#### The deleted neighborhood of a is ____.

The neighborhood of 'a' is an open interval (a - δ, a + δ). The deleted neighborhood in which 'a' doesn't occur is the open interval (a - δ, a + δ) from which 'a' is taken away, i.e. (a - δ, a + δ) - {a}

#### The mathematical expression x → a means ____.

The expression means the variable is approaching a. Consider the variable x → 2. x can reach 2 from 1.9, 1.99, 1.9999,etc. or it can reach from 2.1, 2.01, 2.001,................. Therefore, x ≠ 2.

#### If y = tan x + sec x, then d^{2}y/dx^{2 } is ____.

y = tan x + sec x dy/dx = sec^{2}x + sec xtan x=sec x(sec x+tan x) d^{2}y/dx^{2 }= sec x(sec xtan x+sec^{2}x)+sec xtan x(sec x+tan x) =sec x(dy/dx) + tan x(dy/dx) = dy/dx (sec x+tan x)= (dy/dx)y

#### If y = 2 sin 3x cos x, then d^{20}y/dx^{20} is ____.

We know that sinC + sinD = 2 × sin(C+D)/2 × cos(C−D)/2 Therefore, 2sin 3x cos x can be written as sin 4x + sin 2x y = sin 4x + sin 2x y_{1} = 4 cos 4x + 2 cos 2x y_{2} = -16 sin 4x - 4 sin 2x y_{3} = -64 cos 4x - 8 cos 2x y_{4} = 256 sin 4x + 16 sin 2x y_{4} = 2^{8} sin 4x + 2^{4}sin 2x So, generalizing we get d^{20}y/dx^{20} = y_{20} y_{20} = 2^{40}sin4x+2^{20}sin2x

#### The displacement time relating to a speeding vehicle is given by D(t) = 5 t^{2}- 3t + 30. Then, its speed at the end of 5 seconds is given by ____.

D(t) = 5 t^{2}- 3t + 30 The first derivative of the displacement relation gives speed. Therefore, speed is D_{1} (t) = 10t – 3 Speed at the end of 5 sec is D_{1} (5) = 10 x 5 - 3 = 47 units/ second

#### In a real number line, the neighbourhood of positive real number 'a' with a very small radius δ>0 will be ____.

Around a, the sphere with radius δ > 0 is called the neighborhood of a. But, on a real number line, the intersection will be an open interval with left extreme point being (a - δ) and right extreme point being (a + δ).

#### The ratio of the rate of change of area and the rate of change of radius with respect to given circle is ____.

Given, A = πr^{2} dA/dt = 2πr dr/dt ⇒ (dA/dt) / (dr/dt) = 2πr = Perimeter of the circle

#### The derivative of f(x) = |x| at x = 0 is ____.

f(x) = x, x ≥ 0 f(x) = - x, x < 0 Left hand derivative = - 1 and Right hand derivative = 1 LHD ≠ RHD Therefore, derivative at x = 0 does not exist.

#### In the curve y^{2 }= x^{3}, the slope of the tangent at (4, 8) is ____.

y^{2} = x^{3} Differentiating w.r.t x 2y dy/dx = 3 x^{2} ⇒ dy/dx = 3 x^{2}/2y We know that the slope of the curve is given by the derivative at the point. The slope of the curve at (4, 8) is given by: (dydx)_{(4,8) }= 3(4^{2})/2(8) = 3

#### If f(x) = e^{x }log x, then f'(x) is ____.

f(x) = e^{x} log x Applying the rule for product of two functions, f^{1} (x)= e^{x} logx + e^{x} /x But, f(x) = e^{x} log x. ∴ f^{1} (x) = f(x) + e^{x} /x

#### The number of points in (1, 2) where f(x) = a^{[x2]},a > 1 is not differentiable is ____.

Let g(x) = x^{2}. Then, this is an increasing function. g(1) = 1 and g(2) = 4 [g(x)] = [x^{2}] is discontinuous at x = 2,3 f(x) is not differentiable at 2 points, viz. (1, 2)

#### The differential of (sin x + cos x)^{2}is

(sin x + cos x)^{2} = cos^{2} x + sin^{2} x + 2sin x cos x = 1 + sin 2x The differential of 1 + sin 2x = 2cos 2x, since differential of 1 (a constant) is 0.

#### The slope of the normal at the point θ=π/4 of the curve sin θ - cos θ is ____.

The slope of the tangent at the given point is given by the derivative of the function at that point. The tangent and normal are perpendicular to each other. Therefore, the slope of the normal is given by the negative reciprocal of the derivative of the function at that point. y= sin θ - cos θ dy /dθ = cosθ + sinθ dy/dθ atπ/4 = cosπ/4 + sinπ/4 dy/dθ atπ/4=1/2 + 1/2 = 2 Slope of the normal = -1/(dy/dθ) atπ/4 = -1/2

#### The limit of a function exists only if the left hand limit and the right hand limit coincide with each other.

The limit of a function exists only if the left hand limit and the right hand limit coincide with each other.

#### If y = x^{2} sin 2x, then the first derivative is ____.

y = x^{2} sin 2x dy/dx = x^{2}cos 2x 2x + sin 2x 2x dy/dx = 2x(x cos 2x + sin 2x)

#### The point on the curve x^{2 }+ y^{2} - 2x + 1 = 0, at which the tangent is parallel to Y axis is given by ____.

On differentiating the given function, we get 2x + 2y(dy/dx) - 2 = 0 dy/dx = 1-x/y = the slope of the tangent This is parallel to Y - axis. But slope of Y - axis is infinity. Therefore, dy/dx = 1-x/y = ∞ y should be equal to 0. Substituting, y = 0 in the curve equation, we get x^{2} + 0 - 2x + 1 = 0 ⇒ (x + 1)^{2 }= 0 The required point is (1, 0).

#### The degree of a polynomial function is 5. Then, the degree of it's third derivative is ____.

If y = x^{n}, then, the first derivative is of power n - 1, second derivative is of power n - 2, third derivative is of powre n - 3 and so on. In the given statement, n = 5, therefore, the third derivative is (5 -3) = 2.

#### The displacement of a particle describing a non - linear motion is given by s = (1/6) t^{3 } - 8t. Then, the acceleration at the time when velocity vanishes is ____.

s = (1/6) t^{3 }-8t Then, the velocity is given by the first derivative ds/dt = (3/6) t^{2}-8=0 ⇒(1/2) t^{2} – 8 = 0 ⇒ t^{2} = 16 Derivative of velocity is acceleration. Acceleration at t = 4 seconds is (dv/dt)_{t = 4 }= (2/2) t = 4units / s^{2}

#### If fx = 1 + x + x^{2}/2+x^{3}/3, then f' (x) is ____.

fx = 1 + x + x^{2}/2 + x^{3}/3 f^{1} (x) = 1+ x + x^{2}

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