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

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

Time : 3 Hours
Max. 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 :

1. Answer all the questions.
2. Each question carries two marks.

Question 1.
Transform the equation 3x + 4y + 12 = 0 into normal form.
Solution:
Given straight line equation is 3x + 4y + 12 = 0
⇒ 3x + 4y = -12
⇒ -3x – 4y = 12
⇒ \(\left(\frac{-3}{5}\right) x+\left(\frac{-4}{5}\right) y\) = \(\frac{12}{5}\)
⇒ x cos α + y sin α = \(\frac{12}{5}\)
Where cos α = \(\frac{-3}{5}\) and Sinα = \(\frac{-4}{5}\)
Which is in the normal form.

Question 2.
If the straight lines x + p =0, y + 2 = 0 and 3x + 2y + 5 = 0 are concurrent, then find the value of p.
Solution:
Given straight line equations are
x + P = 0 ….. (1)
y + 2 = 0 ….. (2)
3x + 2y + 5 = 0 ….. (3)
Solving (1) and (2), we get
X = -p, y = -2
∴ The point of intersection of (1) and (2) is (-p, -2)
Since (1), (2), (3) are concurrent The point (-p, -2) lies on (3)
∴ 3(-p) + 2 (-2) + 5 = 0
⇒ -3P – 4 + 5 = 0
⇒ 3P = 1 ⇒ P = \(\frac{1}{3}\).

Question 3.
Find the ratio in which XZ – plane divides the line joining A (-2, 3, 4) and B (1, 2, 3).
Solution:
Given A = (-2, 3, 4)
B = (1, 2, 3)
The ratio in which xz-plane divides the line joining A, B is -y₁ : y₂
= -3 : 2
= 3 : 2 externally

Question 4.
Find the equation of the plane if the foot of the perpendicular from origin to the plane is (2, 3, -5).
Solution:
Let P = (2, 3, -5)
Since P is the foot of the perpendicular from origin to the required plane
∴ \(\overline{O P}\) is perpendicular to the required plane.
∴ D.rs of \(\overline{O P}\) = (2 – 0, 3 – 0, -5 – 0)
= (2, 3, -5)
Hence equation to required plane is
2(x – 2) + 3(y – 3) – 5(z + 5) = 0
⇒ 2x – 4 + 3y – 9 – 5z – 25 = 0
⇒ 2x + 3y – 5z – 38 = 0

Question 5.
Compute

Solution:

Question 6.
Evaluate

Solution:

Question 7.
If y = Cosec^-1 (e^2x+1), find \(\frac{d y}{d x}\).
Solution:

Question 8.
Show that, y = x + Tanx satisfies Cos²x\(\frac{d^2 y}{d x^2}\) + 2x = 2y .
Solution:

Question 9.
Find the approximate value of \(\sqrt[4]{17}\).
Solution:

Question 10.
If 0 ≤ x ≤ \(\frac{\pi}{2}\), then show that x ≥ Sinx.
Solution:
Let f(x) = x – sinx
f’(x) = 1 – Cosx ≥ 0 ∀ x
∴ f is an increasing function for all x
∴ f ≥ 0 6
⇒ x – sinx ≥ 0 – sin 0
⇒ x – sinx ≥ 0
⇒ x ≥ sinx

Section – B

II. Short Answer Type Questions. (5 × 4 = 20)

1. Attempt any five questions.
2. Each question carries four marks.

Question 11.
A (5, 3) and B (3, -2) are two fixed points. Find the equation of the locus of P, so that the area of the triangle PAB is 9.
Solution:
Given A = (5, 3)
B = (3, – 2)
Let P(x₁, y₁) be any point on the locus.
Given geometric condition is area of ∆PAB is 9.

Question 12.
Show that the axes are to be rotated through an angle of \(\frac{1}{2}\)Tan^-1\(\left(\frac{2 h}{a-b}\right)\) so as to remove the xy term from the equation ax² + 2hxy + by² = 0, if a ≠ b and through an angle \(\frac{\pi}{4}\), if a = b.
Solution:
If the axes are rotated through an angle ‘θ’, then
x = x’ cos θ – y’ sin θ
y = x’ sin θ + y’ cos θ
Therefore the given equation transforms as
a (x’ cos θ – y’ sin θ)² + 2h (x‘ cos θ – y’ sin θ) (x’ sin θ + y’ cos θ) + b (x’ sin θ – y’ cos θ)² = 0
To remove x’y’ term from the equation, the coefficient of x’y’ term must be zero.
So, (b – a) sin θ cos θ + h(cos²θ – sin²θ) = 0

Question 13.
Find the value of k, if the angle between the straight lines 4x – y + 7 = 0 and kx – 5y – 9 = 0 is 45°.
Solution:
Given straight line equations are
4x – y + 7 = 0 …….. (1)
kx – 5y – 9 = 0 ……. (2)
The angle between (1) and (2) is 45°.

⇒ 17k² + 425 = 32k² + 80k + 50
⇒ 15k² + 80k – 375 = 0
⇒ 3k² + 16k – 75 = 0
⇒ 3k² + 25k – 9k – 75 = 0
⇒ k(3k + 25) -3 (3k + 25) = 0
⇒ (k – 3)(3k + 25) = 0
⇒ k = 3 (or) \(\frac{-25}{3}\)

Question 14.
Show that

Where a and b are real constants,, is continuous at 0.
Solution:

Question 15.
If xy = e^x-y, then show that \(\frac{d y}{d x}\) = \(\frac{\log x}{(1+\log x)^2}\).
Solution:
Given x^y = e^x-y
Taking logarithms on bothsides, we have

Question 16.
At any point t on the curve x = a (t + Sint), y = a (1 – Cost), find the lengths of tangent and normal.
Solution:
Equation of the curve is x = a (t + sin t), y = a (1 – cos t)

Question 17.
A container is in the shape of an inverted cone has height 8 m and radius 6 m at the top. If it is filled with water at the rate of 2 m³/minute, how fast is the height of water changing when the level is 4 m ?
Solution:
Let OC be the height of the water level at t sec.

Section – C

III. Long Answer Type Questions. (5 × 7 = 35)

1. Attempt any five questions.
2. Each question carries seven marks.

Question 18.
Find the orthocenter of the triangle whose vertices are (-5, -7), (13, 2) and (-5, 6).
Solution:
Let A = (-5, -7)
B = (13, 2)
C = (-5, 6)
Let \(\overline{\mathrm{AD}}\) be the perpendicular drawn from A to \(\overline{\mathrm{BC}}\) and \(\overline{\mathrm{BE}}\) be the perpendicular drawn from B to \(\overline{\mathrm{AC}}\).

From (1)
9x – 2(2) + 31 = 0 ⇒ 9x + 27 = 0
⇒ 9x = -27
⇒ x = -3
∴ x = -3 and y = 2
∴ Ortho center H = (-3, 2)

Question 19.
If (α, β) is the centroid of the triangle formed by the lines ax² + 2hxy + by² = 0 and lx + my =1, then prove that
\(\frac{\alpha}{b l-h m}\) = \(\frac{\beta}{a m-h l}\) = \(\frac{2}{3\left(b l^2-2 h l m+a m^2\right)}\)
Solution:
Let ax² + 2hxy + by² = 0 represent the lines
l₁x + m₁y = 0 ….. (1)
l₂x + m₂y = 0 ……. (2)
Given line is lx + my = 1 ………. (3)
∴ ax² + 2hxy + by² = (l₁x + m₁y) (l₂x + m₂y)
Comparing both sides, we have
l₁l₂ = a, m₁m₂ = b, l₁m₂ + l₂m₁ = 2h
Clearly the point of intersection of (1) and (2) is 0(0, 0)
Let A be the point of intersection of (1) and (3)

General centroid of ∆OAB is (α, β)

Question 20.
Find the value of k, if the lines joining the origin to the points of intersection of the curve 2x² – 2xy + 3y² + 2x – y – 1 = 0 and the line x + 2y = k are mutually perpendicular.
Solution:

Equation of the circle is x² + y² = a² ….. (1)
Equation of AB is lx+ my = 1 ……. (2)
Homogenising (1) with the help of (2)
Combined equation of OA, OB is
x² + y² = a².1²
x² + y² = a² (lx + my)²
= a²(l²x² + m²y² + 2lmxy)
= a²l²x² + a²m²y² + 2a²lmxy
i.e., a²l²x² + 2a²lmxy + a² m²y² – x² – y² = 0
(a²l² – 1) x² + 2a²lmxy + (a²m² – 1) y² = o
Since OA, OB are perpendicular
Co-efficient of x² + co-efficient of y² = 0
a²l² – 1 + a²m² – 1 = 0
a²(l² + m²) = 2
This is the required condition.

Question 21.
Find the direction cosines of two lines which are connected by the relations l + m + n = 0 and mn – 2nl – 2lm = 0.
Solution:
Given l + m + n = 0 …… (1)
mn – 2nl – 2lm = 0 ……… (2)
From (1), l = -(m + n)
Substituting in (2),
mn + 2n(m + n) + 2m(m + n) = 0
mn + 2mn + 2n² + 2m² + 2mn = 0
2m² + 5mn + 2n² = 0
(2m + n) (m + 2n) = 0
2m = -n or m = -2n

Case (i): 2m₁ = -n₁
From l₁ = -m₁ – n₁
= -m₁ + 2m₁ = m₁
\(\frac{l_1}{1}\) = \(\frac{m_1}{1}\) = \(\frac{x_1}{-2}\)
D.Rs of the first line are 1, 1, -2
D.Cs of this line are \(\frac{1}{\sqrt{6}}\), \(\frac{-2}{\sqrt{6}}\), \(\frac{1}{\sqrt{6}}\)

Question 22.
If f(x) = Sin^-1\(\sqrt{\frac{x-\beta}{\alpha-\beta}}\) and g(x) = Tan^-1\(\sqrt{\frac{x-\beta}{\alpha-x}}\) then show that f'(x) = g'(x) (β < x < α).
Solution:
Given f(x) = Sin^-1\(\sqrt{\frac{x-\beta}{\alpha-\beta}}\)
differentiate w.r.to x on both sides, we have

Question 23.
If the tangent at any point on the curve x^2/3 + y^2/3 = a^2/3 intersects the coordinate axes in A and B, then show that the length AB is a constant.
Solution:
Equation of the curve is x^2/3 + y^2/3 = a^2/3
Differentiating w.r.to x

Question 24.
Prove that the radius of the right circular cylinder of greatest curved surface area which can be inscribed in a given cone is half of that of the cone.
Solution:
Let ‘O’ be the centre of the circular base of the cone and its height be h.
Let r be the radius of the circular base of the cone.
Let a cylinder with radius x(OE) be inscribed in the given cone and its height be u.
i.e. PD = RO = QE = u
From big ∆AOC and ∆QEC are similar.
∴ \(\frac{Q E}{O A}\) = \(\frac{E C}{O C}\)
⇒ \(\frac{u}{h}\) = \(\frac{r-x}{r}\)
⇒ u = \(\frac{h(r-x)}{r}\)