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51.	What is the number of critical points for f(x) = max(sinx, cosx) for x belonging to (0, 2π)?(a) 2(b) 5(c) 3(d) 4The question was posed to me in quiz.I'm obligated to ask this question of First Order Derivative topic in section Limits and Derivatives of Mathematics – Class 11
Answer» Right option is (c) 3 To elaborate: We know that in the range of (0, 2π) the graph of SINX and cosx INTERSECTS each other in three points. And we know that these points of intersection are only the CRITICAL points Thus, there are 3 critical points.

52.	What is the value of \(\lim\limits_{y \rightarrow 0}\frac{sin⁡3y}{3y}\)?(a) 0(b) 1(c) 3(d) \(\frac{1}{3}\)This question was posed to me during an interview.My question is taken from Limits of Trigonometric Functions in division Limits and Derivatives of Mathematics – Class 11
Answer» Right answer is (B) 1 To elaborate: We KNOW that \(\lim\limits_{x \rightarrow 0}\frac{sin⁡x}{x}\) = 1. Here x tends to 3Y. Also, since this is of the form \(\frac{0}{0}\), we USE L’Hospital’s rule and differentiate the numerator and denominator separately. = \(\lim\limits_{y \rightarrow 0}\frac{3\, cos\, 3y}{3}\) = 1

53.	What is the value of \(\lim\limits_{y \rightarrow 2} \frac{y^2-4}{y-2}\)?(a) 2(b) 4(c) 1(d) 0I had been asked this question in homework.The query is from Limits in portion Limits and Derivatives of Mathematics – Class 11
Answer» RIGHT option is (b) 4 Best EXPLANATION: y^2 – 4 = (y – 2)(y + 2) Therefore the FRACTION becomes, (y + 2) As y TENDS to 2, the fraction becomes 4

54.	What is the value of (x + y)^2y” if x = e^t sint and y = e^t cost?(a) x/2(y’ + y)(b) x/2(y’ – y)(c) 2(xy’ + y)(d) 2(xy’ – y)I got this question in examination.This key question is from Second Order Derivative in chapter Limits and Derivatives of Mathematics – Class 11
Answer» The CORRECT answer is (d) 2(xy’ – y) The best I can explain: Since, x = e^t sint and y = e^t COST Therefore, dx/dt = e^t sint + e^t cost = y + x And, dy/dt = e^t cost – e^t sint = y – x So, y’ = dy/dx = (dy/dt)/(dx/dt) = (y – x)/(y + x) Thus, y” = [(x + y)(y’ – 1) – (y – x)(y’ + 1)]/(x + y)^2 Or, (x + y)^2y” = (x + y – y + x)y’ – x – y + x = 2xy’ – 2y = 2(xy’ – y)

55.	What will be the domain of the function, if 3^x + 3^f(x) = minimum of Φ(t), where Φ(t) = minimum of (2t^3 – 15t^2 + 36t – 25, 2 + \|sint\|; 2 ≤ t ≤ 4} ?(a) (-∞, 1)(b) (-∞, loge3)(c) (0, loge2)(d) (-∞, loge2)I have been asked this question in a national level competition.This question is from Second Order Derivative topic in chapter Limits and Derivatives of Mathematics – Class 11
Answer» The correct option is (d) (-∞, loge2) Easy explanation: LET, g(t) = 2t^3 – 15t^2 + 36t -25 g’(t) = 6t^2 – 30t + 36 = 0 => 6(t^2 – 5t + 6) = 0 => t = 2, 3 For, 2 ≤ t ≤ 4, at t = 3, g”(t) > 0 So, g(t) is minimum. g(t)min = g(3) = 227 – 159 + 36*3 – 25 = 2 Also, 2 + \|SINT\| ≥ 2 Hence, minimum Φ(t) = 2 Therefore, 3^x + 3^f(x) = 2 =>3^f(x) = 2 – 3^x Thus, 3^f(x) > 0 => 3^x > 0 and 3^x < 2 Therefore, x € (-∞, loge2)

