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Answer is (b) (xy) dx - (x⁴ + y⁴) dy = 0 

For barrel : Height = 50 cm, Radius of base = 20 cm 

∴ Volume of barrel = πr²h = π × (20)² × 50 = 400 × 50 × π

For mug : Height = 15 cm, Diameter of base = 10 cm

∴ Radius of Base = 5 cm

 Volume of mug = πr²h = π × (5)² × 15 = 25 × 15 × π

Volume of barrel / Volume of mug = 400x50x π / 25x 15x π =160 / 3 = 53 1 / 3

When 54^th mug is poured in the barrel it will overflow

Answer is (a) tan^-1 x + c

Answer is (b) [(1,0,0),(0,1,0),(0,0,1)]

Option:  (d) (0,0,1)

Point A(-3, 2) is in second quadrant and point B(12, 0) is on X- axis. 

Answer is (d) 0

d(A, B) = 4 - (-8) = 4 + 8 = 12

Answer is (b) -15

Correct option(b) 2

WY = 2 

OY = 2×5 = 10 cm (Diagonals of parallelogram bisect each other)

Answer is (a) √41

∠ ACD= ∠ B + ∠ A .............. (theorem of remote interior angle) 

= 40 + 70

= 110˚

Correct option(c) √41

∠ RHG= ∠ DHP .................(Opposite angles) 

= 85˚

∠ HGS = ∠ DHP ................ (Corresponding angles) 

= 85˚

Option: (b) 1/4 

Correct option is: (B) –cos25°

Consider \(\frac{d^{2} y}{d x^{2}}=\sqrt{1-\left(\frac{d y}{d x}\right)^{4}} \)

\(\Rightarrow\) d²y/dx² = {1-(dy/dx)⁴}^1/2

\(\Rightarrow\)[d²y/dx²]² = 1-(dy/dx)⁴  

Here the highest order derivative is 2 and its power is 2.

Thus order is 2 and power is 2.

1.employment interview

2.checking reference

3.selection decision

Correct option is: (A) 1

(c) Center

Center is not a type of page margin.

\(\frac{tan^2A}{1+tan^2A}+\frac{cot^2A}{1+cot^2A}=\) \(\frac{tan^2A}{sec^2A}+\frac{cot^2A}{cosec^2A}\) 

= \(\frac{sin^2A}{\frac{cos^2A}{\frac{1}{cos^2A}}}+\frac{cos^2A}{\frac{sin^2A}{\frac{1}{sin^2A}}}\) 

= sin²A + cos²A

=1

Hence Proved.

Correct option is C) keep in touch

Correct option is A) down at heel

Correct option is A) bribe

(b) S N Banerjee

Let the numbers be: a - 3d, a - d, a + d, a + 3d

a - 3d + a - d + a + d + a + 3d = 32

4a = 32

\(\therefore\) a = 8

(a - 3d)² + (a - d)² + (a + d)² + (a + 3d)² = 276

4a² + 20d² = 276

20d² = 276 - 4(8)² = 20

\(\therefore\) d = 1

Therefore the 4 numbers are:

a - 3d, a - d, a + d, a + 3d = 8 - 3, 8 - 1, 8 + 1, 8 + 3 = {5, 7, 9, 11 }

Given

a+ md =2(a+ nd)

then a = (m-2n)d

t_3m+1 = a +3md

= (m -2n)d +3md = 4nd -2nd

= 2(2m-n)n .......(1)

t_m+n+1 = a +(m +n)d

= (m -2n)d + (m +n) d

= (2m -n)d .........(2)

From (1) and (2) t_3m+1 = 2t_m+n+1

f (x) = x² + 2x – 8 

1. f(x) being a polynomial function it is continuous on (-4, 2) 

2. Also f¹ (x) = 2x + 2 hence derivable on (-4, 2) 

3. f(-4) = (-4)² + 2(-4) - 8 = 0 

f(2) = 2² + 2(2) - 8 = 0 

∴ f(-4) = f(2)

∴ All three conditions of Rolle’s theorem are verified. 

f¹(c) = 0 

f¹ (c) = 2c + 2 ⇒ 0 = 2c + 2 

⇒ c = -1 ∈(-4, 2)

1. Draw a circle with O as centre and radius 2 cm 

2. Mark a point P outside the circle such that OP = 4.5 cm 

3. Join OP and bisect it at M 

4. Draw a circle with M as centre and radius equal to MP to intersect the given circle at T and T‟ 

5. Join PT and PT‟ which are required Tangents 

1. Given; sin x = \(\frac{\sqrt{3}}{2}\) = sin \(\frac{\pi}{3}\)

General solution is; x = nπ + (-1)ⁿ \(\frac{\pi}{3}\)

Put n = 0, 1 we get principal solution; x = \(\frac{\pi}{3}\); \(\frac{2\pi}{3}\)

