TS Inter 1st Year Maths 1A Question Paper May 2017

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TS Inter 1st Year Maths 1A Question Paper May 2017

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 A = \(\left\{0, \frac{\pi}{6}, \frac{\pi}{4}, \frac{\pi}{3}, \frac{\pi}{2}\right\}\) and f: A → B is a surjection defined by f(x) = cos x, then find B.
Solution:

Question 2.
If f: R → R, g: R → R are defined by f(x) = 3x – 1, g(x) = x² + 1, then find:
(i) fog (2)
(ii) gof (2a – 3)
Solution:
(i) (fog) (2) = f[g(2)]
= f[2² + 1]
= f(5)
= 3(5) – 1
= 15 – 1
= 14
∴ (fog) (2) = 14
(ii) (gof) (2a – 3) = g[f(2a – 3)]
= g[3(2a – 3) – 1]
= g[6a – 9 – 1]
= g(6a – 10)
= (6a – 10)² + 1
= 36a² – 120a + 100 + 1
= 36a² – 120a + 101
∴ (gof) (2a – 3) = 36a² – 120a + 101

Question 3.
Define the Skew-symmetric matrix and give one example of it.
Solution:
Skew-Symmetric matrix: A square matrix is said to be a skew-symmetric matrix if A^T = -A.
Ex: A = \(\left[\begin{array}{rrr}
0 & a & b \\
-a & 0 & c \\
-b & -c & 0
\end{array}\right]\)

Question 4.
If A = \(\left[\begin{array}{lll}
3 & 0 & 0 \\
0 & 3 & 0 \\
0 & 0 & 3
\end{array}\right]\), then find A⁴.
Solution:

Question 5.
Show that the points whose position vectors are \(-2 \bar{a}+3 \bar{b}+5 \bar{c}\), \(\bar{a}+2 \bar{b}+3 \bar{c}, 7 \bar{a}-\bar{c}\) are collinear when \(\bar{a}, \bar{b}, \bar{c}\) are non-coplanar vectors.
Solution:

Question 6.
Find the unit vector in the direction of the vector \(\bar{a}=2 \bar{i}+3 \bar{j}+\bar{k}\).
Solution:

Question 7.
Find the area of the parallelogram whose diagonals are \(3 \bar{i}+\bar{j}-2 \bar{k}\) and \(\overline{\mathrm{i}}-3 \overline{\mathrm{j}}+4 \overline{\mathrm{k}}\).
Solution:

Question 8.
Find the period of the function f(x) = cos (3x + 5) + 7.
Solution:
Given f(x) = cos (3x + 5) + 7
∴ The period of f(x) = \(\frac{2 \pi}{3}\)

Question 9.
If sin α = \(\frac{1}{\sqrt{10}}\), sin β = \(\frac{1}{\sqrt{5}}\) and α, β are acute, then show that α + β = \(\frac{\pi}{4}\).
Solution:

Question 10.
If cosh x = \(\frac{5}{2}\), then find the values of cosh (2x), sinh (2x).
Solution:
Given cos hx = \(\frac{5}{2}\)

Section – B
(5 × 4 = 20 Marks)

II. Short Answer Type Questions.

• Attempt any five questions.
• Each question carries four marks.

Question 11.
If θ – φ = \(\frac{\pi}{2}\), then show that \(\left[\begin{array}{cc}
\cos ^2 \theta & \cos \theta \sin \theta \\
\cos \theta \sin \theta & \sin ^2 \theta
\end{array}\right]\left[\begin{array}{cc}
\cos ^2 \phi & \cos \phi \sin \phi \\
\cos \phi \sin \phi & \sin ^2 \phi
\end{array}\right]\) = 0
Solution:

Question 12.
Let ABCDEF be a regular hexagon with a center ‘O’. Show that \(\overline{\mathrm{AB}}+\overline{\mathrm{AC}}+\overline{\mathrm{AD}}+\overline{\mathrm{AE}}+\overline{\mathrm{AF}}=3 \overline{\mathrm{AD}}=6 \overline{\mathrm{AO}}\).
Solution:
Given that ABCDEF is a regular hexagon.

Question 13.
If \(\bar{a}=2 \bar{i}+3 \bar{j}+4 \bar{k}, \bar{b}=\bar{i}+\bar{j}-\bar{k}\) and \(\bar{c}=\bar{i}-\bar{j}+\vec{k}\), then compute \(\overline{\mathrm{a}} \times(\overline{\mathrm{b}} \times \overline{\mathrm{c}})\) and verify that it is perpendicular to \(\bar{a}\).
Solution:

Question 14.
Prove that \(\cot \frac{\pi}{16} \cdot \cot \frac{2 \pi}{16} \cdot \cot \frac{3 \pi}{16} \ldots \ldots \ldots \cot \frac{7 \pi}{16}\) = 1.
Solution:

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

Question 16.
If tan^-1 x + tan^-1 y + tan^-1 z = π, then show that x + y + z = xyz.
Solution:
Let tan^-1 x = A ⇒ x = tan A
tan^-1 y = B ⇒ y = tan B
tan^-1 z = C ⇒ z = tan C
Given tan^-1 x + tan^-1 y + tan^-1 z = π
⇒ A + B + C = 180°
⇒ A + B = 180° – C
⇒ tan (A + B) = tan (180° – C)
⇒ \(\frac{\tan A+\tan B}{1-\tan A \tan B}\) = -tan C
⇒ \(\frac{x+y}{1-x y}\) = -z
⇒ x + y = -z + xyz
⇒ x + y + z = xyz

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

Section – C
(5 × 7 = 35 Marks)

III. Long Answer Type Questions.

• Attempt any five questions.
• Each question carries Seven marks.

Question 18.
If f: A → B, g: B → C are bijective functions, then show that gof: A → C is a bijection.
Solution:
∵ f: A → B, g: B → C by composite function gof: A → C
∵ f: A → B, g: B → C are bijections
Let a₁, a₂ ∈ A be such that (gof) (a₁) = (gof) (a₂)
⇒ g(f(a₁)) = g(f(a₂))
⇒ f(a₁) = f(a₂); {∵ g is an injection (bijection)}
⇒ a₁ = a₂; {∵ f is an injection (bijection)}
∴ gof: A → C is an injection
Let c ∈ C, since g: B → C is surjection, (bijection)
there exists b ∈ B such that g(b) = c.
Since f: A → B is surjection, (bijection)
there exists a ∈ A; such that f(a) = b
∴ C = g(b)
= g(f(a))
= (gof) (a)
∴ For each c ∈ C, there exists “a” ∈ A such that (gof) (a) = c
Hence gof: A → C is a surjection
Therefore gof: A → C is bijection.

Question 19.
Show that ∀ n ∈ N, \(\frac{1}{1.4}+\frac{1}{4.7}+\frac{1}{7.10}\) + ……… upto n terms = \(\frac{n}{3 n+1}\) by mathematical induction.
Solution:
In denominator
1, 4, 7,……. are in A.P
nth term = 1 + (n – 1)3
= 1 + 3n – 3
= 3n – 2
4, 7, 10,……. are in A.P
nth term = 4 + (n – 1)3
= 4 + 3n – 3
= 3n + 1
nth term of the series is \(\frac{1}{(3 x-2)(3 x+1)}\)
Let P(n) be the statement that
\(\frac{1}{1.4}+\frac{1}{4.7}+\frac{1}{7.10}+\ldots \ldots .+\frac{1}{(3 n-2)(3 n+1)}=\frac{n}{3 n+1}\)
If n = 1 then
L.H.S = \(\frac{1}{1.4}=\frac{1}{4}\)
R.H.S = \(\frac{1}{3(1)+1}=\frac{1}{4}\)
∴ L.H.S = R.H.S
∴ P(1) is true.
Assume that P(K) is true.

∴ P(k + 1) is true.
By the principle of finite mathematical induction P(n) in for all position integral values of n.

Question 20.
Show that \(\left|\begin{array}{ccc}
a & b & c \\
a^2 & b^2 & c^2 \\
a^3 & b^3 & c^3
\end{array}\right|\) = abc (a – b) (b – c) (-a).
Solution:

Question 21.
Solve the system of equations x + y + z = 3, 2x + 2y – z = 3, x + y – z = 1 by the Gauss-Jordan method.
Solution:
Given system of equations are
x + y + z = 3
2x + 2y – z = 3
x + y – z = 1
The given system of equations can be written as AX = D


Here Rank (A) = 2, Rank ([AD]) = 2 and n = 3
∴ Rank (A) = Rank ([AD]) < n
∴ The given system is consistent and it has infinitely many solutions.
∴ x + y + z = 3
z = 1
If x = k then y = 2 – k
∴ x = k, y = 2 – k, z = 1, k ∈ R.

Question 22.
Find the shortest distance between the skew lines \(\bar{r}=(6 \bar{i}+2 \bar{j}+2 \bar{k})+t(\bar{i}-2 \bar{j}+2 \bar{k})\) and \(\bar{r}=(-4 \bar{i}-\bar{k})+s(3 \bar{i}-2 \bar{j}-2 \bar{k})\).
Solution:
Let \(\overline{\mathrm{a}}=6 \overline{\mathrm{i}}+2 \overline{\mathrm{j}}+2 \overline{\mathrm{k}}\)
\(\overline{\mathrm{b}}=\overline{\mathrm{i}}-2 \overline{\mathrm{j}}+2 \overline{\mathrm{k}}\)
\(\bar{c}=-4 \bar{i}-\bar{k}\)
\(\overline{\mathrm{d}}=3 \overline{\mathrm{i}}-2 \overline{\mathrm{j}}-2 \overline{\mathrm{k}}\)
The equation of the line through \(\bar{a}\) and parallel to \(\bar{b}\) is
\(\overline{\mathrm{r}}=\overline{\mathrm{a}}+\mathrm{t} \overline{\mathrm{b}}\) ……(1)
The equation of the line through \(\bar{c}\) and parallel to \(\bar{d}\) is
\(\overline{\mathrm{r}}=\overline{\mathrm{c}}+s \overline{\mathrm{d}}\) ……(2)
The shortest distance between the Skew lines (1) and (2) is \(\frac{|[\overline{\mathbf{a}}-\overline{\mathbf{c}} \overline{\mathrm{b}} \overline{\mathrm{d}}]|}{|\overline{\mathrm{b}} \times \overline{\mathrm{d}}|}\)

Question 23.
If A + B + C = π, then prove that \(\cos ^2 \frac{A}{2}+\cos ^2 \frac{B}{2}+\cos ^2 \frac{C}{2}=2\left(1+\sin \frac{A}{2} \cdot \sin \frac{B}{2} \cdot \sin \frac{C}{2}\right)\).
Solution:

Question 24.
In a ΔABC, if a = 13, b = 14, c = 15 then show that R = \(\frac{65}{8}\), r = 4, r₁ = \(\frac{21}{2}\), r₂ = 12 and r₃ = 14.
Solution: