TS Inter 1st Year Maths 1A Question Paper March 2020

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TS Inter 1st Year Maths 1A Question Paper March 2020

Time: 3 Hours
Maximum Marks: 75

Note: This question paper consists of three sections A, B, and C.

Section – A
(10 × 2 = 20 Marks)

I. Very Short Answer Type Questions.

• Answer ALL questions.
• Each question carries TWO marks.

Question 1.
If f: R → R is defined by f(x) = \(\frac{1-x^2}{1+x^2}\), then show that f(tan θ) = cos 2θ.
Solution:

Question 2.
Find the domain of the real valued function f(x) = \(\frac{1}{\left(x^2-1\right)(x+3)}\).
Solution:
f(x) = \(\frac{1}{\left(x^2-1\right)(x+3)}\) ∈ R
⇔ (x² – 1) (x + 3) ≠ 0
⇔ (x + 1) (x – 1) (x + 3) ≠ 0
⇔ x ≠ -1, 1, -3
∴ Domain of f is R – {-1, 1, -3}

Question 3.
If \(\left[\begin{array}{ccc}
x-1 & 2 & 5-y \\
0 & z-1 & 7 \\
-1 & 0 & a-5
\end{array}\right]=\left[\begin{array}{lll}
1 & 2 & 3 \\
0 & 4 & 7 \\
1 & 0 & 0
\end{array}\right]\), then find the values of x, y, z and a.
Solution:
Given \(\left[\begin{array}{ccc}
x-1 & 2 & 5-y \\
0 & z-1 & 7 \\
-1 & 0 & a-5
\end{array}\right]=\left[\begin{array}{lll}
1 & 2 & 3 \\
0 & 4 & 7 \\
1 & 0 & 0
\end{array}\right]\)
∴ x – 1 = 1 ⇒ x = 1 + 1 = 2
∴ 5 – y = 3 ⇒ y = 5 – 3 = 2
∴ z – 1 = 4 ⇒ z = 4 + 1 = 5
∴ a – 5 = 0 ⇒ a = 5

Question 4.
Define the Rank of a matrix.
Solution:
Let A be a non-zero matrix, the rank of A is defined as the maximum of the orders of the non-singular square submatrices of A. The rank of a null matrix is zero. The rank of a matrix A is denoted as rank(A) of p(A).

Question 5.
If the vectors \(-3 \overline{\mathrm{i}}+4 \overline{\mathrm{j}}+\lambda \overline{\mathrm{k}}\) and \(\mu \overline{\mathrm{i}}+8 \overline{\mathrm{j}}+6 \overline{\mathrm{k}}\) are collinear vectors, then find λ and µ.
Solution:
If the vectors \(a_1 \bar{i}+b_1 \bar{j}+c_1 \bar{k}\) and \(a_2 \bar{i}+b_2 \bar{j}+c_2 \bar{k}\) are collinear then \(\frac{a_1}{a_2}=\frac{b_1}{b_2}=\frac{c_1}{c_2}\)
⇒ \(\frac{-3}{\mu}=\frac{4}{8}=\frac{\lambda}{6}\)
⇒ \(\frac{-3}{\mu}=\frac{1}{2}\) ⇒ µ = -6
and \(\frac{\lambda}{6}=\frac{1}{2}\) ⇒ λ = 3
∴ λ = 3 and µ = -6

Question 6.
Find the vector equation of the plane passing through the points (0, 0, 0), (0, 5, 0) and (2, 0, 1).
Solution:

Question 7.
Find-the angle between the planes \(\bar{r} \cdot(2 \bar{i}-\bar{j}+2 \bar{k})=3\) and \(\bar{r} \cdot(3 \bar{r}+6 \bar{j}+\bar{k})=4\).
Solution:

Question 8.
Find a cosine function whose period is 7.
Solution:
\(\frac{2 \pi}{|k|}\) = 7
⇒ \(\frac{2 \pi}{7}\) = |k|
∴ cos kx = cos \(\frac{2 \pi}{7}\)x

Question 9.
What is the value of tan 20° + tan 40° + √3 tan 20° tan 40°?
Solution:
consider 20° + 40° = 60°
⇒ tan (20° + 40°) = tan 60°
⇒ \(\frac{\tan 20^{\circ}+\tan 40^{\circ}}{1-\tan 20^{\circ} \tan 40^{\circ}}\) = √3
⇒ tan 20° + tan 40° = √3 – √3 tan 20° tan 40°
⇒ tan 20° + tan 40° + √3 tan 20° tan 40° = √3

