# Class 12 Mathematics Sample Paper Set E

Please see below Class 12 Mathematics Sample Paper Set E with solutions. We have provided Class 12 Mathematics Sample Papers with solutions designed by Mathematics teachers for Class 12 based on the latest examination pattern issued by CBSE. We have provided the following sample paper for Class 12 Mathematics with answers. You will be able to understand the type of questions which can come in the upcoming exams.

## CBSE Sample Paper for Class 12 Mathematics Set E

**SECTION – A**

**1. Write the sum of the order and degree of the given differential equation : d/dx{(dy/dx) ^{3}} = 0 .**

**Solution:**Given d/dx{(dy/dx)

^{3}} = 0 ⇒ 3(dy/dx)

^{2}x d

^{2}y/dx

^{2}= 0 .

Since order and degree of the differential equation is 2 and 1 respectively. So their sum is 3.

**2. Write the integrating factor of (1+y ^{2}) + (2xy – cot y) dy/dx = 0 .**

**Solution:**We’ve (1+y

^{2}) + (2xy – cot y) dy/dx = 0 ⇒ dy/dx(2y/1+y

^{2}) x = cot y/1+y

^{2}

Here P(y) = 2y/1+y

^{2}, Q(y) = cot y/1+y

^{2}

**3. Write the value of **

**Solution:**

**4. Write a unit vector perpendicular to both the vectors a̅ = î + ĵ + k̂and b̅ = î + ĵ .****Solution:**

**5. The equations of a line are 5x -3 =15y + 7 = 3-10z . Write the direction cosines of the line.****Solution:** Given line is 5x -3 =15y + 7 = 3 – 10z i.e., x-3/5/1/5 = y+7/15/1/15 = z-3/10/-1/10

∴ d.r. ’s are 1/5,1/15,-1/10 i.e., 6, 2,-3

So, d.c.’s of the line : ± 6/√36+4+9 , ± 2/√49 , ± (-3/7) i.e., 6/7 , 2/7 , -3/7 or -6/7 , -2/7 , 3/7 .

**6. If a̅ , b̅ and c̅ are mutually perpendicular unit vectors, then find the value of | 2a̅ + b̅ + cˆ |.****Solution:**

**SECTION – B**

**7. Find : **

**Solution:**

**8. Find :**

**Solution:**

**9. **

**Solution:**

**10. Find the equation of a line passing through the point (1, 2,–4) and perpendicular to two lines **

**Solution:** Let the d.r.’s of required line L be a, b, c.

**OR**

**Find the equation of the plane passing through the points (–1, 2, 0), (2, 2, –1) and parallel to the line x-1/1 = 2y+1/2 = z+1/-1****Solution:** OR Let d.r.’s of normal to the required plane through (–1, 2, 0) and (2, 2,–1) be A, B, C.

Equation of plane : A (x + 1) + B (y – 2) + C (z – 0) = 0 …(i)

As (2, 2,–1) lies on (i) so, 3A + 0B – C = 0…(ii)

Also d.r.’s of x-1/1 = 2y+1/2 = z+1/-1 i.e., x-1/1 = y+1/2/1 = z+1/-2 are 1, 1, –1.

Since (i) is parallel to the given line so, A + B – C = 0 …(iii)

By (ii) & (iii), we get : A/1 = B/2 = C/3 i.e., d.r.’s of normal to the plane is 1, 2, 3.

Hence required plane is : 1 (x + 1) + 2 (y – 2) + 3 (z – 0) = 0 i.e., x + 2y + 3z = 3.

**11. Three cards are drawn successively with replacement from a well shuffled pack of 52 cards.****Find the probability distribution of the number of spades. Hence find the mean of distribution.****Solution:** Let X : No. of spades ∴ possible values of X : 0, 1, 2, 3. Let E : spade card is drawn.

Total playing cards = 52, No. of spades = 13, P(E) = ^{13}C_{1}/^{52}C_{1} = 13/52 = 1/4 = p, q = 1-p = 3/4

**OR**

**For 6 trials of an experiment, let X be a binomial variate which satisfies the relation 9 P(X = 4) = P(X = 2) . Find the probability of success.****Solution:** Here n = 6. Let p be the probability of success and q be that of failure so, p + q = 1.

