2nd PUC Maths Previous Year Question Paper March 2017

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Karnataka 2nd PUC Maths Previous Year Question Paper March 2017

Time: 3 Hrs 15 Min
Max. Marks: 100

Instructions

• The question paper has five parts namely A, B, C, D, and E. Answer all the parts.
• Use the graph sheet for the question on Linear programming in Part – E

Part – A

Answer ALL the following questions: (10 × 1 = 10)

Question 1.
Let * be the binary operation on N, given by a * b = LCM of a and b. Find 20 * 16.
Solution:
20 * 16 = LCM of 20 and 16 = 80

Question 2.
Find the principal value of cosec^-1 (-√2)
Solution:
cosec^-1 (-√2) = -cosec √2 = \(-\pi / 4\)

Question 3.
Construct a 2 × 2 matrix, A = [a_ij], where elements are given by, a_ij = \(\frac{i}{j}\)
Solution:
a₁₁ = \(\frac{1}{1}\) = 1
a₁₂ = \(\frac{1}{2}\)
a₂₁ = \(\frac{2}{1}\) = 2
a₂₂ = \(\frac{2}{2}\) = 1
∴ A = \(\left[\begin{array}{ll}1 & 2 \\1 / 2 & 1\end{array}\right]\)

Question 4.
If A is a square matrix with |A| = 8 then find the value of |AA’|.
Solution:
|AA’| = |A| |A’| = (8) (8) = 64

Question 5.
If y = cos√x, find \(\frac{d y}{d x}\)
Solution:

Question 6.
Find \(\int\left(\sqrt{x}+\frac{1}{\sqrt{x}}\right) d x\)
Solution:

Question 7.
Define collinear vectors.
Solution:
Collinear vectors Two or moe non-zero vectors are said to be collinear if they are parallel to the same line. Two vectors a and b are collinear if a = λb for some scalar λ.

Question 8.
Find the direction cosines of a line which makes equal angles with the positive co-ordinates axes.
Solution:
α = β = r

Question 9.
Define a feasible region in a linear programming problem.
Solution:
The region containing the set of points satisfying all the constraints of an LPP is called the feasible region.

Question 10.
If A and B are independent events, P(A) = \(\frac{3}{5}\) and P(B) = \(\frac{1}{5}\) then find P(A∩B).
Solution:
P(A∩B) = P(A) . P(B) = \(\frac{3}{5} \cdot \frac{1}{5}=\frac{3}{25}\)

Part – B

Answer any TEN questions: (10 × 2 = 20)

Question 11.
If f : R → R defined by f(x) = 1 + x², then show that f is neither 1-1 nor onto.
Solution:
f(1) = 1 + 1² = 2
f(-1) = 1 + (-1)² = 2
∴ f(1) = f(-1) But 1 ≠ -1
∴ f is not one-one
Let y ∈ R ∃ x ∈ R such that
f(x) = y
⇒ 1 + x² = y
⇒ x² = y – 1
⇒ x = \(\sqrt{y-1}\)
if y = 0 then x = \(\sqrt{-1} \notin \mathrm{R}\)
∴ 0 has no pre-image
∴ f is not onto.

Question 12.

Solution:

Question 13.
Solve the equaton \(\tan ^{-1}\left(\frac{1-x}{1+x}\right)=\frac{1}{2} \tan ^{-1} x,(x>0)\)
Solution:

Question 14.
Find the values of k, if the area of the triangle is 4 sq. units and vertices are (k, 0), (4, 0) and (0, 2) using determinant.
Solution:

⇒ -2k + 8 = 8
⇒ 2k =
⇒ k = 0
on taking -ve sign we get -2k + 8 = -8
⇒ 2k = 16
⇒ k = 8
∴ k = 0, 8.

