(6.1) f ( x ) = 1 + x 1 − x f(x) = \frac{1+x}{1-x} f ( x ) = 1 − x 1 + x f ( 1 − x 1 + x ) = 1 + 1 − x 1 + x 1 − 1 − x 1 + x = ( 1 + x ) + ( 1 − x ) 1 + x ( 1 + x ) − ( 1 − x ) 1 + x = 2 2 x = 1 x f\left(\frac{1-x}{1+x}\right) = \frac{1 + \frac{1-x}{1+x}}{1 - \frac{1-x}{1+x}} = \frac{\frac{(1+x)+(1-x)}{1+x}}{\frac{(1+x)-(1-x)}{1+x}} = \frac{2}{2x} = \frac{1}{x} f ( 1 + x 1 − x ) = 1 − 1 + x 1 − x 1 + 1 + x 1 − x = 1 + x ( 1 + x ) − ( 1 − x ) 1 + x ( 1 + x ) + ( 1 − x ) = 2 x 2 = x 1 উত্তর: 1 x \frac{1}{x} x 1
(6.2) ∫ 0 1 x d x 9 − x 2 \int_0^1 \frac{x\,dx}{\sqrt{9-x^2}} ∫ 0 1 9 − x 2 x d x u = 9 − x 2 u = 9 - x^2 u = 9 − x 2 d u = − 2 x d x du = -2x\,dx d u = − 2 x d x x d x = − d u 2 x\,dx = -\frac{du}{2} x d x = − 2 d u x = 0 ⇒ u = 9 x=0 \Rightarrow u=9 x = 0 ⇒ u = 9 x = 1 ⇒ u = 8 x=1 \Rightarrow u=8 x = 1 ⇒ u = 8 = − 1 2 ∫ 9 8 u − 1 / 2 d u = 1 2 ∫ 8 9 u − 1 / 2 d u = [ u ] 8 9 = 3 − 2 2 = -\frac12\int_9^8 u^{-1/2}du = \frac12\int_8^9 u^{-1/2}du = \left[\sqrt{u}\right]_8^9 = 3 - 2\sqrt2 = − 2 1 ∫ 9 8 u − 1/2 d u = 2 1 ∫ 8 9 u − 1/2 d u = [ u ] 8 9 = 3 − 2 2 উত্তর: 3 − 2 2 3 - 2\sqrt2 3 − 2 2
(6.3) 4 × n C 1 = n C 2 4 \times {}^nC_1 = {}^nC_2 4 × n C 1 = n C 2 4 n = n ( n − 1 ) 2 4n = \frac{n(n-1)}{2} 4 n = 2 n ( n − 1 ) 8 n = n ( n − 1 ) 8n = n(n-1) 8 n = n ( n − 1 ) 8 = n − 1 8 = n - 1 8 = n − 1 n ≠ 0 n \neq 0 n = 0 n = 9 n = 9 n = 9
(6.4) A = [ 3 6 2 2 ] A = \begin{bmatrix}3&6\\2&2\end{bmatrix} A = [ 3 2 6 2 ] ∣ A ∣ = 3 × 2 − 6 × 2 = − 6 ≠ 0 |A| = 3\times2 - 6\times2 = -6 \neq 0 ∣ A ∣ = 3 × 2 − 6 × 2 = − 6 = 0 adj A = [ 2 − 6 − 2 3 ] \text{adj}A = \begin{bmatrix}2&-6\\-2&3\end{bmatrix} adj A = [ 2 − 2 − 6 3 ] A − 1 = 1 − 6 [ 2 − 6 − 2 3 ] = [ − 1 3 1 1 3 − 1 2 ] A^{-1} = \frac{1}{-6}\begin{bmatrix}2&-6\\-2&3\end{bmatrix} = \begin{bmatrix}-\frac13&1\\ \frac13&-\frac12\end{bmatrix} A − 1 = − 6 1 [ 2 − 2 − 6 3 ] = [ − 3 1 3 1 1 − 2 1 ] উত্তর: A − 1 = [ − 1 3 1 1 3 − 1 2 ] A^{-1} = \begin{bmatrix}-\frac13&1\\ \frac13&-\frac12\end{bmatrix} A − 1 = [ − 3 1 3 1 1 − 2 1 ]
