#### If f: [a, b] → R is continuous on [a, b] and differentiable on (a, b), then there exists some c in (a, b) such that f'(c) = f(b) – f(a) /k. The value of k is ____.

#### If a function f (x) = x^{3} - 3x for x ∈ [– 4, 2] is verified by Rolle’s theorem, then the number of all possible values of c in (-4, 2) is/are ____.

Rolle’s Theorem states: If a function f: [a, b] → R is continuous on [a, b] and differentiable on (a, b) such that f (a) = f (b), then there exists some c in (a, b) such that f ’(c) = 0. The given function is f (x) = x^{3} - 3x and the derivative is f ‘(x) = 3 x^{2}- 3 derivative at c is f ‘(c) = 3 c^{2} – 3 f ‘(c) = 3 c^{2} - 3 = 0 (By Rolle’s Theorem) ⇒ c^{2} - 1 = 0 ⇒ c = -1and 1 ⇒ (c + 1)(c - 1) = 0 So, we get two values of c (-4, 2). Hence, the number of all possible values of c in (-4, 2) is/are 2.

#### A function f is verified by Mean Value Theorem at c in(a, b), if f (x) = 3 x^{2} – 4x in the interval [a, b], where a = 1 and b = 4. Then, the value of f'(c) is ____.

Mean Value Theorem states: If f: [a, b] → R is continuous on [a, b] and differentiable on (a, b), then there exists some c in (a, b) such that f '(c) = f(b) – f(a) /b-a. Here, f(b) = f(4) = (3 × 4^{2}) - (4 × 4) = 32 f(a) = f(1) = (3 ×1^{2}) - (4 × 1) = -1 Now, f '(c) = f(b) – f(a)/b-a =32—(1) /(4 - 1) = 11 Hence, the value of f ‘(c) is 11.

#### If a function f (x) = cos (2x) for x ∈ [0, π/2] is verified by Mean Value theorem, then all possible values of c in (0, π/2) is/are ____.

Mean Value Theorem states: If f : [a, b] → R is continuous on [a, b] and differentiable on (a, b), then there exists some c in (a, b) such that f'(c)=f(b) – f(a)/b-a. The given function is f (x) = cos (2x) a = 0, b = π/2 f‘(x) = -2sin (2x) f‘(c) = -2sin (2c) Here, f(b) = f(π/2) = cos(2 × π/2) = -1 f(a) = f(0) = cos (2 × 0) = 1 Now, f '(c) = f(b) – f(a)/b-a -2sin(2c) =(-1-1)/(π/2-0) sin2c = 2/π c=1/2arcsin2/π Hence, all possible values of c in (0, π/2) is/are 1/2arcsin2/π.

#### If y = sin x, then y'' = ____ y.

Given function is y = sin x Differentiate both sides with respect to x, we get y'= d(sinx)/dx ⇒ y' = cos x Again, differentiate both sides with respect to x, we get y' ' = d(cosx)/dx ⇒y'' = -sin x ⇒ y'' = -y

#### If volume of a cylindrical container of radius R and height H is V, then the second derivative of V with respect to R is ____ H.

Volume of a cylindrical container of radius R and height H is V = πR2H Differentiate both sides with respect to R, we get V' = 2πRH Again, differentiate both sides with respect to R, we get V'' = 2πH Hence, the second derivative of V with respect to R is 2πH.

#### If y = log x, then x^{2 }D^{2}y + x Dy = ____.

Given function is y = log x. Differentiate both sides with respect to x, we get Dy = 1/x --------- (1) Differentiate equation (1) both sides with respect to x, we get D^{2}y=-1/x^{2} ----------- (2) Now, x^{2}D^{2}y + xDy = x^{2}(-1/x^{2}) + x(1/x) ( from equations (1) and (2)) = -1 + 1 = 0 ⇒ x^{2 }D^{2}y + x Dy = 0 Hence, x^{2 }D^{2}y + x Dy = 0.

