Plus Two Maths Previous Year Question Paper March 2018

Kerala Plus Two Maths Previous Year Question Paper March 2018 with Answers

BoardSCERT
ClassPlus Two
SubjectMaths
CategoryPlus Two Previous Year Question Papers

Time : 2½ Hours
Cool off time : 15 Minutes
Maximum : 80 Score

General Instructions to Candidates :

  • There is a ‘Cool off time’ of 15 minutes in addition to the writing time.
  • Use the ‘Cool off time’ to get familiar with questions and to plan your answers.
  • Read questions carefully before you answering.
  • Read the instructions carefully.
  • When you select a question, all the sub-questions must be answered from the same question itself.
  • Calculations, figures and graphs should be shown in the answer sheet itself.
  • Malayalam version of the questions is also provided.
  • Give equations wherever necessary.
  • Electronic devices except non programmable calculators are not allowed in the Examination Hall.

Question 1 to 7 carry 3 scores each. Answer any six questions. (6 × 3 = 18)

Question 1.
If f(x) = \(\frac{x}{x-1}\), x ≠ 1
i) Find fof (x)
ii) Find the inverse of f.
Answer:
Plus Two Maths Previous Year Question Paper March 2018, 1

Question 2.
Using elementary row operations, find the inverse of the matrix \(\left[\begin{array}{cc}
1 & 2 \\
2 & -1
\end{array}\right]\)
Answer:
Plus Two Maths Previous Year Question Paper March 2018, 2

Question 3.
i) f(x) is a strictly increasing function, if f'(x) is ………..
a) positive
b) negative
c) 0
d) None of these,
ii) Show that the function f given by f(x) = x3 – 3x2 + 4x, x ∈ R is strictly increasing.
Answer:
i) a) Positive.
ii) f(x) = x3 – 3x2 + 4x
= 3 (x2 – 2x + 1 – 1 + \(\frac{4}{3}\)) = 3((x + 1)2 + \(\frac{1}{3}\)) > 0

Question 4.
i) \(\int_{0}^{a}\)f(a – x)dx = …………
Plus Two Maths Previous Year Question Paper March 2018, 3
Answer:
Plus Two Maths Previous Year Question Paper March 2018, 4

Question 5.
Find the area of the region bounded by the curve y2 = x’, x-axis and the lines x = 1 and x = 4.
Answer:
Plus Two Maths Previous Year Question Paper March 2018, 5

Question 6.
Find the general solution of the differential equation x \(\frac{d y}{d x}\) + 2y = x2 log x
Plus Two Maths Previous Year Question Paper March 2018, 6
Solution is
Plus Two Maths Previous Year Question Paper March 2018, 7

Question 7.
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. It takes 3 hours on Machine A and 1 hour on Machine B to produce a package of bolts. He earns profit of Rs. 17.50 per package on nuts and Rs. 7.00 per package on bolts. Formulate the above LPp if the Machine operates for at most 12 hours a day.
Answer:
Let x packet of nuts and y packets of bolts.
Maximise: Z=17. 5x + 7y
Subject to
x + 3y ≤ 12; 3x + y ≤ 12; x, y ≥ O

Questions 8 to 17 carry 4 scores each. Answer any 8. (8 × 4 = 32)

Question 8.
Let A = N × N and ‘*’ be a binary operation on A defined by
(a, b)*(c, d)=(a + c, b + d)
i) Find (1, 2) * (2, 3)
ii) Prove that * is commutative.
iii) Prove that * is associative.
Answer:
i) (1, 2)*(2, 3)=(1 + 2, 2 + 3) = (3, 5)
ii) (c, d)*(a, b) = (c + a, d + b)
= (a + c, b + d) = (a, b)* (c, d)
iii) (a, b)*[(c, d)*(e, f)] = (a, b)*(c + e, d + f)
= (a + c + e, b + d + f)
[(a, b) * (c, d)] * (e, f) = (a + c, b + d) * (e, f)
= (a + c + e, b + d + f)

Question 9.
i) Identify the function from the above graph
Plus Two Maths Previous Year Question Paper March 2018, 8
a) tan-1x
b) sin-1x
c) cos-1x
d) cos ec-1x
ii) Find the domain and range of the function represented in above graph.
iii) Prove that \(\tan ^{-1} \frac{1}{2}+\tan ^{-1} \frac{2}{11}=\tan ^{-1} \frac{3}{4}\)
Answer:
i) sin-1x
Plus Two Maths Previous Year Question Paper March 2018, 9

Question 10.
i) \(\frac{d\left(a^{x}\right)}{d x}\) = ………..
a) ax
b) log(ax)
c) ax log a
d) xax-1
ii) Find \(\frac{d y}{d x}\) if xy = yx
Answer:
i) ax loga
ii) Given; yx = xy, taking log on both sides;
x log y = y log x,
Differentiating with respect to x;
Plus Two Maths Previous Year Question Paper March 2018, 10

Question 11.
i) Find the slope of the tangent to the curve y = (x – 2)2 at x = 1.
ii) Find a point at which the tangent to the curve y = (x – 2)2 is parallel to the chord joining the point A (2, 0) and B (4, 4)
iii) Find the equation of the tangent to the above curve and parallel to the line AB.
y2 = 4ax, a > 0 and x2 = 4ay, a > 0
Answer:
i) \(\frac{d y}{d x}\) = 2(x – 2) ⇒ Slope = 2(1 – 2) = -2
ii) Slope of AB = \(\frac{y_{2}-y_{1}}{x_{2}-x_{1}}=\frac{4-0}{4-2}=2\)
Here 2(x – 2) = 2 ⇒ x = 3, y = 1
iii) Equation of the tangent line is
y – 1 = 2 (x – 3) ⇒ 2x – y = 5

Question 12.
\(\int_{0}^{2}\)(x2 +1)dx as the limit of a sum.
Answer:
Here; f(x) = x2 + 1; a = 0, b = 2 ⇒ h = \(\frac{2}{n}\)
\(\int_{0}^{2}\)(x2 +1)dx
\(\lim _{h \rightarrow 0}\) h[f(0) + f(0 + h) + f(0 + 2h)+…+ f(0 + (n – 1)h)]
\(\lim _{h \rightarrow 0}\) h (1 + (h2 + 1)+((2h) + 1)+…+[(n – 1)h2] + 1]
Plus Two Maths Previous Year Question Paper March 2018, 11

Question 13.
Consider the following figure:
Plus Two Maths Previous Year Question Paper March 2018, 12
i) Find the point of intersection ‘P’ of the circle x2 + y2 = 50 and the line y = x
ii) Find the area of the shaded region.
Answer:
i) point of intersection is x2 + x2 = 50 ⇒ x2 = 25 ⇒ ±5; y = ±5
point P is (5, 5)
Plus Two Maths Previous Year Question Paper March 2018, 13

Question 14.
i) The degree of the differential equation \(x y\left(\frac{d^{2} y}{d x^{2}}\right)^{2}+x^{4}\left(\frac{d y}{d x}\right)^{3}-y \frac{d y}{d x}=0\) is ………
a) 4
b)3
c)2
d)1
ii) Find the general solution of the differential equation
sec2 x tan ydx + sec2 y tan xdy = 0
Answer:
i) c
ii) Given; sec2x tan ydx + sec2y tan xdy = 0
Plus Two Maths Previous Year Question Paper March 2018, 14
log tanx = -log tany + c
log tan x + log tan y = c

Question 15.
i) Prove that for any vector \(\bar{a}, \bar{b}, \bar{c}\)
\([\bar{a}+\bar{b}, \bar{b}+\bar{c}, \bar{c}+\bar{a}]=2\left[\begin{array}{lll}
\bar{a} & \bar{b} & \bar{c}
\end{array}\right]\)
ii) Show that if \(\bar{a}+\bar{b}, \bar{b}+\bar{c}, \bar{c}+\bar{a}\) are coplanar then \(\bar{a}, \bar{b}, \bar{c}\) are also coplanar.
Answer:
Plus Two Maths Previous Year Question Paper March 2018, 15

Question 16.
i) Find the equation of a plane which makes x,y,z intercepts respectively as 1, 2, 3.
ii) Find the equation of a plane passing through the point (1, 2, 3) which is parallel to above plane.
Answer:
\(\frac{x}{1}+\frac{y}{2}+\frac{z}{3}\) ⇒ 6x + 3y + 2z = 6
ii) Equation of the parallel plane passing
6(x – 1) + 3( y – 2) + 2(z – 3) = 0
⇒ 6x + 3y + 2z – 18 = 0

Question 17.
Solve the LPP graphically:
Minimise z = -3x + 4y
Subject to constraints:
x + 2y ≤ 8; 3x + 2y ≥ l2; x, y ≥ 0
Answer:
x + 2y = 8

X08
Y40

3x + 12y = 12

X04
Y60

Plus Two Maths Previous Year Question Paper March 2018, 16
The corner points are O(0, 0), A(4, 0), B(2, 3), C(0, 4)

Corner pointZ = -3x + 4y
O(0, 0)Z = 0 + 0 = 0
A(4, 0)Z = -12 + 0 = -12
B(2, 3)Z = -6 + 12 = 6
C(0, 4)Z = 0 + 16 = 16

Z attains minimum at (4,0).