(6.5) y = x 2 − 1 x 2 + 1 y = \frac{x^2-1}{x^2+1} y = x 2 + 1 x 2 − 1 y ′ = 2 x ( x 2 + 1 ) − ( x 2 − 1 ) ⋅ 2 x ( x 2 + 1 ) 2 = 2 x [ ( x 2 + 1 ) − ( x 2 − 1 ) ] ( x 2 + 1 ) 2 = 4 x ( x 2 + 1 ) 2 y' = \frac{2x(x^2+1) - (x^2-1)\cdot 2x}{(x^2+1)^2} = \frac{2x[(x^2+1)-(x^2-1)]}{(x^2+1)^2} = \frac{4x}{(x^2+1)^2} y ′ = ( x 2 + 1 ) 2 2 x ( x 2 + 1 ) − ( x 2 − 1 ) ⋅ 2 x = ( x 2 + 1 ) 2 2 x [( x 2 + 1 ) − ( x 2 − 1 )] = ( x 2 + 1 ) 2 4 x উত্তর: f ′ ( x ) = 4 x ( x 2 + 1 ) 2 f'(x) = \frac{4x}{(x^2+1)^2} f ′ ( x ) = ( x 2 + 1 ) 2 4 x
(6.6) f ( x ) = 2 3 x 3 + 1 2 x 2 − 6 x + 8 f(x) = \frac23x^3 + \frac12x^2 - 6x + 8 f ( x ) = 3 2 x 3 + 2 1 x 2 − 6 x + 8 f ′ ( x ) = 2 x 2 + x − 6 = ( 2 x − 3 ) ( x + 2 ) = 0 f'(x) = 2x^2 + x - 6 = (2x-3)(x+2) = 0 f ′ ( x ) = 2 x 2 + x − 6 = ( 2 x − 3 ) ( x + 2 ) = 0 x = 3 2 , − 2 x = \frac32,\, -2 x = 2 3 , − 2 f ′ ′ ( x ) = 4 x + 1 f''(x) = 4x + 1 f ′′ ( x ) = 4 x + 1 x = − 2 x=-2 x = − 2 f ′ ′ = − 7 < 0 f'' = -7 < 0 f ′′ = − 7 < 0 x = 3 2 x=\frac32 x = 2 3 f ′ ′ = 7 > 0 f'' = 7 > 0 f ′′ = 7 > 0 সর্বোচ্চ মান হবে x = − 2 x = -2 x = − 2 (মান f ( − 2 ) = 50 3 f(-2) = \frac{50}{3} f ( − 2 ) = 3 50
(6.7) সংজ্ঞা অনুযায়ী এটি f ( x ) = x 1 / 2 f(x)=x^{1/2} f ( x ) = x 1/2 lim h → 0 x + h − x h = lim h → 0 h h ( x + h + x ) = 1 2 x \lim_{h\to0}\frac{\sqrt{x+h}-\sqrt x}{h} = \lim_{h\to0}\frac{h}{h(\sqrt{x+h}+\sqrt x)} = \frac{1}{2\sqrt x} lim h → 0 h x + h − x = lim h → 0 h ( x + h + x ) h = 2 x 1 উত্তর: 1 2 x \frac{1}{2\sqrt x} 2 x 1
(6.8) x x 2 − 3 x + 1 = 1 \frac{x}{x^2-3x+1} = 1 x 2 − 3 x + 1 x = 1 x 2 − 3 x + 1 = x x^2 - 3x + 1 = x x 2 − 3 x + 1 = x x 2 − 4 x + 1 = 0 x^2 - 4x + 1 = 0 x 2 − 4 x + 1 = 0 x ≠ 0 x \neq 0 x = 0 x x x x − 4 + 1 x = 0 x - 4 + \frac1x = 0 x − 4 + x 1 = 0 x + 1 x = 4 x + \frac1x = 4 x + x 1 = 4 উত্তর: ৪
(6.9) ∫ e − 5 x d x = e − 5 x − 5 + C = − 1 5 e − 5 x + C \int e^{-5x}dx = \frac{e^{-5x}}{-5} + C = -\frac{1}{5}e^{-5x} + C ∫ e − 5 x d x = − 5 e − 5 x + C = − 5 1 e − 5 x + C উত্তর: − 1 5 e − 5 x + C -\frac15 e^{-5x} + C − 5 1 e − 5 x + C
(6.10) খুঁটি ভেঙে ভাঙা অংশ (দৈর্ঘ্য ১৬ মি) ভূমির সাথে ৩০° কোণ করেছে। অবশিষ্ট দণ্ডায়মান অংশের দৈর্ঘ্য h = 16 sin 30 ∘ = 16 × 1 2 = 8 h = 16\sin30^\circ = 16\times\frac12 = 8 h = 16 sin 3 0 ∘ = 16 × 2 1 = 8 = 8 + 16 = = 8 + 16 = = 8 + 16 = ২৪ মিটার ।