Question 10.
For any x ∈ R, prove that cosh⁴x – sinh⁴x = cosh(2x).
Solution:
L.H.S = cosh⁴x – sinh⁴x
= (cosh²x)² – (sinh²x)²
= [cosh²x – sinh²x] [cosh²x + sinh²x]
= (1) cosh (2x)
= cosh (2x)
∴ cosh⁴x – sinh⁴x = cosh (2x)

Section – B
(5 × 4 = 20 Marks)

II. Short Answer Type Questions.

• Answer ANY FIVE questions.
• Each question carries FOUR marks.

Question 11.
If A = \(\left[\begin{array}{lll}
3 & 0 & 0 \\
0 & 3 & 0 \\
0 & 0 & 3
\end{array}\right]\), then find A⁴.
Solution:

Question 12.
If the points whose position vectors are \(3 \overline{\mathrm{i}}-2 \overline{\mathrm{j}}-\overline{\mathrm{k}}, 2 \overline{\mathrm{i}}+3 \overline{\mathrm{j}}-4 \overline{\mathrm{k}}\), \(-\bar{i}+\bar{j}+2 \bar{k}\) and \(4 \bar{i}+5 \bar{j}+\lambda \bar{k}\) are coplanar, then show that λ = \(\frac{-146}{17}\).
Solution:
Let ‘O’ be the origin and Let A, B, C, and D be the given points. Then

Question 13.
If |\(\bar{a}\)| = 13, |\(\bar{b}\)| = 5 and \(\overrightarrow{\mathrm{a}} \cdot \overline{\mathrm{b}}\) = 60, then find \(|\overline{\mathrm{a}} \times \overline{\mathrm{b}}|\).
Solution:
We have \(|\bar{a} \times \bar{b}|^2=|\bar{a}|^2|\bar{b}|^2-(\bar{a} \cdot \bar{b})^2\)
= (13)² (5)² – (60)²
= 169 × 25 – 3600
= 4225 – 3600
= 625
∴ \(|\bar{a} \times \bar{b}|=\sqrt{625}\) = 25

Question 14.
Prove that \(\cos \frac{2 \pi}{7} \cos \frac{4 \pi}{7} \cos \frac{8 \pi}{7}=\frac{1}{8}\).
Solution:

Question 15.
Solve the equation √3 sin θ – cos θ = √2.
Solution:

Question 16.
Prove that \(\sin ^{-1} \frac{4}{5}+2 \tan ^{-1} \frac{1}{3}=\frac{\pi}{2}\).
Solution:

Question 17.
In ΔABC, show that \(b \cos ^2 \frac{C}{2}+c \cos ^2 \frac{B}{2}\) = s.
Solution:

Section – C
(5 × 7 = 35 Marks)

III. Long Answer Type Questions.

• Answer ANY FIVE questions.
• Each question carries SEVEN marks.

Question 18.
If f: A → B, g: B → C are bijections, then show that (g o f)^-1 = f^-1 o g^-1.
Solution:
f: A → B, g: B → C are bijections
⇒ gof: A → C is a bijection
Also g^-1: C → B and f^-1: B → A are bijections
⇒ f^-1 o g^-1: C → A is a bijection.
Let c be any element of C.
Then ∃ an element b ∈ B such that g(b) = c
⇒ b = g^-1(c)
Also ∃ an element A such that f(a) = b
⇒ a = f^-1(b)
Now (g o f) (a) = g(f(a)
= g(b)
= c
⇒ a = (g o f)^-1 (c)
⇒ (g o f)^-1 (c) = a
Also (f^-1 o g^-1) (c) = f^-1 (g^-1 (c))
= f^-1 (b)
= a
From (1) and (2),
(g o f)^-1 (c) = (f^-1 o g^-1 ( c))
⇒ (g o f)^-1 = f^-1 o g^-1