Questions from 18 to 24 carry 6 scores each. Answer any 5. (5 × 6 = 30)

Question 18.
i) Find x and y if \(x\left[\begin{array}{l}
2 \\
3
\end{array}\right]+y\left[\begin{array}{c}
-1 \\
1
\end{array}\right]=\left[\begin{array}{c}
10 \\
5
\end{array}\right]\)
ii) Express the matrix \(\left[\begin{array}{ccc}
2 & -2 & -4 \\
-1 & 3 & 4 \\
1 & -2 & -3
\end{array}\right]\) as the sum of a symmetric and a skew symmetric matrices.
Answer:
Given; 2x – y = 10; 3x + y = 5
5x = 15 ⇒ x = 3; y = -4
Plus Two Maths Previous Year Question Paper March 2018, 17
Plus Two Maths Previous Year Question Paper March 2018, 18

Question 19.
i) Prove that \(\left|\begin{array}{ccc}
a & b & c \\
a+2 x & b+2 y & c+2 z \\
x & y & z
\end{array}\right|=0\)
ii) If A = \(\left[\begin{array}{ccc}
1 & -1 & 2 \\
0 & 2 & -3 \\
3 & -2 & 4
\end{array}\right]\), B = \(\left[\begin{array}{ccc}
-2 & 0 & 1 \\
9 & 2 & -3 \\
6 & 1 & -2
\end{array}\right]\)
a) Prove that B = A-1
b) Using Asolve the system of linear equations given below:
x – y + 2z = 1; 2y – 3z = 1; 3x – 2y + 4z = 2
Answer:
Plus Two Maths Previous Year Question Paper March 2018, 19
iii) Matrix form of the system of linear equation is
Plus Two Maths Previous Year Question Paper March 2018, 20
Plus Two Maths Previous Year Question Paper March 2018, 21

Question 20.
i) Prove that the function defined by f(x) = cos x2 is a continuous function.
ii) a) If y = \(e^{a \cos ^{-1} x}\), -1 ≤ x ≤ 1, show that
\(\frac{d y}{d x}=\frac{-a e^{a \cos ^{-1} x}}{\sqrt{1-x^{2}}}\)
b) Hence prove that
\(\left(1-x^{2}\right) \frac{d^{2} y}{d x^{2}}-x \frac{d y}{d x}-a^{2} y=0\)
Answer:
i) f(x) = cosx; g(x) = x2 both are continuous.
Composition two continuous function is again continuous.
fog(x) = f(g(x)) = cos(x2)
Plus Two Maths Previous Year Question Paper March 2018, 22

Question 21.
Evaluate the following:
Plus Two Maths Previous Year Question Paper March 2018, 23
Answer:
Plus Two Maths Previous Year Question Paper March 2018, 24
Plus Two Maths Previous Year Question Paper March 2018, 25

Question 22.
i) lf \(\bar{a}\) = 3i + 2j + 2k, \(\bar{b}\) = i + 2j – 2k
a) Find \(\bar{a}\) + \(\bar{b}\), \(\bar{a}\) – \(\bar{b}\)
b) Find a unit vector perpendicular to both \(\bar{a}\) + \(\bar{b}\), \(\bar{a}\) – \(\bar{b}\)
ii) Consider the points A(1, 2, 7); B(2, 6, 3); C (3, 10, -1)
a) Find \(\overline{A B}, \overline{B C}\)
b) Prove that A, B, C are collinear points.
Answer:
Plus Two Maths Previous Year Question Paper March 2018, 26

Question 23.
i) Find the angle between the lines \(\frac{x-2}{2}=\frac{y-1}{5}=\frac{z+3}{-3}\) and \(\frac{x+2}{-1}=\frac{y-4}{8}=\frac{z-5}{4}\)
ii) Find the shortest distance between the pair of lines
\(\bar{r}\) = (i + 2j + 3k) + λ(i – 3j + 2k)
\(\bar{r}\) = (4i + 5j +6k) + μ(2i + 3j + k)
Answer:
Plus Two Maths Previous Year Question Paper March 2018, 27
Plus Two Maths Previous Year Question Paper March 2018, 28

Question 24.
i) The probability distribution of a random variable is given by P(x). What is Σp(x)?
ii) The following is a probability distribution function of a random variable.

X-5-4-3-2-10
P(x)k2k3k4k5k7k
X12345
P(x)8k9k10k11k12k

a) Find k
b) P(x > 3)
c) P(-3 < x < 4)
d) P(x < -3)
Answer:
Σp(x) = 1

ii) a) k + 2k + 3k + 4k + 5k + 7k
+ 8k + 9k + 10k + 11k + 12k = 1
72k = 1 ⇒ k = \(\frac{1}{72}\)
b) P(x > 3) = P(4) + P(5) = 23k = \(\frac{23}{72}\)
c) P(-3 < x < 4) = P(-2) + P(-1) + P(0) + P(1) + P > (2)+ P > (3) = 43k = \(\frac{43}{72}\)
d) P(x < -3) = P(-5) + P(-4) = 3k = \(\frac{3}{72}\)

Plus Two Maths Previous Year Question Papers and Answers