#### If e^{n} + e^{-n} = 2, then the value of n is ____.

Given that e^{n} + e^{-n} = 2 By multiplying both sides by ℮^{n}, we get (e^{n})^{2} + (e^{n})(e^{-n}) = 2e^{n} ⇒ (e^{n})^{2} + 1 = 2e^{n} ⇒ (e^{n})^{2} - 2e^{n}+ 1 = 0 ⇒ (e^{n} - 1)^{2}= 0 ⇒ e^{n} = 1 ⇒ e^{n} = e^{0} ⇒ n = 0 Hence, the value of n is 0.

#### If A = 2 log sin θ and B = 2 log cos θ, then the value of ℮A + ℮B is ____.

Given A = 2 log sin θ = log sin^{2} θ ⇒ e^{A} = sin^{2 }θ --------- (1) Given B = 2 log cos θ = log cos^{2} θ ⇒ e^{B }= cos^{2}θ -------- (2) By adding equations (1) and (2), we get ℮^{A }+ ℮^{B} = sin^{2} θ + cos^{2} θ ⇒ ℮^{A} + ℮^{B} = 1 Hence, the value of ℮^{A} + ℮^{B} is 1.

#### If f(x) = (3 - 2x)^{2}, then f''(x) = ____.

Given function is f(x) = (3 - 2x)^{2} Differentiate both sides with respect to x, we get ⇒f'x = d(3-2x)^{2}/dx = d(3-2x)^{2}/d(3-2x) × d(3-2x/dx) = 2(3 - 2x) × (-2) = -4(3 - 2x) ⇒ f'(x) = 8x – 12 -----(i) Differentiate equation (i) with respect to x ⇒ f''(x) = 8 Hence, f''(x) = 8.

#### If a function f : [a, b] → R is continuous on [a, b] and differentiable on (a, b) and verified by Rolle’s Theorem, there exists some c in (a, b), such that f ‘(c) = 0,. Then, the value of f(a)/f(b) is ____.

Rolle's Theorem states: If a function f : [a, b] → R is continuous on [a, b] and differentiable on (a, b) such that f (a) = f (b), then there exists some c in (a, b) such that f ’(c) = 0. Therefore f (a) = f (b). ⇒f(a)/f(b) = 1 Hence, the value of f(a)/f(b) is 1

#### If area of a circle of radius R is A, then the second derivative of A with respect to R is ____.

Area of a circle of radius R is A = π R^{2} Differentiate both sides with respect to R, we get A' = 2πR Again, differentiate both sides with respect to R, we get A'' = 2π Hence, the second derivative of A with respect to R is 2π.

#### If y = 3log x – 5x, then D2y = ____.

Given function is y = 3log x – 5x. Differentiate both sides with respect to x. We get Dy = (3/x) - 5. Again, differentiate both sides with respect to x. We get D^{2}y= -3/ x^{2}. Hence, D^{2}y = -3/ x^{2}.

#### If A = e^{2n} - e^{n }and B = e^{n}, then the value of (A - B^{2})^{2} is ____.

Given that A = e^{2n} - e^{n} and B = e^{n}, Now A - B^{2} = (e^{2n} - e^{n}) - (e^{n})^{2} = e^{2n }- e^{n }- e^{2n} = - e^{n} And (A - B^{2})^{2} = (-e^{n})^{2} = e^{2n} Hence the value of value of (A - B^{2})^{2} is e^{2n}.

#### If a function f: [a, b] → R is continuous on [a, b] and differentiable on (a, b) such that f (a) = f (b), then there exists some c in (a, b) such that f ’(c) = 0. This is called ____ theorem.

Rolle's Theorem states: If a function f: [a, b] →R is continuous on [a, b] and differentiable on (a, b) such that f (a) = f (b), then there exists some c in (a, b) such that f ’(c) = 0.

#### If a function f (x) = sin (2x) for x ∈ [0, π/2] is verified by Rolle’s theorem, then all possible values of c in (0, π/2) is/are ____.

Rolle’s Theorem states: If a function f : [a, b] → R is continuous on [a, b] and differentiable on (a, b) such that f (a) = f (b), then there exists some c in (a, b) such that f ’(c) = 0. The given function is f (x) = sin (2x) And the derivative is f ‘(x) = 2cos (2x) f ‘(x) = 2cos (2x) = 0 (By Rolle’s Theorem) ⇒ cos (2x) = 0 ⇒ cos (2x) = cos π/2 ⇒2x=2nπ ± π/2 ⇒nπ ± π/4 So, the possible value of x in (0, π/2) is π/4 Hence, all possible values of c in (0, π/2) is/are π/4.

#### If y = 7x^{3} - 5x^{2}+ 1 then y'' = ____.

Given y = 7x^{3} - 5x^{2} +1 Differentiate both sides with respect to x, we get y' = 21 x^{2} - 10x Again, differentiate both sides with respect to x, we get y'' = 42x – 10 Hence, y'' = 42x – 10

#### If Volume of a spherical ball of radius R is V, then the second derivative of A with respect to R is ____ R.

Volume of a Spherical Ball of radius R is V = 4/3πR3. Differentiate both sides with respect to R, we get V' = 3 × 4/3π R^{2} = 4π R^{2} Again, differentiate both sides with respect to R, we get V'' = 8πR Hence, the second derivative of V with respect to R is 8πR.

#### If n = log 2 + log sin A + log cos A, then the value of ℮^{n} is ____.

Given n = log 2 + log sin A + log cos A ⇒ n = log (2 sin A cos A) ⇒ n = log sin(2A) ⇒ ℮^{n} = sin 2A Hence, the value of ℮^{n} is sin 2A.

#### A polynomial function is differentiable at ____.

A polynomial function is differentiable at every point of real number R, i.e. the polynomial function is differentiable at R. Hence, a polynomial function is differentiable at (-∞, ∞).

#### If a function f (x) = x^{3} - 7 for x ∈ [–4, 2] is verified by Rolle’s theorem, the possible value/values of c in (-4, 2) is/are ____.

Rolle’s Theorem states: If a function f: [a, b] → R is continuous on [a, b] and differentiable on (a, b) such that f (a) = f (b), then there exists some c in (a, b) such that f ’(c) = 0. The given function is f (x) = x^{3} - 7. and the derivative is f ‘(x) = 3 x^{2}. derivative at c is f ‘(c) = 3 c^{2} f ‘(c) = 3 c^{2} = 0 (By Rolle’s Theorem) Hence, c = 0.

#### If e^{(log x)}= 1/3, then the value of x is ____.

Given e^{(log x)}= 1/3 ⇒log(1/3) = logx ⇒x =1/3 Hence, the value of x is 1/3.

#### If f: [a, b] → R is continuous on [a, b] and differentiable on (a, b). Then there exists some c in (a, b) such that f'(c) = f(b) – f(a)/b-a and it is called__________theorem.

Mean Value Theorem states: If f: [a, b] → R is continuous on [a, b] and differentiable on (a, b). Then there exists some c in (a, b) such that f'(c) = f(b) – f(a)/b-a

#### If e^{n} + e^{-n} = x, then the value of e^{2n} + e^{-2n}is ____.

Given that e^{n} + e^{-n} = x By squaring on both sides, we get (e^{n} + e^{-n})^{2} = x^{2} ⇒ (e^{n})^{2} + (e^{-n})^{2} + 2 (e^{n})(e^{-n}) = x^{2} ⇒ e^{2n} + e^{-2n }+ 2 = x^{2} ⇒ e^{2n} + e^{-2n }= x^{2} – 2 Hence, the value of e^{2n} + e^{-2n }is x^{2} - 2.

#### A function is said to be differentiable in an interval [a, b] if it is differentiable at ____.

From the definition, a function is said to be differentiable in the interval [a, b], if it is differentiable at every point of [a, b]. Hence, a function is said to be differentiable in an interval [a, b], if it is differentiable at every point of [a, b].

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