Doubtify – JEE Ki Taiyaari, Simple Bhasha Mein

 ❓ Question Evaluate the integral: ∫ ( 1 x + 1 x 3 ) 3 x − 24 + x − 26 3   d x = − α 3 ( α + 1 ) ( 3 x β + x γ ) α / ( α + 1 ) + C , where C C  is the constant of integration. Find α + β + γ \alpha + \beta + \gamma . 🖼️ Question Image ✍️ Short Solution Factor inside the cube root: 3 x − 24 + x − 26 = x − 26 ( 3 x 2 + 1 ) Then, cube root: 3 x − 24 + x − 26 3 = x − 26 ( 3 x 2 + 1 ) 3 = x − 26 / 3 ( 3 x 2 + 1 ) 1 / 3 . Simplify the squared bracket: ( 1 x + 1 x 3 ) = x 2 + 1 x 3 . \left(\frac{1}{x} + \frac{1}{x^3}\right) = \frac{x^2+1}{x^3}. Raise to first power (as given), then multiply with cube root: ( 1 x + 1 x 3 ) 3 x − 24 + x − 26 3 = x 2 + 1 x 3 ⋅ x − 26 / 3 ( 3 x 2 + 1 ) 1 / 3 = ( x 2 + 1 ) x − 3 − 26 / 3 ( 3 x 2 + 1 ) 1 / 3 . \left(\frac{1}{x} + \frac{1}{x^3}\right) \sqrt[3]{3x^{-24} + x^{-26}} = \frac{x^2+1}{x^3} \cdot x^{-26/3} (3x^2 +1)^{1/3} = (x^2+1) x^{-3-26/3} (3x^2+1)^{1/3}. Compute exponent: − 3 − 26 / 3 = − 35 / 3 -3 - 26/3 = -35/3 . So: (...

 ❓ Question: For t > − 1 t > -1 , let α t \alpha_t ​ and β t \beta_t ​ be the roots of the equation: ( ( t + 2 ) 1 / 6 − 1 ) x 2 + ( ( t + 2 ) 1 / 6 − 1 ) x + ( ( t + 2 ) 1 / 21 − 1 ) = 0. If lim ⁡ t → − 1 + α t = a and lim ⁡ t → − 1 + β t = b , then find 72 ( a + b ) 2 . 🖼️ Question Image ✍️ Short Solution Write the quadratic in standard form: A ( t ) x 2 + B ( t ) x + C ( t ) = 0 where A ( t ) = ( t + 2 ) 1 / 6 − 1 , B ( t ) = ( t + 2 ) 1 / 6 − 1 , C ( t ) = ( t + 2 ) 1 / 21 − 1. Observe the limits as t → − 1 + t \to -1^+ : A ( t ) → ( 1 ) − 1 = 0 , B ( t ) → 0 , C ( t ) → 2 1 / 21 − 1 but we must use L’Hospital-type analysis since coefficients vanish. Step 1: Factor out A ( t ) A(t)  from the quadratic x 2 + x + C ( t ) A ( t ) = 0 ⇒ x 2 + x + ( t + 2 ) 1 / 21 − 1 ( t + 2 ) 1 / 6 − 1 = 0. Step 2: Evaluate the limit Set h = t + 2 h = t+2 , then as t → − 1 + t\to -1^+ , h → 1 + h\to 1^+ . Then: ( t + 2 ) 1 / 21 − 1 ( t + 2 ) 1 / 6 − 1 = h ...

❓ Question:  Find the area of the region { ( x , y ) :    1 + x 2 ≤ y ≤ min ⁡ {   x + 7 ,    11 − 3 x   } } . If this area is A A , compute 3 A 3A . 🖼️ Question Image ✍️ Short Solution Find where the two lines cross each other. Compare x + 7 x+7  and 11 − 3 x 11-3x : x + 7 ≤ 11 − 3 x    ⟺    4 x ≤ 4    ⟺    x ≤ 1. So on ( − ∞ , 1 ] (-\infty,1]  the upper boundary is x + 7 x+7 ; on [ 1 , ∞ ) [1,\infty)  the upper boundary is 11 − 3 x 11-3x . Both meet at x = 1 x=1  with value 8 8 . Find intersection points of parabola with each line. With y = x + 7 y=x+7 : solve 1 + x 2 = x + 7 ⇒ x 2 − x − 6 = 0 1+x^{2}=x+7 \Rightarrow x^{2}-x-6=0  → x = − 2 ,   3 x=-2,\,3 . With y = 11 − 3 x y=11-3x : solve 1 + x 2 = 11 − 3 x ⇒ x 2 + 3 x − 10 = 0 →  x = − 5 ,   2 x=-5,\,2 . Determine the x-range where the parabola lies below the relevant line. For the branch where upper = x + 7 x+7  (valid for x ≤ 1 x\le1 ), the parabola is below this...