56.	What will be the value of dy/dx if x = asec^2θ and y = atan^3θ at θ = π/4?(a) 1/2(b) 3/4(c) 3/2(d) 1/4The question was asked in a job interview.This question is from First Order Derivative in section Limits and Derivatives of Mathematics – Class 11
Answer» Correct ANSWER is (C) 3/2 Explanation: Since, X = asec^2θ, Therefore, dx/dθ = ad/dx(sec^2θ) = 2asecθsecθ tanθ Again, dy/dθ = ad/dx(TAN^3θ) = a 3 tan^2θ * d/dθ(tanθ) = 3a tan^2θ sec^2θ Therefore, dy/dx = (dy/dθ)/(dx/dθ) = 3a tan^2θ sec^2θ/2asecθ*secθ tanθ Thus, at θ = π/4 we have, dy/dx = 3/2(tan π/4) = 3/2

57.	What is the value of \(\frac{d}{dx}\)(cos^2⁡ xtan⁡ x) at x = 1?(a) -1(b) 0(c) -2(d) 1The question was posed to me during an interview.My doubt is from Derivatives in portion Limits and Derivatives of Mathematics – Class 11
Answer» Right option is (d) 1 For EXPLANATION I would SAY: We need to use product rule in both the terms to get the answer. \(\FRAC{d}{dx}\) (f.G) = g.\(\frac{d}{dx}\) (f) + f.\(\frac{d}{dx}\) (g) Here f = cos^2⁡ x and g = tan ⁡x To differentiate f, we need to use chain rule. \(\frac{d}{dx}\) (cos^2 ⁡x tan⁡ x) = tan ⁡x.\(\frac{d}{dx}\) (cos^2 x) + cos^2 x.\(\frac{d}{dx}\) (tan⁡ x) \(\frac{d}{dx}\) (e^x tan⁡x) = tan⁡ x.(-2 cos⁡ x sin⁡ x) + cos^2 ⁡x.sec^2 x At x = 1 we get, = tan⁡0.(-2 cos⁡ 0 sin⁡ 0) + cos^2⁡ 0.sec^2 ⁡0 = 1

58.	The derivative of ln⁡ e^x = 1. Is the statement true or false?(a) True(b) FalseThe question was posed to me in an international level competition.I would like to ask this question from Derivatives in chapter Limits and Derivatives of Mathematics – Class 11
Answer» Correct option is (a) True The BEST I can EXPLAIN: We KNOW that ln⁡ e^x = x \(\frac{d}{dx}\) (x) = 1

59.	What is the value of the \(\lim\limits_{x \rightarrow \frac{3\pi}{2}}\frac{cos⁡ x sin⁡ x}{sin⁡2x}\)?(a) \(\frac{-1}{2}\)(b) \(\frac{1}{2}\)(c) \(\frac{1}{4}\)(d) \(\frac{-1}{4}\)I got this question during an internship interview.The question is from Limits of Trigonometric Functions topic in chapter Limits and Derivatives of Mathematics – Class 11
Answer» Correct option is (b) \(\frac{1}{2}\) To explain I WOULD say: \(\lim\limits_{x \RIGHTARROW \frac{3\pi}{2}}\frac{COS⁡ x sin⁡ x}{sin⁡2x}\) =\(\lim\limits_{x \rightarrow \frac{3\pi}{2}}\frac{cos⁡ x sin⁡ x}{2 cos x sin⁡ x}\) = \(\frac{1}{2}\)

60.	What is the value of \(\lim\limits_{y \rightarrow \pi/2}\frac{sin⁡ x}{x}\)?(a) \(\frac{2}{\pi}\)(b) \(\frac{\pi}{2}\)(c) 1(d) 0The question was posed to me in an internship interview.Enquiry is from Limits of Trigonometric Functions topic in portion Limits and Derivatives of Mathematics – Class 11
Answer» The correct option is (a) \(\frac{2}{\pi}\) To explain I WOULD say: SIN ⁡\(\frac{\pi}{2}\) = 1 \(\lim\limits_{y \rightarrow \pi/2}\frac{sin⁡x}{x} = \frac{sin⁡\frac{π}{2}}{\frac{\pi}{2}}\) = \(\frac{1}{\frac{\pi}{2}}\) = \(\frac{2}{\pi}\)