Thoroughly analyzing TS Inter 1st Year Maths 1A Model Papers and TS Inter 1st Year Maths 1A Question Paper May 2017 helps students identify their strengths and weaknesses.

## 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) = x^{2} + 1, then find:

(i) fog (2)

(ii) gof (2a – 3)

Solution:

(i) (fog) (2) = f[g(2)]

= f[2^{2} + 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

= 36a^{2} – 120a + 100 + 1

= 36a^{2} – 120a + 101

∴ (gof) (2a – 3) = 36a^{2} – 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 A^{4}.

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 a_{1}, a_{2} ∈ A be such that (gof) (a_{1}) = (gof) (a_{2})

⇒ g(f(a_{1})) = g(f(a_{2}))

⇒ f(a_{1}) = f(a_{2}); {∵ g is an injection (bijection)}

⇒ a_{1} = a_{2}; {∵ 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, r_{1} = \(\frac{21}{2}\), r_{2} = 12 and r_{3} = 14.

Solution: