CBSE Class 11 Mathematics Limits And Derivatives Worksheet Set 04

CBSE Class 11 Mathematics Worksheet - Limits and Derivatives. The questions in the worksheets have been specifically designed by best teachers so that the students can practise them to clear their concepts and get better marks in tests and examinations. Students can download these worksheets and practice them. This will help them to get better marks in examinations. Also refer to other worksheets for the same chapter and other subjects too. Use them for better understanding of the subjects.

Question. If \( \lim_{x\to0} \left( \sin(mx) \cot \frac{x}{\sqrt{3}} \right) = 2 \). Find \( m \)
Answer: We have \( \lim_{x\to0} \left( \sin(mx) \cot \frac{x}{\sqrt{3}} \right) = 2 \)
\( \implies \lim_{x\to0} \left( \frac{\sin(mx)}{\tan\left(\frac{x}{\sqrt{3}}\right)} \right) = 2 \)
\( \implies \lim_{x\to0} \left( \frac{\frac{\sin(mx)}{mx} \times mx}{\frac{\tan\left(\frac{x}{\sqrt{3}}\right)}{\left(\frac{x}{\sqrt{3}}\right)} \times \frac{x}{\sqrt{3}}} \right) = 2 \)
\( \implies \frac{1 \times m}{1 \times \frac{1}{\sqrt{3}}} = 2 \)
\( \implies \sqrt{3} m = 2 \)
\( \implies m = \frac{2}{\sqrt{3}} \) ans.

Question. If \( f(x) = 1 - x + x^2 - x^3 \dots - x^{99} + x^{100} \), find \( f'(1) \).
Answer: We have \( f(x) = 1 - x + x^2 - x^3 \dots - x^{99} + x^{100} \)
Differentiate both sides w.r.t \( x \)
\( f'(x) = 0 - 1 + 2x - 3x^2 \dots - 99x^{98} + 100x^{99} \)
\( f'(1) = -1 + 2 - 3 \dots - 99 + 100 \) (put \( x = 1 \))
\( f'(1) = -(1 + 3 + 5 \dots + 99) + (2 + 4 + 6 \dots + 100) \)
For the first A.P.: \( a = 1, d = 2, n = 50 \)
For the second A.P.: \( a = 2, d = 2, n = 50 \)
\( = -\frac{50}{2}[2(1) + (49)2] + \frac{50}{2}[2(2) + 49 \times 2] \)
\( = -25(100) + 25(102) \)
\( = -2500 + 2550 \)
\( f'(1) = 50 \) ans.

Question. Let \( f(x) = \begin{cases} x^2 - 1 & : \ a < x < 2 \\ 2x + 3 & : \ 2 \le x < 3 \end{cases} \) find the quadratic curve whose roots are \( \lim_{x\to2^-} f(x) \) and \( \lim_{x\to2^+} f(x) \).
Answer: \( \lim_{x\to2^-} f(x) = \lim_{x\to2^-} (x^2 - 1) \)
Put \( x = 2 - h \) & \( h \to 0 \)
\( = \lim_{h\to0} ((2 - h)^2 - 1) = 4 - 1 = 3 \)
Now \( \lim_{x\to2^+} (2x + 3) \)
Put \( x = 2 + h \) & \( h \to 0 \)
\( \lim_{h\to0} (2(2 + h) + 3) = 4 + 3 = 7 \)
Given 3 & 7 are the roots of the quadratic curve: \( x^2 - (\text{sum of roots}) \times x + \text{product of roots} = 0 \)
\( x^2 - (3 + 7)x + 21 = 0 \)
\( x^2 - 10x + 21 = 0 \) ans.