Question 15.
If ax + by² = cos y, find \(\frac{d y}{d x}\)
Solution:

Question 16.
Verify Rolle’s theorem for the function f(x) = x² + 2x – 8, x ∈ [-4, 2]
Solution:
f(x) is a polynomial in x. Flence it is continuous over {-4, 2} and differentiable over (-4, 2).
f(-4) = (-4)² + 2(-4) – 8 = 16 – 8 – 8 = 0
f(2) = 4 + 4 – 8 = 0
∴ f(-4) = f(2)
∴ All the conditions of the Rolle’s theorem are satisfied.
∴ there exists a c ∈ [-4, 2] such that
f'(c) = 0
f'(x) = 2x + 2
f'(c) = 2c + 2
f'(c) = 0
⇒ 2c + 2 = 0
⇒2c = -2
⇒ c = -1 ∈ [-4, 2]
∴ Rolle’s theorem is verified.

Question 17.
Find the approximate change in the volume of a cube of side x metres caused by increasing the side by 3%.
Solution:
v = x³
∆x = 3% of x = (0.03)x
\(\frac{d v}{d x}\) = 3x²
∆v = \(\frac{d v}{d x}\) . ∆x
= (3x²) (0.03) x
= 0.09 x³ m³

Question 18.
Integrate \(\frac{\tan ^{4} \sqrt{x} \sec ^{2} \sqrt{x}}{\sqrt{x}}\) with respect to x.
Solution:

Question 19.
Evaluate \(\int_{0}^{2 / 3} \frac{d x}{4+9 x^{2}}\)
Solution:

Question 20.
Find the order and degree of the differential equation \(\left( \frac { dy }{ dx } \right) ^{ 2 }+\frac { dy }{ dx } { sin }^{ 2 }y=0\)
Solution:
Order = 1, Degree = 2.

Question 21.
Find the position vector of a point R which divides the lipe joining two points P and Q whose position vectors are \(\hat{i}+2 \hat{j}-\hat{k}\) and \(-\hat{i}+\hat{j}+\hat{k}\) respectively in theratio 2 : 1
(i) Internally.
(ii) Externally.
Solution:
The position vector of a point R divided the line segment joining two points P and Q in the ratio m : n is given by

Question 22.
Find the area of the parallelogram whose adjacent sides are determined by the vector \(\vec{a}=\hat{i}-\hat{j}+3 \hat{k}\) and \(\vec{b}=2 \hat{i}-7 \hat{j}+\hat{k}\)
Solution:

Area of the parallelogram = \(|\vec{a} \times \vec{b}|\) = √450 sq units.

Question 23.
Find the vector and Cartesian equation of the line that passes through the points (3, -2, 5) and (3, -2, 6).
Solution:
Let a and b be the position vectors of points (3, -2, -5), and (3, -2, 6) respectively.
\(\vec{a}=3 \hat{i}-2 \hat{j}-5 \hat{k}\) and \(b=3 \hat{i}-2 \hat{j}+6 \hat{k}\)
We know that the vector equation of a line passing through the points having position vectors a and b is \(r=\vec{a}+\lambda(\vec{b}-\vec{a})\)

Question 24.
Find the probability distribution of the number of heads in two tosses of a coin.
Solution:
When one coin is tossed twice, the sample space is
S = {HH, HT, TH, TT}.
Let X denotes, the number of heads in any outcome in S,
X(HH) = 2, X(HT) = 1, X(TH) = 1 and X(TT) = 0
Therefore, X can take the value of 0, 1 or 2. It is known that
P(HH) = P(HT) = P(TH) = P(TT) = \(\frac{1}{4}\)
P(X = 0) = P (tail occurs on both tosses) = P({TT}) = \(\frac{1}{4}\)
P(X = 1) = P(one head and one tail occurs) = P({TH, HT}) = \(\frac{2}{4}\) = \(\frac{1}{2}\)
and P(X = 2) = P (head occurs on both tosses) = P({HH}) = \(\frac{1}{4}\)
Thus, the required probability distribution is as follows.

Part – C

Anstver any TEN questions: (10 × 3 = 30)

Question 25.
Show that the relation R in R (set of real numbers) is defined as R = {(a, b) : a ≤ b} is reflexive and transitive but not symmetric.
Solution:
a ≤ a is always true
∴ (a, a) R a R
∴ R is reflexive
Let (a, b) ∈ R ⇒ a ≤ b which does not imply b ≤ a
∴ (b, a) ∉ R
∴ R is not symmetric
Let (a, b) ∈ R and (b, c) ∈ R
⇒ a ≤ b and b ≤ c
⇒ a ≤ c
⇒ (a, c) ∈ R
∴ R is transitive

Question 26.
Write \(\tan ^{-1}\left(\frac{\sqrt{1+x^{2}}-1}{x}\right)\), x ≠ 0 in the simplest form.
Solution:

Question 27.
If A and B are symmetric matrices of the same order, then show that AB is symmetric if and only if AB = BA.
Solution:
Let A and B are symmetric
∴ A’ = A and B’ = B
Let AB is symmetric
∴ (AB)’ = AB, B’A’ = AB, BA = AB
Conversely, Let AB = BA
(AB)’ = B’A’ = BA = AB
∴ AB is symmetric

Question 28.
Differentiate (log x) cos x with respect to x.
Solution:
Let cos x
y = (log x) cos x
log y = log (log x)
log y = cos x (log x)

Question 29.
Differentiate sin² x with respect to e^{cos x}.
Solution:

Question 30.
Find two positive numbers x and y such that x + y = 60 and xy³ is maximum.
Solution:
Let the two numbers be x, y and P = xy³
Given x + y = 60 ⇒ x = 60 – y
On putting this value in P = xy³, we get
P = (60 – y) y³ ⇒ P = 60y³ – y⁴
On differentiating twice w.r.t. y, we get
\(\frac{d P}{d y}\) = 180y² – 4y³
and \(\frac{d^{2} P}{d y^{2}}\) = 360y – 12y²
For maxima, we must have \(\frac{d P}{d y}\) = 0
⇒ 180y² – 4y³ = 0
⇒ 4y²(45 – y) = 0
⇒ y = 0, 45
But y ≠ 0, so y = 45
At y = 45, \(\left(\frac{d^{2} P}{d y^{2}}\right)_{y=45}\) = 360 × 45 – 12 × (45)²
=16200 – 24300
= -8100 < 0
⇒ P has afocal maxima at y = 45
∴ By second derivative test, x = 45 is a point of local maxima of P. Thus, the function xy³ is maximum when y = 45 and x = 60 – 45 = 15.
Hence, the required numbers are 15 and 45.

Question 31.
Evaluate \(\int \frac{2 x}{x^{2}+3 x+2} d x\)
Solution:

Question 32.
Evaluate ∫e^x sin x dx.
Solution:

Question 33.
Find the area of the region bounded by the curve y² = 4x, y-axis and the line y = 3.
Solution:

Question 34.
Form the differential equation of the family of circles having a centre on y-axis and radius 3 units.
Solution:
The equation of the family circles having centre on y-axis and radius 3 unit is
x² + (y – b)² = 9 …….(i)
On differentiating Eq. (i) w.r.t x, we get
2x + 2(y – b) y’ = 0
⇒ y – b = \(-\frac{x}{y^{\prime}}\) ……. (ii)
On substituting this value of (y – b) in Eq. (i) we get
\(x^{2}+\left(-\frac{x}{y^{\prime}}\right)^{2}=9\)
⇒ x²[(y’)² + 1] = 9(y’)²
⇒ (x² – 9)(y’)² + x² = 0
which is the required differential equation.

Question 35.
Find x, such that the four points A (3, 2, 1), B(4, x, 5), C(4, 2, -2) and D (6, 5, -1) are coplanar.
Solution:

Question 36.

Solution:

Question 37.
Find the shortest distance between the lines \(\vec{r}=\hat{i}+2 \hat{j}+\hat{k}+\lambda(\hat{i}-\hat{j}+\hat{k})\) and \(\vec{r}=2 \hat{i}-\hat{j}-\hat{k}+\mu(2 \hat{i}+\hat{j}+2 \hat{k})\)
Solution:

Question 38.
Given that the two numbers appearing on throwing two dice are different. Find the probability of the event ‘the sum of numbers on the dice is 4’.
Solution:
When dice is thrown, number of observations in the sample space S = 6 × 6 = 36 (equally likely sample events)
i.e., n(S) = 36
Let E : set of numbers in which numbers appearing on the two dice are different
Then, E = {(1, 3), (2, 2), (3, 1)}
⇒ n(E) = 3

n(F) = 30
Here, F contains all points of S except {(1, 1), (2, 2), (3, 3),(4, 4), (5, 5), (6, 6)}
E ∩ F = {(1, 3), (3, 1)}

Part – D

Answer any SIX questions: (6 × 5 = 30)

Question 39.
Let f : N → R be a function defined as f(x) = 4x² + 12x + 15. Show that f : N → S, where S is the range of f, is invertible. Find the inverse of f.
Solution:

Question 40.
If A = \(\left[\begin{array}{lll}1 & 0 & 2 \\0 & 2 & 1 \\2 & 0 & 3\end{array}\right]\) prove that A³ – 6A² + 7A + 2I = 0.
Solution:

Question 41.
Solve the following system of linear equations by matrix method.
x – y + 2z = 1, 2y – 3z = 1 and 3x – 2y + 4z = 2.
Solution:

Question 42.
If y = (tan^-1 x)². Show that (x² + 1)² y₂ + 2x(x² + 1) y₁ = 2.
Solution:

Question 43.
The length x of a rectangle is decreasing at the rate of 5 cm/minute and the width y is increasing at the rate of 4 cm/minute when x = 8 cm and y = 6 cm, find the rate of change of
(i) The perimeter and
(ii) The are of the rectangle.
Solution:
At any instant of time t, let length, breadth, perimeter and area of the rectangle are x, y, P and A respectively, then
P = 2(x + y) and A = xy …… (i)
It is given that \(\frac{d x}{d t}\) = -5 cm/mm and \(\frac{d y}{d t}\) = -5 = 4 cm/min
(-ve sign shows that the length is decreasing)
(i) Now, P = 2(x + y).
On differentiating w.r.t. t, we get Rate of change of pert meter
\(\frac{d P}{d t}=2\left(\frac{d x}{d t}+\frac{d y}{d t}\right)\)
= 2(-5 + 4) cm/min
= -2 cm/min [∴ \(\frac{d x}{d t}\) = -5 and \(\frac{d y}{d t}\) = 4]
Hence, perimeter of the rectangle is decreasing (-ve sign) at the rate of 2 cm/min.
(ii) Here, area of rectangle A = xy.
On differentiating w.r.t. t, we get
Rate of change \(\frac{d A}{d t}=x \frac{d y}{d t}+y \frac{d x}{d t}\)
= 8 × 4 + 6 × (5) [∴ \(\frac{d x}{d t}\) = -5 and \(\frac{d y}{d t}\) = 4]
= 32 – 30
= 2 cm²/min
Hence, area of the rectangle is increasing at the rate of 2 cm²/min.
Note: It rate of change is increasing, we take positive sign and if the rate of change is decreasing, then we take a negative sign.

Question 44.
Find the integral of \(\sqrt{x^{2}-a^{2}}\) with respect to x and hence evaluate \(\int \sqrt{x^{2}-8 x+7}\) dx
Solution:

Question 45.
Using integration find the area of the triangular region whose sides have the equations y = 2x + 1, y = 3x + 1 and x = 4.
Solution:
Given equation of sides of the triangle are y = 2x + 1, y = 3x + 1 and x = 4.
On solving these equations, we obtain the vertices of the triangle as A(0, 1), B(4, 13) and C(4, 9).
Required area (shown in the shaded region)
= Area (OLBAO) – Area (OLCAO)

= 8 sq.units

Question 46.
Solve the differential equation \(\cos ^{2} x \frac{d y}{d x}+y=\tan x\left(0 \leq x<\frac{\pi}{2}\right)\)
Solution:

Question 47.
Derive the equation of a plane perpendicular to a given vector and passing through a given point both in vector and Cartesian form.
Solution:
Vector Form
Let a plane pass through a point A with position vector \(\overrightarrow{\mathrm{a}}\) and perpendicular to the vector \(\overrightarrow{\mathrm{N}}\)
Let \(\overrightarrow{\mathrm{r}}\) be the position vector of any point P(x, y, z) in the plane.
Then the point P lies in the plane if and only if \(\overline{\mathrm{AP}}\) is perpendicular to \(\overrightarrow{\mathrm{N}}\)
i.e. \(\overline{\mathrm{AP}} \cdot \overline{\mathrm{N}}=0\)
But \(\overline{\mathrm{AP}}=\overrightarrow{\mathrm{r}}-\overrightarrow{\mathrm{a}}\)
Therefore \((\overrightarrow{\mathbf{r}}-\overrightarrow{\mathrm{a}}) \cdot \overrightarrow{\mathrm{N}}=0\) …….. (1)
This is the vector equation of the plane.

Cartesian form
Let the given point A be (x₁, y₁, z₁), P be (x, y, z) and direction ratios of
\(\overline{\mathrm{N}}\) are A, B and C. Then
\(\overrightarrow{\mathrm{a}}=x_{1} \hat{\mathrm{i}}+\mathrm{y}_{1} \hat{\mathrm{j}}+\mathrm{z}_{1} \hat{\mathrm{k}}, \quad \overrightarrow{\mathrm{r}}=x \hat{\mathrm{i}}+\mathrm{y} \hat{\mathrm{j}}+\mathrm{z} \hat{\mathrm{k}}\) and \(\overline{\mathrm{N}}=\mathrm{A} \hat{\mathrm{i}}+\mathrm{B} \hat{\mathrm{j}}+\mathrm{C} \hat{\mathrm{k}}\)

Question 48.
The probability that a bulb produced by a factory will fuse after 150 days of use is 0.05. Find the probability that out of 5 such bulbs.
(i) none
(ii) not more than one
(iii) more than one will fuse after 150 days of use.
Solution:
Let X represents the number of bulbs that will fuse after 150 days of use in an experiment of 5 trials. The trials are Bernoulli trials, n = 5p = p (success) = 0.05, q = 1 – p = 1 – 0.05 = 0.95

Part – E

Answer ary ONE question: (1 × 10 = 10)

Question 49(a).

Solution:

Question 49(b).
Show that \(\left|\begin{array}{ccc}x & x^{2} & y z \\y & y^{2} & z x \\z & z^{2} & x y\end{array}\right|\) = (x – y) (y – z) (z – x) (xy + yz + zx)
Solution:


= (y + x)(y – x)(z – x)(z³ + x² + xz) – (z + x)(z – x)(y – x)(y² + x² + xy)
= (y – x)(z – x)[(y – x)(z³ + x² + xz) – (z + x)(y² + x² + xy)]
= (y – x)(z – x) [(yz² + yx² + xyz + xz² + x³ + x²z – zy² – zx² – xyz – xy² – x³ – x²y]
= (y – x)(z – x)[yz² + zy² + xz² – xy²]
= (y – x) (z – x) [yz (z – y) + x (z – y)(z + y)]
= (y – x) (z – x) [(z – y) (xy + yz + zx)]
= (x – y) (y – z) (z – x) (xy + yz + zx)
= RHS
Hence proved.

Question 50(a).
Minimize and maximize z = 600x + 400y
Subject to the constraints:
x + 2y ≤ 12
2x + y ≤ 12
4x + 5y ≥ 20 and x ≥ 0, y ≥ 0 by graphical method.
Solution:
x + 2y = 12
y = 0 ⇒ x = 12 ∴ P (12, 0)
x = 0 ⇒ y = 6 ∴ D (0, 6)
2x + 2y = 12
y = 0 ⇒ x = 6 ∴ B (6, 0)
x = 0 ⇒ y = 12 ∴ F (0, 12)
4x + 5y = 20
y = 0 ⇒ x = 5 ∴ A (5, 0)
x = 0 ⇒ y = 4 ∴ E (0, 4)

Question 50(b).
Find the value of k, if

Solution: