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

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TS Inter 1st Year Maths 1A Question Paper March 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 surjection defined by f(x) = cos x then find B.
Solution:

Question 2.
If f(x) = 2, g(x) = x², h(x) = 2x for all x ∈ R, then find (fo(goh))(x).
Solution:
Given f(x) = 2, g(x) = x² and h(x) = 2x
[fo(goh)] (x) = f[(goh)(x)]
= f[g{h(x)}]
= f[g(2x)]
= f[(2x)²]
= f(4x²)
= 2
Hence [fo(goh)] (x) = 2

Question 3.
If A = \(\left[\begin{array}{ccc}
3 & 2 & -1 \\
2 & -2 & 0 \\
1 & 3 & 1
\end{array}\right]\), B = \(\left[\begin{array}{ccc}
-3 & -1 & 0 \\
2 & 1 & 3 \\
4 & -1 & 2
\end{array}\right]\) and X = A + B then find X.
Solution:

Question 4.
If A = \(\left[\begin{array}{rr}
-1 & 2 \\
0 & 1
\end{array}\right]\) then find AA’.
Solution:

Question 5.
a = 2i + 5j + k and b = 4i + mj + nk are collinear vectors then find m and n.
Solution:

Question 6.
Find the vector equation of the line passing through the point \(2 \bar{i}+3 \bar{j}+\bar{k}\) and parallel to the vector \(4 \overline{\mathrm{i}}-2 \overline{\mathrm{j}}+3 \overline{\mathrm{k}}\).
Solution:

Question 7.
Find the angle between the vectors i + 2j + 3k and 3i – j + 2k.
Solution:

Question 8.
If sin θ = \(\frac{4}{5}\) and θ is not in the first quadrant, find the value of cos θ.
Solution:
Given sin θ = \(\frac{4}{5}\), θ is not in the first quadrant.
sin θ = \(\frac{4}{5}\) > 0
∴ ‘θ’ lies in second quadrant.
cos²θ = 1 – sin²θ
= 1 – \(\left(\frac{4}{5}\right)^2\)
= 1 – \(\frac{16}{25}\)
= \(\frac{9}{25}\)
∴ cos θ = ±\(\frac{3}{5}\)
since ‘θ’ lies in the second quadrant.
∴ cos θ = \(\frac{3}{5}\)

Question 9.
Prove that cos 48° . cos 12° = \(\frac{3+\sqrt{5}}{8}\).
Solution:
L.H.S = cos 48° . cos12°
= \(\frac{1}{2}\) [2 cos 48° . cos 12°]
= \(\frac{1}{2}\) [cos (48° + 12°) + cos (48° – 12°)]
= \(\frac{1}{2}\) [cos 60° + cos 36°]
= \(\frac{1}{2}\left[\frac{1}{2}+\frac{\sqrt{5}+1}{4}\right]\)
= \(\frac{1}{2}\left[\frac{2+\sqrt{5}+1}{4}\right]\)
= \(\frac{3+\sqrt{5}}{8}\)
= R.H.S
∴ L.H.S = R.H.S
Hence cos 48° . cos 12° = \(\frac{3+\sqrt{5}}{8}\)

Question 10.
If cosh x = \(\frac{5}{2}\), find the values of (i) cosh (2x) and (ii) sinh (2x).
Solution:

Section – B
(5 × 4 = 20 Marks)

II. Short Answer Type Questions

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

Question 11.
Show that \(\left|\begin{array}{lll}
1 & a & a^2 \\
1 & b & b^2 \\
1 & c & c^2
\end{array}\right|\) = (a – b) (b – c) (c – a).
Solution:

Question 12.
If a, b, c are non-coplanar find the points of intersection of the line passing through the points 2a + 3b – c, 3a + 4b – 2c with the line joint the points a – 2b + 3c, a – 6b + 6c.
Solution:
The vector equation of the line passing through the points

If (1) and (2) intersect then
2 + t = 1 ⇒ t = -1
3 + t = -2 – 4s
⇒ 3 + (-1) = -2 – 4s
⇒ 2 = -2 – 4s
⇒ 4 = -4s
⇒ s = -1
∴ t = -1, s = -1
∴ point of intersection of (1) and (2) is
\(\bar{r}\) = [2 + (-1)] \(\bar{a}\) + [3 + (-1)] \(\bar{b}\) + [-(-1) – 1] \(\bar{c}\)
= (2 – 1) \(\bar{a}\) + (3 – 1) \(\bar{b}\) + (1 – 1) \(\bar{c}\)
= \(\bar{a}\) + 2\(\bar{b}\)

Question 13.
If a = 2i + j – k, b = -i + 2j – 4k and c = i + j + k then find (a × b) . (b × c).
Solution:

Question 14.
(i) Find the range of 13 cos x + 3√3 sin x – 4.
Solution:
Given function 13 cos x + 3√3 sin x – 4
Here a = 13, b = 3√3 and c = -4

∴ Range = [-18, 10]

(ii) Evaluate \(\sin ^2 82 \frac{1}{2}^{\circ}-\sin ^2 22 \frac{1}{2}^{\circ}\).
Solution:

Question 15.
Solve 1 + sin²θ = 3 sin θ . cos θ.
Solution:
Given 1 + sin²θ = 3 sin θ cos θ
⇒ \(\frac{1}{\cos ^2 \theta}+\frac{\sin ^2 \theta}{\cos ^2 \theta}=\frac{3 \sin \theta \cos \theta}{\cos ^2 \theta}\)
⇒ sec²θ + tan²θ = 3 tan θ
⇒ 1 + tan²θ + tan²θ = 3 tan θ
⇒ 2 tan²θ – 3 tan θ + 1 = 0
⇒ 2 tan²θ – 2 tan θ – tan θ + 1 = 0
⇒ 2 tan θ (tan θ – 1) – 1(tan θ – 1) = 0
⇒ (tan θ – 1) (2 tan θ – 1) = 0
⇒ tan θ – 1 = 0 (or) 2 tan θ – 1 = 0
∴ tan θ = 1
⇒ tan θ = tan \(\frac{\pi}{4}\)
⇒ θ = nπ + \(\frac{\pi}{4}\), n ∈ z
∴ 2 tan θ = 1
⇒ tan θ = \(\frac{1}{2}\)
⇒ tan θ = tan α
where α is principal value of tan θ = \(\frac{1}{2}\)
∴ θ = nπ + α, n ∈ z.
∴solution set X = {nπ +\(\frac{\pi}{4}\)/n ∈ z} ∪ {nπ + α/n ∈ z}

Question 16.
Show that \(\cot \left(\sin ^{-1} \sqrt{\frac{13}{17}}\right)=\sin \left(\tan ^{-1} \frac{2}{3}\right)\).
Solution:

Question 17.
In ΔABC, if \(\frac{1}{a+c}+\frac{1}{b+c}=\frac{3}{a+b+c}\), show that C = 60°.
Solution:
Given \(\frac{1}{a+c}+\frac{1}{b+c}=\frac{3}{a+b+c}\)
⇒ \(\frac{1}{a+c}+\frac{1}{b+c}-\frac{3}{a+b+c}\) = 0
⇒ \(\frac{(b+c)(a+b+c)+(a+c)(a+b+c)-3(a+c)(b+c)}{(a+c)(b+c)(a+b+c)}\) = 0
⇒ (b + c) (a + b + c) + (a + c) (a + b + c) – 3(a + c) (b + c) = 0
⇒ (b + c + a + c) (a + b + c) – 3(a + c) (b + c) = 0
⇒ (a + b + 2c) (a + b + c) – 3(ab + ac + bc + c²) = 0
⇒ a² + ab + ac + ba + b² + bc + 2ac + 2bc + 2c² – 3ab – 3ac – 3bc – 3c² = 0
⇒ a² + b² – c² – ab = 0
⇒ 2ab cos c = ab {∵ By cosine rule c² = a² + b² – 2ab cos c}
⇒ cos c = \(\frac{1}{2}\)
⇒ cos c = cos 60°
⇒ c = 60°

Section – C
(5 × 7 = 35 Marks)

III. Long Answer Type Questions

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

Question 18.
(i) If f: Q → Q is defined by f(x) = 5x + 4, ∀ x ∈ Q, show that f is a bijection and find f^-1.
Solution:
Given f: Q → Q is defined by f(x) = 5x + 4 ∀ x ∈ Q.
Let x₁, x₂ ∈ Q and f(x₁) = f(x₂)
⇒ 5x₁ + 4 = 5x₂ + 4
⇒ 5x₁ = 5x₂
⇒ x₁ = x₂
∴ f is one-one.
Let y ∈ Q (codomain) then there exists x = \(\frac{y-4}{5}\)
such that f(x) = f(\(\frac{y-4}{5}\))
= 5(\(\frac{y-4}{5}\)) + 4
= y – 4 + 4
= y
∴ f is onto
∴ f is one-one and onto
∴ f is a bijection.
∴ f^-1; Q → Q is also a bijection.
Let f(x) = y
⇒ 5x + 4 = y
⇒ 5x = y – 4
⇒ x = \(\frac{y-4}{5}\)
⇒ f^-1(y) = \(\frac{y-4}{5}\)
⇒ f^-1(x) = \(\frac{x-4}{5}\)

(ii) If f = {(4, 5), (5, 6), (6, -4)} and g = {(4, -4), (6, 5), (8, 5)} then find f + g and fg.
Solution:
Given f = {(4, 5), (5, 6), (6, -4)}
g = {(4, -4), (6, 5), (8, 5)}
∴ Domain of f is A = {4, 5, 6}
Domain of g is B = {4, 6, 8}
Domain of f + g is A ∩ B = {4, 6}
∴ f + g = {(4, 5 – 4), (6, -4 + 5)} = {(4, 1), (6, 1)}
Domain of fg is A ∩ B = {4, 6}
∴ fg = {(4, (5)(-4), (6, (-4)(5)} = {(4, -20), (6, -20)}

Question 19.
Using mathematical induction, prove 1 . 2. 3 + 2 . 3 . 4 + 3 . 4 . 5 + …… upto n terms = \(\frac{n(n+1)(n+2)(n+3)}{4}\), ∀ n ∈ N.
Solution:
1, 2, 3 …… are in A.P.
n^th term = 1 + (n – 1) 1 = n
2, 3, 4 ……… are in A.P.
n^th term = 2 + (n – 1) 1 = n + 1
3, 4, 5 ……… are in A.P.
n^th term = 3 + (n – 1) 1 = n + 2
∴ n^th term of the series = n (n + 1) (n + 2)
Let S(n) be the statement that
1.2.3 + 2.3.4 + 3.4.5 +……. + n(n+1)(n+2) = \(\frac{n(n+1)(n+2)(n+3)}{4}\)
If n = 1 then L.H.S. = 1.2.3 = 6
RHS = \(\frac{1(1+1)(1+2)(1+3)}{4}\) = 6
∴ L.H.S. = R.H.S.
∴ S(1) is true.
Assume that S(k) is true.
1.2.3 +2.3.4 + 3.4.5+ ……. + k(k+1) (k+2) = \(\frac{k(k+1)(k+2)(k+3)}{4}\)
Adding (k + 1) (k + 2) (k + 3) on both sides, we have
1.2.3 + 2.3.4 +…..+ k(k + 1) (k + 2) + (k + 1) (k + 2) (k + 3)
= \(\frac{k(k+1)(k+2)(k+3)}{4}\) + (k + 1)(k + 2)(k + 3)
= \(\frac{k(k+1)(k+2)(k+3)+4(k+1)(k+2)(k+3)}{4}\)
= \(\frac{(k+1)(k+2)(k+3)(k+4)}{4}\)
∴ S(k + 1) is true.
∴ By the principle of finite mathematical induction S(n) is true for all n ∈ N.