Question. Evaluate \( \lim_{x\to\frac{\pi}{3}} \left( \frac{\sqrt{1-\cos(6x)}}{\sqrt{2}\left(\frac{\pi}{3}-x\right)} \right) \)
Answer: We have \( \lim_{x\to\frac{\pi}{3}} \left( \frac{\sqrt{1-\cos(6x)}}{\sqrt{2}\left(\frac{\pi}{3}-x\right)} \right) \)
\( = \lim_{x\to\frac{\pi}{3}} \left( \frac{\sqrt{2 \sin^2(3x)}}{\sqrt{2} \left(\frac{\pi}{3}-x\right)} \right) \)
\( = -\lim_{x\to\frac{\pi}{3}} \left( \frac{\sqrt{2} \sin(3x)}{\sqrt{2} \left(\frac{\pi}{3}-x\right)} \right) \)
Put \( x = \frac{\pi}{3} + h \) & \( h \to 0 \)
\( = -\lim_{h\to0} \left( \frac{\sin\left(3\left(\frac{\pi}{3}+h\right)\right)}{\frac{\pi}{3}+h-\frac{\pi}{3}} \right) \)
\( = -\lim_{h\to0} \left( \frac{\sin(\pi+3h)}{h} \right) \)
\( = -\lim_{h\to0} \left( \frac{-\sin(3h)}{h} \right) \)
\( = \lim_{h\to0} \left( \frac{\sin(3h)}{3h} \right) \times 3 \)
\( = 1 \times 3 = 3 \) ans.

Question. Evaluate \( \lim_{x\to\frac{\pi}{6}} \left( \frac{\cot^2 x-3}{\text{cosec } x-2} \right) \)
Answer: We have \( \lim_{x\to\frac{\pi}{6}} \left( \frac{\cot^2 x-3}{\text{cosec } x-2} \right) \)
\( = \lim_{x\to\frac{\pi}{6}} \left( \frac{\text{cosec}^2 x - 1 - 3}{\text{cosec } x - 2} \right) \)
\( = \lim_{x\to\frac{\pi}{6}} \left( \frac{\text{cosec}^2 x - 4}{\text{cosec } x - 2} \right) \)
\( = \lim_{x\to\frac{\pi}{6}} \left( \frac{(\text{cosec } x + 2)(\text{cosec } x - 2)}{\text{cosec } x - 2} \right) \)
\( = \text{cosec}\left(\frac{\pi}{6}\right) + 2 = 2 + 2 = 4 \) ans.

Question. Evaluate \( \lim_{x\to1} \left( \frac{x^7-2x^5+1}{x^3-3x^2+2} \right) \)
Answer: Since \( \lim_{x\to1} \), \( \therefore (x - 1) \) is the factor of N & D.
Using polynomial division:
\( (x^7 - 2x^5 + 1) \div (x - 1) = x^6 + x^5 - x^4 - x^3 - x^2 - x - 1 \)
\( (x^3 - 3x^2 + 2) \div (x - 1) = x^2 - 2x - 2 \)
\( \therefore \lim_{x\to1} \frac{(x-1)(x^6+x^5-x^4-x^3-x^2-x-1)}{(x-1)(x^2-2x-2)} \)
\( = \frac{1+1-1-1-1-1-1}{1-2-2} = \frac{2-5}{-3} = \frac{-3}{-3} = 1 \) ans.

Question. Evaluate \( \lim_{x\to0} \left( \frac{|\sin x|}{x} \right) \)
Answer: L.H.L. \( \lim_{x\to0^-} \left( \frac{|\sin x|}{x} \right) \)
Put \( x = 0 - h = -h \) & \( h \to 0 \)
\( \lim_{h\to0} \left( \frac{|\sin(-h)|}{-h} \right) = -1 \)
R.H.L. \( \lim_{x\to0^+} \left( \frac{|\sin x|}{x} \right) \)
Put \( x = 0 + h = h \)
\( \lim_{h\to0} \left( \frac{|\sin h|}{h} \right) = \lim_{h\to0} \left( \frac{\sin h}{h} \right) = 1 \)
\( \therefore \lim_{x\to0} f(x) \) does not exists as L.H.L \(\neq\) R.H.L.