Question 19.
Using mathematical induction, prove the statement:
a + ar + ar² + ……..upto n terms = \(\frac{a\left(r^n-1\right)}{(r-1)}\), r ≠ 1.
Solution:
Let p(n) be the statement:
a + ar + ar² + …….. + a . r^n-1 = \(\frac{a\left(r^n-1\right)}{(r-1)}\), r ≠ 1
and let S(n) be the sum on the L.H.S
Since S(1) = a = \(\frac{a\left(r^1-1\right)}{r-1}\) = a
∴ p(1) is true
Assume that the statement is true for n = k
(i.e) S(k) = a + ar + ar² + …….. + a . r^k-1 = \(\frac{a\left(r^k-1\right)}{(r-1)}\)
We show that the statement is true for n = k + 1


∴ The statement holds for n = k + 1
∴ By the principle of mathematical induction, p(n) is true for all n ∈ N.
∴ a + ar + ar² + …….. + a . r^n-1 = \(\frac{a\left(r^n-1\right)}{(r-1)}\), r ≠ 1

Question 20.
If \(\left|\begin{array}{lll}
a & a^2 & 1+a^3 \\
\text { b } & b^2 & 1+b^3 \\
c & c^2 & 1+c^3
\end{array}\right|\) = 0 and \(\left|\begin{array}{lll}
a & a^2 & 1 \\
b & b^2 & 1 \\
c & c^2 & 1
\end{array}\right|\) ≠ 0, then show that abc = -1.
Solution:
Hint: If each element in a row (column) of a square matrix is the sum of two numbers, then its discriminant can be expressed as the sum of discriminants of two square matrices.

= (1 + abc) \(\left|\begin{array}{lll}
1 & a & a^2 \\
1 & b & b^2 \\
1 & c & c^2
\end{array}\right|\) = 0
∴ 1 + abc = 0
⇒ abc = -1

Question 21.
Solve the system of equations x + y + z = 1, 2x + 2y + 3z = 6, x + 4y + 9z = 3 by using Cramer’s rule.
Solution:
Cramer’s rule:
Δ = \(\left|\begin{array}{lll}
1 & 1 & 1 \\
2 & 2 & 3 \\
1 & 4 & 9
\end{array}\right|\)
= 1(18 – 12) – 1(18 – 3) + 1(8 – 2)
= 6 – 15 + 6
= -3
Δ₁ = \(\left|\begin{array}{lll}
1 & 1 & 1 \\
6 & 2 & 3 \\
3 & 4 & 9
\end{array}\right|\)
= 1(18 – 12) – 1(54 – 9) + 1(24 – 6)
= 6 – 45 + 18
= -21
Δ₂ = \(\left|\begin{array}{lll}
1 & 1 & 1 \\
2 & 6 & 3 \\
1 & 3 & 9
\end{array}\right|\)
= 1(54 – 9) – 1(18 – 3) + 1(6 – 6)
= 45 – 15
= 30
Δ₃ = \(\left|\begin{array}{lll}
1 & 1 & 1 \\
2 & 2 & 6 \\
1 & 4 & 3
\end{array}\right|\)
= 1(6 – 24) – 1(6 – 6) + 1(8 – 2)
= -18 – 0 + 6
= -12
x = \(\frac{\Delta_1}{\Delta}=\frac{-21}{-3}\) = 7
y = \(\frac{\Delta_2}{\Delta}=\frac{30}{-3}\) = -10
z = \(\frac{\Delta_3}{\Delta}=\frac{-12}{-3}\) = 4
∴ Solution is x = 7, y = -10, z = 4

Question 22.
Find the volume of the tetrahedron whose vertices are (1, 2, 1), (3, 2, 5), (2, -1, 0) and (-1, 0, 1).
Solution:
Let ‘O’ be the given A, B, C, D be the vertices of the ten tetrahedron. Then

Question 23.
If A, B, C are angles in a triangle, then prove that sin A + sin B + sin C = \(4 \cos \frac{A}{2} \cos \frac{B}{2} \cos \frac{C}{2}\).
Solution:
∴ A, B, C are angles of a triangle
⇒ A + B + C = 180°
L.H.S = (sin A + sin B) + sin C

Question 24.
If r : R : r₁ = 2 : 5 : 12, then prove that the triangle is right-angled at A.
Solution:
r : R : r₁ = 2 : 5 : 12
then r = 2k, R = 5k and r₁ = 12K
r₁ – r = 12k – 2k = 10k = 2(5k) = 2R

Hence, the triangle is right-angled at A.