Question 20.
Solve the following system of equations by using Cramer’s rule:
2x – y + 3z = 9
x + y + z = 6
x – y + z = 2
Solution:
Given system of equations are
2x – y + 3z = 9
x + y + z = 6
x – y + z = 2
The given system of equations can be expressed as AX = B
where A = \(\left[\begin{array}{ccc}
2 & -1 & 3 \\
1 & 1 & 1 \\
1 & -1 & 1
\end{array}\right]\), X = \(\left[\begin{array}{l}
x \\
y \\
z
\end{array}\right]\) and B = \(\left[\begin{array}{l}
9 \\
6 \\
2
\end{array}\right]\)
Δ = \(\left[\begin{array}{ccc}
2 & -1 & 3 \\
1 & 1 & 1 \\
1 & -1 & 1
\end{array}\right]\)
= 2(1 + 1) – (-1) (1 – 1) + 3(-1 – 1)
= 2(2) + 1(0) + 3(-2)
= 4 – 6
= -2
Δ1 = \(\left[\begin{array}{ccc}
9 & -1 & 3 \\
6 & 1 & 1 \\
2 & -1 & 1
\end{array}\right]\)
= 9(1 + 1) + 1(6 – 2) + 3(-6 – 2)
= 18 + 4 – 24
= -2
Δ2 = \(\left[\begin{array}{lll}
2 & 9 & 3 \\
1 & 6 & 1 \\
1 & 2 & 1
\end{array}\right]\)
= 2(6 – 2) – 9(1 – 1) + 3(2 – 6)
= 8 – 0 – 12
= -4
Δ3 = \(\left[\begin{array}{ccc}
2 & -1 & 9 \\
1 & 1 & 6 \\
1 & -1 & 2
\end{array}\right]\)
= 2(2 + 6) + 1(2 – 6) + 9(-1 – 1)
= 16 – 4 – 18
= -6
x = \(\frac{\Delta_1}{\Delta}=\frac{-2}{-2}\) = 1
y = \(\frac{\Delta_2}{\Delta}=\frac{-4}{-2}\) = 2
z = \(\frac{\Delta_3}{\Delta}=\frac{-6}{-2}\) = 3
∴ x = 1, y = 2, z = 3

Question 21.
(i) Show that A = \(\left[\begin{array}{lll}
1 & 2 & 1 \\
3 & 2 & 3 \\
1 & 1 & 2
\end{array}\right]\) is non-singular and find A^-1.
Solution:
Given A = \(\left[\begin{array}{lll}
1 & 2 & 1 \\
3 & 2 & 3 \\
1 & 1 & 2
\end{array}\right]\)
|A| = \(\left[\begin{array}{lll}
1 & 2 & 1 \\
3 & 2 & 3 \\
1 & 1 & 2
\end{array}\right]\)
= 1(4 – 3) – 2(6 – 3) + 1(3 – 2)
= 1 – 6 + 1
= -4 ≠ 0
∴ A in non-singular.
∴ A^-1 exists.

(ii) If A = \(\left[\begin{array}{lll}
1 & 2 & 2 \\
2 & 1 & 2 \\
2 & 2 & 1
\end{array}\right]\) then show that A² – 4A – 5I = 0.
Solution:

Question 22.
Find the shortest distance between the skew lines r = (6i + 2j + 2k) + t(i – 2j + 2k) and r = (-4i – k) + s(3i – 2j – 2k).
Solution:

Question 23.
If A + B + C = 2S, then prove that cos (S – A) + cos (S – B) + cos C = -1 + 4 cos \(\frac{S-A}{2}\) cos \(\frac{S-B}{2}\) cos \(\frac{C}{2}\).
Solution:
Given A + B + C = 2s
⇒ \(\frac{A + B + C}{2}\) = s

Question 24.
Show that in a ΔABC, \(\frac{1}{r^2}+\frac{1}{r_1^2}+\frac{1}{r_2^2}+\frac{1}{r_3^2}=\frac{a^2+b^2+c^2}{\Delta^2}\).
Solution: