by Rahul Anand | May 9, 2023 | PROBABILITY
Solution We have A = (1,4), (2,3), (3,2), (4,1) B = (1,4), (2,4), (3,4), (4,4), (5,4), (6,4), (4,1), (4,2), (4,3), (4,5), (4,6) \( P(A|B) = \dfrac {P(A∩B)}{P(B)} \) \( A∩B = (1,4), (4,1) \) The sample space comprises of 6×6 = 36 eventsHence,\( P(A∩B) = \dfrac{2}{36} =...
by Rahul Anand | May 9, 2023 | CALCULUS
We have \( \lim_{\theta\to0} { \sin\theta \over \theta } \) = 1 Consider the below diagram. We have r = radius of the circle.A = centre of the circle.The sector ⌔ formed by the arc BD subtends an angle θ at the centre. Case 1 : θ > 0 i.e. θ is +ve Let 0 ≤ θ ≤ \(...
by Rahul Anand | May 9, 2023 | CALCULUS
To prove : lim\( _{x \to a} { x^n – a^n \over x – a } = na^{n-1} \) where n is a rational number Proof: Let \( x = a + h \) Then as \(x \to a \), we have \(h \to 0 \) Now, \( \lim_{x \to a} { x^n – a^n \over x – a }...
by Rahul Anand | May 9, 2023 | CALCULUS
Proof : We have, lim\(_{θ\to 0} { \dfrac {\mathrm tan \mathrm θ}{ \mathrm θ} } \) = lim\(_{θ\to 0} { \dfrac {\mathrm \sin \mathrm θ} {\mathrm θ \mathrm \cos\mathrm θ} } \) \( \{∵ \tan\theta = \dfrac...
by Rahul Anand | May 9, 2023 | CALCULUS
As θ → 0, we have cosθ → 1 Proof : When θ = 0, We have, lim\(_{θ\to 0} \cos \)θ = cos0 = 1 { ∵ cos0 = 1 } Hence, lim\(_{θ\to 0} \cos \)θ = 1