**12. Find the adjoint of the matrix **

**and hence show that A. (adj A) = | A|I3 .****Solution:**

**13. Show that the function f (x) = | x -1| + | x +1|, for all x ∈ R, is not differentiable at the points x = -1 and x =1.****Solution: **

**14. If y = e ^{msin-1} x , then show that (1-x^{2}) d2y/dx^{2} -x dy/dx – m^{2} y= 0 .**

**Solution:**

**15. If f (x) = √x ^{2}+1 ; g(x) = x+1/x^{2}+1 and h(x) 2x -3 , then find f ‘[h'{g'(x)}] .**

**Solution:**

**16. Evaluate : ∫(3 – 2x)√2 + x – x ^{2} dx **

**Solution:**Let I = ∫(3 – 2x)√2 + x – x

^{2}dx Put 3 – 2x = A d/dx (2+x – x

^{2}) + B = A (1-2x) + B

On equating the coefficients of like terms, we get : A = 1, B = 2.

**OR**

**Evaluate : ∫ x ^{2}+x+1/(x^{2}+1)(x+2)**

**Solution:**

**17. To promote the making of toilets for women, an organization tried to generate awareness through (i) house calls (ii) letters, and (iii) announcements. The cost for each mode per attempt is given below :**

(i) ₹ 50

(ii) ₹ 20

(iii) ₹ 40

The number of attempts made in three villages X, Y and Z are given below :

Find the total cost incurred by the organization for the three villages separately, using matrices.

Write one value generated by the organization in the society.**Solution:** The given information can be expressed as :

Hence the cost incurred by organization for villages X, Y and Z respectively are 30000, 23000 and 39000 (in ₹).

Value of cleanliness/ women welfare is generated by the organization in society

**18. Solve for x : tan ^{-1}(x+1) + tan^{-1}(x-1) = tan^{-1} 8/31**

**Solution:**We have tan-1(x+1) + tan-1(x-1) = tan-1 8/31

**OR**

**Prove the following :****cot-1(xy+1/x-y) + cot ^{-1}(yz+1/y-z) + cot^{-1} (zx+1/z-x) = 0 ; (0 < xy, yz, zx < 1) .**

**:**

**Solution****19. Using properties of determinants, prove the following : **

**Solution:**

**SECTION – C**

**20. A company manufactures three kinds of calculators : A, B and C in its two factories I and II.**

The company has got an order for manufacturing at least 6400 calculators of kind A, 4000 of kind B and 4800 of kind C. The daily output of factory I is of 50 calculators of kind A, 50 calculators of kind B, and 30 calculators of kind C. The daily output of factory II is of 40 calculators of kind A, 20 of kind B and 40 of kind C. The cost per day to run factory I is ₹ 12000 and of factory II is ₹ 15000. How many days do the two factories have to be in operation to produce the order with the minimum cost? Formulate this as an LPP and solve it graphically.**Solution:** Let factory I and II respectively work for x and y number of days respectively.

To minimize : Z = ₹ (12000x + 15000y)

Subject to the constraints :50x + 40y ≥ 6400, 50x + 20y ≥ 4000, 30x + 40y ≥ 4800, x ≥ 0, y ≥ 0 That is, 5x + 4y ≥ 640, 5x + 2y ≥ 400, 3x + 4y ≥ 480, x ≥ 0, y ≥ 0

As the feasible region is unbounded;

therefore 1860000 may or may not be the minimum value of Z. In order to check,

we plot the graph of inequality

12000x + 15000y < 1860000 i.e., 4x + 5y < 620 .

It can be seen easily that the feasible region has no common point with 3x + 5y < 26 therefore,

the minimum value of Z is ₹1860000.

Hence the factory I and II respectively work for 80 and 60 number of days respectively.

**21. In a factory which manufactures bolts, machines A, B and C manufacture respectively 30%, 50% and 20% of the bolts. Of their output 3, 4 and 1 percent respectively are defective bolts. A bolt is drawn at random from the product and is found to be defective. Find the probability that this is not manufactured by machine B.****Solution:** Let D be the event that the bolt drawn is defective. Let A, B and C be the event that the balls are manufactured by machines A, B and C respectively.

So, P(A) = 30/100 , P(B) = 50/100 , P(C) = 20/100 ; P(D|A) = 3/100 , P(D|B) = 4/100 , P(D|C) = 1/100

By using Bayes’ Theorem,

(P|B|D) = (P|B|P|D|B)/P(A) P (D|A) + P(B) P + (D|B) + P(C) P (D|C)

**22. Consider f : R _{+} →[ -9, ∞) given by f (x) = 5x^{2} + 6x – 9 . Prove that f is invertible with f-1(y) = √54 + 5y – 3/5**

**Solution:**Let y be an arbitrary element of range of function i.e., y∈[-9,∞) .

Then y = 5x

^{2}+ 6x – 9, for some x∈R

_{+}, which implies that

**OR**

**A binary operation * is defined on the set X = R -{-1} by x * y = x + y + xy, ∀x, y ∈ X.****Check whether * is commutative and associative. Find the identity element and also find the inverse of each element of X.****Solution: **We have x * y = x + y + xy, ∀x, y ∈ X where X = R – {-1}.

Commutativity : Let Let x, y∈X ⇒ x * y = x + y + xy = y + x + yx = y* x

Hence * is commutative.

Associativity : Let x, y, z ∈ X ⇒ (x * y)*z = (x + y + xy)*z = x + y + z + xy + yz + zx + xyz

Also, x *(y*z) = x *(y + z + yz) = x + y + z + xy + yz + zx + xyz

That implies, x *(y*z) = (x * y)*z so, * is associative as well.

Let e be the identity element.

So, x *e = x ⇒ x + e + xe = x ⇒ e = 0/1+x = 0 ∈ R – {-1} [∴ x ≠ -1]

Let inverse of an element in X be a. So, x *a = e ⇒ x + a + ax = 0

⇒ a = -x/1+x [∴ x ≠ -1]

**23. Find the value of p for when the curves x ^{2} = 9p(9 – y) and x^{2} = p(y +1) cut each other at right angles.**

**Solution:**Given x

^{2}= 9p(9 – y)…(i) and, x

^{2}= p(y +1)…(ii)

By (i) & (ii), p(y +1) = 9p(9 – y) ⇒ y = 8, x

^{2}= 9p…(A)

On diff. (i) & (ii) w.r.t. x both sides, dy/dx = -2x/9p and dy/dx = 2x/p

Since the curves (i) & (ii) cut each other at right angles so, (2x/p)(-2x/9p) = -1 ⇒ 4x

^{2}= 9p

^{2}

By (A), 4x9p = 9p

^{2}⇒ p(p – 4) = 0 ∴ p = 0, 4

We shall reject p = 0 since it doesn’t fit into the given conditions so, p = 4 .

**24. Show that the differential equation dy/dx = y ^{2}/xy-x^{2} is homogeneous and also solve it.**

**Solution:**We have dy/dx = y2/xy-x

^{2}Let f (x, y) = y

^{2}/xy-x

^{2}

**OR**

**Find the particular Solution: of the differential equation (tan-1 y – x)dy = (1+ y2 )dx , given that x =1 when y = 0 .****Solution:** Given (tan^{-1} y – x)dy = (1+ y^{2} )dx ⇒ dy/dx + x/1+y2 = tan-1 y/1+y^{2}

This is linear differential equation of the form dy/dx + xP(y) = Q(y) , where P(y) = 1/1+y2 and Q(y) = tan^{-1} y/1+y^{2} . Now Integrating Factor, I.F. = e ∫dx/1+y^{2} = e tan^{-1}y

**25. Find the distance of the point P(3, 4, 4) from the point, where the line joining the points A(3,-4,-5) and B(2, –3, 1) intersects the plane 2x + y + z = 7 .****Solution:** Equation of the line joining the points A(3,-4,-5) and B(2, –3, 1) is :

x-3/2-3 = y+4/-3+4 = z+5/1+5 i.e., x-3/-1 = y+4/1 = z+5/6 = λ (say)

Coordinates of any random point on this line is M(3- λ,λ – 4,6λ – 5)

When the given line intersects the plane 2x + y + z = 7 , we’ve :

2(3- λ) + λ – 4 + 6λ -5 = 7 ∴ λ = 2 So, M(1,-2,7)

Therefore, MP = √(3-1)^{2} + (4+2)^{2} (4-7)2 = √4+36+9 = 7 units

**26. Using integration, prove that the curves y2 = 4x and x2 = 4y divide the area of the square bounded by x = 0, x = 4, y = 4 and y = 0 into three equal parts.****Solution:** Let y^{2} = 4x…(i) and x^{2} = 4y…(ii)

On solving (i) & (ii) we get the point of intersection as (0, 0) and (4, 4).

Now, the area of the region OAQBO bounded by

From (A), (B) and (C), it can be concluded that the area of the region OAQBO = area of the region OPQAO = area of the region OBQRO, i.e., area bounded by parabolas (i) and (ii) divides the area of the square in three equal parts.