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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 :

- Answer all the questions.
- 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 -y_{1} : y_{2}

= -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 Cos^{2}x\(\frac{d^2 y}{d x^2}\) + 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)

- Attempt any five questions.
- 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(x_{1}, y_{1}) 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 ax^{2} + 2hxy + by^{2} = 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(cos^{2}θ – sin^{2}θ) = 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°.

⇒ 17k^{2} + 425 = 32k^{2} + 80k + 50

⇒ 15k^{2} + 80k – 375 = 0

⇒ 3k^{2} + 16k – 75 = 0

⇒ 3k^{2} + 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 m^{3}/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)

- Attempt any five questions.
- 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 ax^{2} + 2hxy + by^{2} = 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 ax^{2} + 2hxy + by^{2} = 0 represent the lines

l_{1}x + m_{1}y = 0 ….. (1)

l_{2}x + m_{2}y = 0 ……. (2)

Given line is lx + my = 1 ………. (3)

∴ ax^{2} + 2hxy + by^{2} = (l_{1}x + m_{1}y) (l_{2}x + m_{2}y)

Comparing both sides, we have

l_{1}l_{2} = a, m_{1}m_{2} = b, l_{1}m_{2} + l_{2}m_{1} = 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 2x^{2} – 2xy + 3y^{2} + 2x – y – 1 = 0 and the line x + 2y = k are mutually perpendicular.

Solution:

Equation of the circle is x^{2} + y^{2} = a^{2} ….. (1)

Equation of AB is lx+ my = 1 ……. (2)

Homogenising (1) with the help of (2)

Combined equation of OA, OB is

x^{2} + y^{2} = a^{2}.1^{2}

x^{2} + y^{2} = a^{2} (lx + my)^{2}

= a^{2}(l^{2}x^{2} + m^{2}y^{2} + 2lmxy)

= a^{2}l^{2}x^{2} + a^{2}m^{2}y^{2} + 2a^{2}lmxy

i.e., a^{2}l^{2}x^{2} + 2a^{2}lmxy + a^{2} m^{2}y^{2} – x^{2} – y^{2} = 0

(a^{2}l^{2} – 1) x^{2} + 2a^{2}lmxy + (a^{2}m^{2} – 1) y^{2} = o

Since OA, OB are perpendicular

Co-efficient of x^{2} + co-efficient of y^{2} = 0

a^{2}l^{2} – 1 + a^{2}m^{2} – 1 = 0

a^{2}(l^{2} + m^{2}) = 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 + 2n^{2} + 2m^{2} + 2mn = 0

2m^{2} + 5mn + 2n^{2} = 0

(2m + n) (m + 2n) = 0

2m = -n or m = -2n

Case (i): 2m_{1} = -n_{1}

From l_{1} = -m_{1} – n_{1}

= -m_{1} + 2m_{1} = m_{1}

\(\frac{l_1}{1}\) = \(\frac{m_1}{1}\) = \(\frac{x_1}{-2}\)

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 = \(\frac{h(r-x)}{r}\)