2. Given; cos x = \(\frac{1}{2}\)= cos \(\frac{\pi}{3}\) 

General solution is; x = 2nπ ± \(\frac{\pi}{3}\), n ∈ Z 

Put n = 0, 1 we get principal solution; 

n = 0 

⇒ x = \(\frac{\pi}{3}\) ; n = 1 

⇒ x = 2π – \(\frac{\pi}{3}\) = \(\frac{5 \pi}{3}.\)

3. Given; tan x = \(\sqrt{3}\) = tan \(\frac{\pi}{3}\) 

General solution is; ⇒ x = nπ + \(\frac{\pi}{3}\), n ∈ Z 

Put n = 0, 1 we get principal solution;

n = 0 \(\Rightarrow\) \(\frac{\pi}{3}\); n = 1 \(\Rightarrow\) x = π + \(\frac{\pi}{3}\) = 4\(\frac{\pi}{3}\).

4. Given; cosec x = -2 

⇒ sin x = \(\frac{-1}{2}\) = – sin \(\frac{\pi}{6}\)= sin(-\(\frac{\pi}{6}\)) 

General solution is; x = nπ – (-1) , n ∈ Z 

Put n = 1, 2 we get principal solution;

⇒ n = 1 ⇒ x = \(\pi + \frac{\pi}{6} = \frac{7\pi}{6}\)

⇒ n = 2 ⇒ x = 2\(\pi + \frac{\pi}{6} = \frac{11\pi}{12}\)

(i) (b) \(\frac{\pi}{4}\)

(ii) sin 100° = sin(l 80 – 80) = sin 80°

sin 200° = sin(l 80° + 20°) = -sin 20°

The ascending order is

sin 200°, sin 0°, sin 50°, sin 100°

(iii) sin2x + sin6x – sin4x = 0

⇒ 2sin 4x cos2x – sin 4x = 0

⇒ sin 4x(2 cos 2x – 1) = 0

⇒ sin4x = 0 or (2cos2x – 1) = 0

⇒ 4x = nπ or cos2x = \(\frac{1}{2}\)

⇒ \(x = \frac{n \pi}{4}\)

⇒ cos 2x = \(\frac{\pi}{3}\)

⇒ \(x = \frac{n \pi}{4}\)

⇒ 2x = 2nπ ± \(\frac{\pi}{3}\)

⇒ \(x = \frac{n \pi}{4}\)

⇒ x = nπ ± \(\frac{\pi}{6}\)

1. \(\frac{2 \pi}{3}\) = \(\frac{2 \pi}{3}\) x \(\frac{180}{\pi}\) = 120°

2. LHS

\(\frac{sin\, 5x + sin\, 3x}{cos\, 5x + cos \,3x}\)

= \(\frac{2sin 4x \,cos x}{2 cos 4x \,cos x}\)

= \(\frac{sin\, 4x}{cos \, 4x}\)

= tan 4x

The correct option (1) 2   

Explanation:

sinx + sin3x = sin2x ⇒ 2sin2x cosx = sin2x

⇒ sin2x = 0 or cosx = 1/2

⇒ 2x ∈ {0}    x ∈ {π/3}  

correct option:

(B) 1 

(c) Statement (I) is true but Statement (II) is false 

Non-dimensional cures are defined for all homologous machines and defines general characteristic of machines. 

It is vapour pressure not viscosity which causes cavitation in suction side i.e. low pressure region.

Given:

Let Ram's and Sham's 1 day's work be 1/a and 1/b respectively.

⇒ 1/a + 1/b = 1/4.5

⇒ 1/a + 1/b = 10/45 = 2/9

Then,

⇒ b = 3a

Solving,

a = 6 and b = 18

∴ Sham will complete work in 18 days.

X + 1/X =  25/12

X² +1 = 25X/12

X² -25X/12  = -1

X² -25X/12 + (25/24)² =-1 + (25/24)²

(X - 25/24)²  = 625/576 -1 = 49/576 = (7/24)²

x- 25/24 = + or - 7/24

X = 25/24 + or - 7/24  =  18/24 or 32/24

X = 3/4 or 4/3

4/3 + 3/4 = 16/12 + 9/12 = 25/12

Let the length of rectangle be x and breadth be b.

Area of triangle = l*b

40 = l*b

b=40/l cm

Perimeter of triangle = 2(l+b)

28 = 2(l+40/l)

14=(l²+40)/l

14l=l²+40

l²-14l+40=0

l²-10l-4l+40=0

l(l-10)-4(l-10)=0

(l-4)(l-10)=0

l=4cm or l=10cm

b=40/l = 10cm or 4cm

\(\lim\limits_{x \to 0} \) \(\cfrac{tan(sinx)-x}{tanx^3}\) \(\)(0/0 - type)

= \(\lim\limits_{x \to 0} \) \(\cfrac{sec^2(sinx).cosx-1}{sec^2(x^3).3x^2}\) (BY D.L.H. Rule)

= \(\cfrac{\lim\limits_{x\to 0}sec^2sinx\lim\limits_{x\to 0}cosx-1}{\lim\limits_{x\to 0}sec^2x^3.\lim\limits_{x\to 0}3x^2}\)

= \(\lim\limits_{x \to 0} \) = \(\cfrac{sec^2sinx-1}{3x^2}\) (\(\because\) \(\lim\limits_{x \to 0} \) cosx = 1 and \(\lim\limits_{x \to 0} \) sec²x³ = 1)

(0/0. case)=

= \(\lim\limits_{x \to 0} \)\(\cfrac{2sec^2(sinx)tan(sinx).cosx}{6x}\)

= \(\cfrac26\) \(\lim\limits_{x \to 0} \) sec² (sinx) . cosx \(\lim\limits_{x \to 0} \) \(\cfrac{tan(sinx)}x\) (0/0 type)

=  \(\cfrac13\) sec² 0. cos 0 \(\lim\limits_{x \to 0} \) sec² (sinx) cos x

= \(\cfrac13\) x 1 x 1 sec² 0 . cos 0

= \(\cfrac13\) x 1 x 1 = \(\cfrac13\) 

Alternative = \(\lim\limits_{x \to 0} \) \(\cfrac{tan(sinx)-x}{tanx^3}\) 

= \(\lim\limits_{x \to 0} \) \(\cfrac{tanx-x}{x^3}\) (0/0case)(\(\because\) \(\theta\) is very small then sin \(\theta\) ≈ \(\theta\) and tan   \(\theta\) ≈ \(\theta\) )

= \(\lim\limits_{x \to 0} \) \(\cfrac{sec^2x-1}{3x^2}\) (0/0case)(By  D.L.H Rule)

= \(\lim\limits_{x \to 0} \) \(\cfrac{2sec^2xtanx}{6x}\) 

= \(\cfrac26\) \(\lim\limits_{x \to 0} \) sec² x \(\lim\limits_{x \to 0} \) \(\cfrac{tanx}x\) 

= \(\cfrac13\) x sec² 0 x 1 (\(\because\) \(\lim\limits_{x \to 0} \) \(\cfrac{tanx}x\) = 1)

= \(\cfrac13\) 

\(\therefore\) \(\lim\limits_{x \to 0} \) \(\cfrac{tan(sinx)-x}{tanx^3}\) = \(\cfrac13\) 

\(\lim\limits_{x\to 0}\frac{\int\limits_0^{x^2}sin\sqrt t dt}{x^3}\) (\(\frac00\) case)

= \(\lim\limits_{x\to 0}\frac{2xsin\sqrt{x^2}-0}{3x^2}\)(By using D.L.H Rule)

=  \(\lim\limits_{x\to 0}\frac{sinx}x=\frac23\) (\(\because\lim\limits_{x\to 0}\frac{sin x}x=1\))

\(\therefore\) \(\lim\limits_{x\to 0}\frac{\int\limits_0^{x^2}sin\sqrt t dt}{x^3}\) = 2/3

Correct option is (C) Only two minima

f(x) = 2x² - ln |x|

f'(x) = 0 gives

4x - 1/x = 0

⇒ 4x² = 1

⇒ x = \(\pm\frac12\)

Also, f"(x) = 4 + \(\frac1{x^2}>0\)

\(\therefore\) Critical points x = -1/2 & x = 1/2 are point of minima of function f.

Correct option is (C) y = h(f(x)) is not differentiable at 3 points

f(x) = 2|x + 1| - |x - 1| + 3|x - 2| + 2x + 1

 = \(\begin{cases}-2(x+1)+x-1-3(x-2)+2x+1;&x\leq -1\\2(x+1)+x-1-3(x-2)+2x+1;&-1\leq x\leq1\\2(x+1)-(x-1)-3(x-2)+2x+1;&1\leq x\leq2\\2(x+1)-(x-1)+3(x-2)+2x+1;&x\geq2\end{cases}\)

=\(\begin{cases}-2x+4&;&x\leq -1\\2x+8&;&-1\leq x\leq1\\10&;&1\leq x\leq2\\6x-2&;&x\geq2\end{cases}\)

\(\because\) f(x) is continuous

\(\therefore\) h(f(x)) is discontinuous only where h is discontinuous and h(x) can be discontinuous only at x = 3

Hence, h(f(x)) is discontinuous only when f(x) = 3

when x \(\leq\) - 1, 2x + 4 = 3 ⇒ 2x = 4 - 3 = 1 

⇒ x = 1/2 (Not satisfied)

when -1 \(\leq\) x \(\leq\) 1; 2x + 8 = 3 ⇒ 2x = 3 - 8 = -5

⇒ x = -5/2 (not satisfied)

when 1 \(\leq\) x \(\leq\) 2 ; f(x) = 10 \(\neq3\)

when x \(\geq\) 2, 6x - 2 = 3

⇒ 6x = 2 + 3 = 5

⇒ x = 5/6 \(\ngeq\)2 (not satisfied)

Hence, h(f(x)) is always continuous.

Also f(x) \(\geq\) 6 \(\forall\) x \(\in\) R

\(\therefore\) h(f(x)) = 2f(x) which is always continuous but not differentiable at x = 1, -1, & 2.

\(\lim\limits_{x\to 0}\frac{cosecx-cot x}x\) 

 = \(\lim\limits_{x\to 0}\frac{1-cos x}{xsin x}\) [0/0 case]

= \(\lim\limits_{x\to 0}\frac{sin x}{xcos x+sin x}\) [By using D.L.H. Rule]

 = \(\lim\limits_{x\to 0}\cfrac{x\frac{sin x}x}{x(cos x+\frac{sin x}x)}\) 

 = \(\lim\limits_{x\to 0}\cfrac{\lim\limits_{x\to 0}\frac{sin x}x}{cos x+\lim\limits_{x\to 0}\frac{sin x}x}\) 

= \(\lim\limits_{x\to 0}\) \(\frac1{cos x+1}\) (\(\because\) \(\lim\limits_{x\to 0}\frac{sin x}x=1\))

= \(\frac1{1+cos 0}=\frac1{1+1} = \frac12\)

\(\frac{tan20^o}{cot 70^o}+\frac{cot 50^o}{tan 40^o}+\frac{sin^220^o+sin^270^o}{sin\theta cos(90^o-\theta)}\) + cos θ sin (90°- θ)

 = \(\frac{tan20^o}{cot(90^o-20^o)}+\frac{cot(90^o-40^o)}{tan40^o}\) + \(\frac{sin^220^o+sin^2(90^o-20^o )}{sin\theta.sin\theta}\) + cos θ. cos θ

(∵ sin (90° - θ) - cos θ, cos(90° - θ) = sin θ )

 = \(\frac{tan20^o}{tan 20^o}+\frac{tan 40^o}{tan 40^o}+\frac{sin^220^o+cos^220^o}{sin^2\theta}\) + cos² θ

(∵ cot (90° - θ) = tan θ, & sin (90° -  θ) = cos θ )

 =  1 + 1 + \(\frac1{sin^2\theta}+cos^2\theta\) ( ∵ sin²θ + cos²θ = 1)

 = 2 + cosec² θ + cos² θ ( ∵ \(\frac1{sin\theta}\) = cosec θ)

y = sin^-1\((\frac{x^2-y^2}{x^2+y^2})\)

Differentiate both sides w.r.t. x we get

\(\frac{dy}{dx}\) = \(\cfrac{1}{\sqrt{1-(\frac{x^2-y^2}{x^2+y^2})^2}}\) \(\times\cfrac{(x^2+y^2)(2x-2y\frac{dy}{dx})(x^2-y^2)(2x+2y\frac{dy}{dx})}{(x^2+y^2)^2}\) 

\(=\frac{x^2+y^2}{\sqrt{(x^2+y^2)^2-(x^2-y^2)^2}}\)\(\times\cfrac{(2x^3+2xy^2-2x^3+2xy^2)-2y\frac{dy}{dx}(x^2+y^2+x^2-y^2)}{(x^2+y^2)^2}\)

\(=\frac1{\sqrt{4x^2y^2}}\times\cfrac{4xy^2-4x^2y\frac{dy}{dx}}{x^2+y^2}\)

\(=\frac1{2xy}\times\cfrac{2xy(2y-2x\frac{dy}{dx})}{x^2+y^2}\) 

⇒ \(\frac{dy}{dx}=\cfrac{2y-2x\frac{dy}{dx}}{x^2+y^2}\)

⇒ \(\frac{dy}{dx}+\frac{2x}{x^2+y^2}\frac{dy}{dx}=\frac{2y}{x^2+y^2}\)

⇒ \(\frac{dy}{dx}(1 + \frac{2x}{x^2+y^2})=\frac{2y}{x^2+y^2}\)

⇒ \(\frac{dy}{dx}=\frac{2y}{x^2+y^2+2x}\)

P × Q = {(a, r),(b, r),(e, r)}

B = {1,2,3,4,5 }

Inverse of element 1 under x₅ (multiplication modulo 18) = 1

Inverse of element 5 = 1

Inverse of element 7 = 13

Inverse of element 11 = 5

Inverse of element 13 = 7

Inverse of element 17 = 17