Doubtify – JEE Ki Taiyaari, Simple Bhasha Mein

  Question Let y = y ( x ) y = y(x) be the solution of the differential equation ( x 2 + 1 ) y ′ − 2 x y = ( x 4 + 2 x 2 + 1 ) cos ⁡ x , with y ( 0 ) = 1 y(0) = 1 . Then evaluate: ∫ − 3 3 y ( x )   d x Question Image Short Solution Rewrite the ODE in standard linear form: y ′ − 2 x x 2 + 1 y = ( x 4 + 2 x 2 + 1 ) cos ⁡ x x 2 + 1​ Find the integrating factor : μ ( x ) = e ∫ − 2 x x 2 + 1 d x = e − ln ⁡ ( x 2 + 1 ) = 1 x 2 + 1​ Multiply the equation by μ ( x ) \mu(x) : d d x ( y x 2 + 1 ) = cos ⁡ x Integrate: y x 2 + 1 = sin ⁡ x + C Apply y ( 0 ) = 1 y(0)=1 : 1 1 = 0 + C    ⟹    C = 1 Hence: y ( x ) = ( x 2 + 1 ) ( sin ⁡ x + 1 ) Compute the definite integral: I = ∫ − 3 3 y ( x )   d x = ∫ − 3 3 ( x 2 + 1 ) ( sin ⁡ x + 1 ) d x Image Solution Conclusion Break the integral: I = ∫ − 3 3 ( x 2 + 1 ) sin ⁡ x   d x + ∫ − 3 3 ( x 2 + 1 ) d x First term: ( x 2 + 1 ) sin ⁡ x (x^{2}+1)\sin x  is an odd function (because x 2 + 1 x^{2}+1  is eve...

  Question If the range of the function f ( x ) = 5 − x x 2 − 3 x + 2 , x ≠ 1 , 2 is ( − ∞ , α ] ∪ [ β , ∞ ) (-\infty, \alpha] \cup [\beta, \infty) , then find the value of α 2 + β 2 \alpha^{2}+\beta^{2} . Question Image Short Solution Let y = 5 − x x 2 − 3 x + 2 y = \dfrac{5 - x}{x^{2} - 3x + 2} ' Multiply through to get: y ( x 2 − 3 x + 2 ) = 5 − x Rearrange to a quadratic in x x : y x 2 − 3 y x + 2 y − 5 + x = 0 or y x 2 + ( − 3 y + 1 ) x + ( 2 y − 5 ) = 0 For x x  to be real, discriminant Δ \Delta  must be non-negative: Δ = ( − 3 y + 1 ) 2 − 4 y ( 2 y − 5 ) ≥ 0 Simplify Δ \Delta : Δ = 9 y 2 − 6 y + 1 − 8 y 2 + 20 y = y 2 + 14 y + 1 Solve Δ = 0 \Delta =0 : y = − 14 ± 196 − 4 2 = − 14 ± 192 2 = − 14 ± 8 3 2 y = \frac{-14\pm \sqrt{196-4}}{2} = \frac{-14\pm \sqrt{192}}{2} = \frac{-14\pm 8\sqrt3}{2} ​ ​ y = − 7 ± 4 3 y = -7 \pm 4\sqrt3 Because the quadratic in x x  is valid for Δ ≥ 0 \Delta\ge0 , the range of y y  is: ( − ∞ ,...