following equivalence relations can be possible in the given conditions these are,

{(1,1), (2, 2),(3,3), (1,2),(2,1)} and {(1,1),(2, 2),(3,3), (1,2),(1,3),(2,1),(2,3), (3,1), (3, 2)}.

Clearly, only two equivalence relations are there.

**6. If e ^{y} (x +1) =1, show that dy/dx = −e^{y} .**

**Sol.**Here e

^{y}(x +1) =1 ⇒ ey + (x+1)e

^{y}+ 1 × dy/dx = 0 ∴ dy/dx = −e

^{y}

**7. Let l _{i}, m_{i} , n_{i} ; i =1, 2, 3 denote the direction cosines of three mutually perpendicular vectors in the space. Show that AA^{T} = I_{3} where **

**Sol. **

**8. Find the sum of the order and the degree of the differential equation : **

**Sol. **

Clearly order and degree here are 2 and 3 respectively. Therefore, their required sum is 5.

**9. Find the Cartesian and vector equations of the line which passes through the point (–2, 4, –5) and parallel to the line given by 10(x + 3) = 6(y − 4) = 5(z −8) .****Sol.**

Also the vector eq. : r̅ = −2î + 4ĵ−5 k̂ + λ(3î + 5ĵ+ 6 k̂ ) .

**10. Solve the following Linear Programming Problem graphically :**** Maximize Z = 3x + 4y , Subject to x + y ≤ 4, x ≥ 0 and y ≥ 0 .****Sol.**

**11. A couple has 2 children. Find the probability that both are boys, if it is known that (i) one of them is a boy (ii) the older child is a boy.****Sol. **Here S = {B_{1} B_{2} , B_{1}G_{2} ,G_{1}B_{2} ,G_{1} G_{2} }, B_{1} and G_{1} are the older boy and girl respectively.

Let E_{1} : both the children are boys, E_{2} : one of the children is a boy and E_{3} : the older child is a boy. Then,

**12. The sides of an equilateral triangle are increasing at the rate of 2 cm/sec. Find the rate at which its area increases, when side is 10 cm long.****Sol. **

**SECTION C**

**13. If A + B +C = π, then find the value of **

**Sol.**

**OR**

**Using properties of determinants, prove that **

**Sol.**

**14. It is given that for the function f (x) = x ^{3} − 6x^{2} + ax + b Rolle’s theorem holds in [1, 3] with c = 2 + 3^{−1/2} . Find the value of a and b.**

**Sol.**We have f (x) = x

^{3}− 6x

^{2}+ ax + b, 1≤ x ≤ 3

Since Rolle’s Theorem holds for f (x), so f (1) = f (3)

i.e., 1

^{3}− 6×1

^{2}+ a ×1+ b = 3

^{3}− 6×3

^{2}+ a ×3 + b ⇒ −5 + a = −27 + 3a ∴ a =11

Also f ‘(x) = 3x

^{2}−12x + a = 3x

^{2}−12x +11. As f ‘(c) = 0 ⇒ 3c

^{2}−12c +11 = 0

**15. Determine for what values of x, the function f (x) = x ^{3} + x^{−3}, x ≠ 0 is strictly increasing or decreasing.**

**Sol.**

If x >1 or x < −1 then f ‘(x) > 0. Also if −1< x <1 then f ‘(x) < 0 .

Therefore, f (x) is strictly increasing for x >1 or x < −1; and strictly decreasing for (−1,1) − 0 .

**OR**

**Find point(s) on y = x ^{3} −11x + 5 at which the tangent is y = x −11.**

**Sol.**Here y = x

^{3}−11x + 5 ⇒ dy/dx = 3x

^{2}− 11

Slope of given tangent line y = x −11 is 1 so, 3x

^{2}−11=1 ⇒ x = 2,−2

When x = 2, y = 2

^{3}−11×2 + 5 = −9 , when x = −2, y = (−2)

^{3}−11×(−2) + 5 =19 .

As the point (2, –9) satisfies y = x −11 but (–2, 19) doesn’t satisfy y = x −11.

So only (2, –9) lies both on the curve and the tangent. Therefore the required point is (2,−9) .

