Limits and Derivatives MCQs

Ans. (c)8

Explanation: The numerator is represented as the product of two terms.

⇒ \(\frac{(x + 4)(x – 4) }{ x – 4}\)

⇒ 4 + 4

Therefore the result is 8

Ans. (a) 2x

Explanation: According to the first law of derivatives: \(f'x = \frac{f(x+h) − f(x) }{ h}\)

⇒ Now for the required f(x) = x².

⇒ f'x = (x+h)² − (x)² / h

⇒ f'x = x² + h² + 2xh / h

⇒ f'x = h + 2x(x+h)

⇒ Substituting h = 0 we get,

Therefore f’x = 2x

Ans. (b) 3

Explanation: Given, \(lim_{x→3} (sin\; \frac{3x}{x})\)

⇒ We can write it as;

⇒ \(lim_{x→3} (sin \;\frac{3x}{3x}) × 3\)

⇒ Since, \(lim_{x→3} (sin \frac{x}{x}) = 1\)

⇒ 1 × 3 

Therefore the result is 3

Ans. (a) lim_x→a(f(x) = L

Explanation: The limit of a function is given as: lim_x→a(f(x) = L

Ans. (a) f′(x)=lim_h→0(f(x+h)−f(x))/h

Explanation: The derivative of a function is given as: f′(x)=lim_h→0(f(x+h)−f(x))/h

Ans. (d) 0

Explanation: \(lim_{x→4} \frac{(x−4)(x+4) }{ (x−3)(x+4)}\)

⇒ Simplify the phrase to obtain:

⇒ \(lim_{x→4} \frac{x-4 }{ x-3}\)

Therefore the result is 0

Ans. (b) 62

Explanation: \(lim_{x→4}(6x^2+8x-3)=lim_{x→-4} 6x^2+lim_{x→-4}(8x)-lim_{x→-4}(2)\)

⇒ 6(-4)²+8(-4)-2

⇒ 96-32-2

Therefore the result is 62

Ans. (c)2x sin x + x² cos x

Explanation: d/dx(x² sin x)

⇒ Using the formula of derivative which goes as: d/dx [f(x) g(x)] = f(x) [d/dx g(x)] + g(x) [d/dx f(x)]

⇒ d/dx(x² sin x) = x² [d/dx (sin x)] + sin x [d/dx x²]

⇒ x²(cos x) + sin x (2x)

Therefore the result is 2x sin x + x² cos x

Ans. (c) No value

Explanation: Evaluate the RHS and LHS of the equation.

⇒ According to R.H.S: \(lim_{x→0}+ \frac{|sin x|}{ x} = lim_{x→0}+ sin \;\frac{x}{x} = 1\)

⇒ According to L.H.S: \(lim_{x→0-} \frac{|sin x|}{ x} = lim_{x→0-} -sin \;\frac{x}{x} = -1\)

⇒ R.H.S ≠ L.H.S

So no value

Ans. (c) 1

Explanation: Given in the question, f(x) = x sin x

⇒ f'(x) = x[d/dx sin x] + sin x [d/dx (x)]

⇒ x cos x + sin x

⇒ f′(π/2) = (π/2) cos π/2 + sin π/2

⇒ (π/2) (0) + 1

Therefore the result is 1

Ans. (d) 1/e

Explanation: f(x) = logx (logx)

⇒ log (log x) / logx

⇒ f′(x)= 1 / x − 1 / x log (logx) / (logx)²

⇒ f′(e) = [1 / e − 0] / 1

Therefore the result is 1 / e

Ans. (a) -2 sin x

Explanation: Let the given function be f(x) = 2 cos x + 31

Now, taking the derivative,

⇒ d/dx f(x) = d/dx (2 cos x + 31)

⇒ d/dx (2 cos x) + d/dx (31)

⇒ 2 (-sin x) + 0

Therefore the result is -2 sin x

Ans. (a) 10

Explanation: The numerator is represented as the product of two terms.

⇒ (x + 5)(x – 5) / x – 5

⇒ 5 + 5

Therefore the result is 10

Ans. (a) 10x

Explanation: According to the first law of derivatives: f'x = f(x+h) − f(x) / h

Now for the required f(x) = 5x².

⇒ f'x = (x+h)² − (x)² / h

⇒ f'x = x² + h² + 10xh / h

⇒ f'x = h + 10x(x+h)

⇒ Substituting h = 0 we get,

Therefore the result is f’x = 10x

Ans. (b) 9

Explanation: Given, \(lim_{x→9} (sin \frac{9x}{x})\)

⇒ We can write it as;

⇒ \(lim_{x→9} (sin\; \frac{9x}{9x}) × 9\)

Since, \(lim_{x→9} (sin \;\frac{x}{x}) \)

⇒ 1 × 9 

Therefore the result is 9