**16. Evaluate**

**Sol.**

**17. Find the area of the region bounded by the y-axis, y = cos x and y = sin x, 0 ≤ x ≤ π/2.****Sol.** We have y = sin x…(i), y = cos x…(ii), x = 0 (y axis) when 0 ≤ x ≤ π/2

**18. Can y = ax + b/a be a solution of the differential equation y = x dy/dx + b/dy/dx ? If no, then solve it.****Sol. **

**OR**

**Check whether the differential equation x ^{2}y’ − xy =1+ cos(y/x), x ≠ 0 is homogeneous or not. Find the general solution of the D.E. using substitution y = vx .**

**Sol.**

**19. If the vectors p̅ = aî + ĵ+ k̂, q̅ = î + bĵ+ k̂ and r̅ = î + ĵ+ c k̂ are coplanar, then for a, b,c ≠1 show that **

**Sol.** If the vectors p̅ = aî + ĵ+ k̂, q̅ = î + bĵ+ k̂ and r̅ = î + ĵ+ c k̂ are coplanar, then for a, b,c ≠1 show that

**20. A plane meets the coordinate axes in A, B and C such that the centroid of ΔABC is the point (α,β, γ) . Show that the equation of the plane is x/α + y/β + z/γ = 3.****Sol. **Let the plane meets the coordinate axes in points A(a, 0, 0), B(0, b, 0) and C(0, 0, c)

respectively. Also the centroid of the triangle formed is G (α, β, γ) .

**21. If a 20 year old girl drives her car at 25 km/h, she has to spend Rs 4/km on petrol. If she drives her car at 40 km/h, the petrol cost increases to Rs 5/km. She has Rs 200 to spend on petrol and wishes to find the maximum distance she can travel within one hour. Express the above problem as a linear programming problem. Write any one value reflected in the problem.****Sol. **Let the distance covered with speed of 25 km/h be x km and the distance covered with speed of

40km/h be y km.

To maximize : Z = x + y

Subject to constraints : x ≥ 0, y ≥ 0, 4x + 5y ≤ 200 and 8x + 5y ≤ 200.

Value : We should save petrol.

**22. The random variable X has a probability distribution P(X) of the following form : **

**(i) Find the value of k (ii) Find P(X < 2) (iii) Find P(X ≤ 2) (iv) Find P(X ≥ 2) .****Sol.**

(i) Since ∑P(X) =1 ⇒ P(0) + P(1) + P(2) =1 ⇒ 6k =1 ∴ k =1/6

(ii) P(X < 2) = P(X = 0) + P(X =1) = 3k = 3/6

(iii) P(X ≤ 2) = P(X = 0) + P(X =1) + P(X = 2) = 6k =1

(iv) P(X ≥ 2) = P(X = 2) = 3k = 3/6 .

**23. A bag contains 2n + 1 coins. It is known that n of these coins have a head on both of its sides whereas the rest of the coins are fair. A coin is picked up at random from the bag and is tossed. If the probability that the toss results in a head is 31/42, find the value of n.****Sol. **Total number of coins = 2n + 1, no. of biased coins = n and no. of fair coins = n + 1.

Let E : the toss resulted in Head, E_{1} : Head appeared on biased coins and E_{2} : Head appeared on

fair coins.

**SECTION D**

**24. Using properties of integral, evaluate**

**Sol. **

**OR**

**Find **

**Sol.**

**25.**

**have any solutions? If yes, then solve.****Sol. **

Hence x = 2 does not satisfy the condition (i).

Therefore there is no solution to the given equation

**OR**

**Determine whether the operation * defined below on Q is binary operation or not :****a *b = ab +1. If yes, check the commutative and associative properties. Also check the existence ****of identity element and the inverse of all elements in Q.****Sol.** Here * is defined on Q as a *b = ab +1.

Let a ∈ Q, b ∈ Q then, ab ∈ Q and, ab +1 ∈ Q i.e., a *b ∈ Q.

Hence * is binary operation on Q.

Commutativity : As a *b = ab +1 = ba +1 = b*a . So, * is commutative.

Associativity : As (a *b)*c = (ab +1)*c = abc + c +1…(i)

And, a *(b*c) = a *(bc +1) = abc + a +1…(ii)

By (i) and (ii), a *(b*c) ≠ (a *b)*c .

So, * is not associative.

Identity element : Let e ∈ Q be the identity element, then for every a ∈ Q we must have

a *e = a = e*a .

Consider a *e = a ⇒ ae + 1 = a ⇒ e = a − 1/a.

As e is not unique since it is not defined for a = 0 ∈ Q.

Therefore, identity element doesn’t exist for *.

Inverse of an element : Since there is no identity element hence, there is no inverse.

**26. Find the value of x, y and z, if **

**Sol.**

**OR**

**Verify : **

**Sol.**

**27. Find **

**Sol. **

**28. Find the shortest distance between the line x − y +1= 0 and the curve y ^{2} = x .**

**Sol.**Let (t

^{2}, t) be any point on the curve y

^{2}= x .

**29. Define skew lines. Using only vector approach, find the shortest distance between the following skew lines : r̅ = (8 + 3λ)î − (9 +16λ)ĵ+ (10 + 7λ) k̂ and r̅ =15î + 29ĵ+ 5 k̂ + μ (3î + 8ĵ−5 k̂).****Sol.** Skew lines: Two straight lines in space which are neither parallel nor intersecting are known as

the skew lines. They lie in different planes and are non-coplanar.