Doubtify – JEE Ki Taiyaari, Simple Bhasha Mein

  Question Let p p  be the number of all triangles that can be formed by joining the vertices of a regular polygon P P  of n n  sides, and q q  be the number of all quadrilaterals that can be formed by joining the vertices of P P . If p + q = 126 , then find the eccentricity of the ellipse x 2 16 + y 2 n = 1. Question Image Short Solution Triangles from n n n vertices: p = ( n 3 ) Quadrilaterals from n n n vertices: q = ( n 4 ) Given: ( n 3 ) + ( n 4 ) = 126 Solve for n n . Substitute n n n as the denominator of y 2 y^{2}  in the ellipse: x 2 16 + y 2 n = 1 Identify a a  and b b , then find eccentricity: e = 1 − b 2 a 2​ Image Solution Conclusion Step 1: Solve for n n n ( n 3 ) = n ( n − 1 ) ( n − 2 ) 6 , ( n 4 ) = n ( n − 1 ) ( n − 2 ) ( n − 3 ) 24​ So: n ( n − 1 ) ( n − 2 ) 6 + n ( n − 1 ) ( n − 2 ) ( n − 3 ) 24 = 126 Multiply by 24: 4 n ( n − 1 ) ( n − 2 ) + n ( n − 1 ) ( n − 2 ) ( n − 3 ) = 3024 Factor n ( ...

  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: ( − ∞ ,...

  Question: Consider the lines L 1 : x − 1 = y − 2 = z L_{1} : x - 1 = y - 2 = z and L 2 : x − 2 = y = z − 1 L_{2} : x - 2 = y = z - 1 Let the feet of the perpendiculars from the point P ( 5 ,   1 ,   − 3 ) P(5,\,1,\,-3)  on the lines L 1 L_{1} ​ and L 2 L_{2} ​ be Q Q  and R R  respectively. If the area of the triangle P Q R PQR  is A A , then find 4 A 2 4A^{2} . Question Image Short Solution To find 4 A 2 4A^{2} , follow these steps: Write lines in parametric form L 1 :    x = 1 + t ,    y = 2 + t ,    z = t L 2 :    x = 2 + s ,    y = s ,    z = 1 + s Find the foot of the perpendicular from P P  to each line Use the formula for foot of perpendicular on a line: Q = A + ( P − A ) ⋅ d ∣ d ∣ 2 d Q = A + \dfrac{(P-A)\cdot d}{|d|^{2}} d where A A  is a point on the line and d d  its direction vector. Get coordinates of Q Q  and R R  by solving the dot product condition. Find the area of triangle P Q R P...

Question: Let the system of equations { x + 5 y − z = 1 4 x + 3 y − 3 z = 7 24 x + y + λ z = μ , λ , μ ∈ R \begin{cases} x + 5y - z = 1 \\[1em] 4x + 3y - 3z = 7 \\[1em] 24x + y + \lambda z = \mu,\quad \lambda, \mu \in \mathbb{R} \end{cases} ​ have infinitely many solutions. Then, the number of solutions of this system, if x , y , z x, y, z  are integers and satisfy 7 ≤ x + y + z ≤ 77 , is: Question Image Short Solution Idea: For infinitely many solutions, the third equation must be a linear combination of the first two. Find λ \lambda  and μ \mu . Reduce to two equations in x , y , z x, y, z , find the relation among variables. Put x + y + z = t x + y + z = t , derive bounds 7 ≤ t ≤ 77 Count integer solutions. Image Solution

❓ Question  If the equation of the line passing through the point  ( 0 , − 1 2 , 0 ) (0, -\dfrac{1}{2}, 0)  and perpendicular to the lines r = λ ( i ^ + a j ^ + b k ^ ) and r = ( i ^ − j ^ − 6 k ^ ) + μ ( − b i ^ + a j ^ + 5 k ^ ) is x − 1 − 2 = y + 4 d = z − c − 4 , then a + b + c + d a + b + c + d  is equal to: Question Image Short Solution We are given: A point P ( 0 , − 1 / 2 , 0 ). Two lines L 1 L_1 ​ and L 2 L_2 ​ . Required line is perpendicular to L 1 L_1 L 1 ​ and L 2 L_2 L 2 ​ and has equation x − 1 − 2 = y + 4 d = z − c − 4 \frac{x-1}{-2} = \frac{y+4}{d} = \frac{z-c}{-4} Steps: Direction ratios (DRs) of L 1 L_1 : 1 , a , b 1, a, b . DRs of L 2 L_2 : − b , a , 5 -b, a, 5 . The required line’s DRs are proportional to the cross product of these (since it is ⟂ to both). Use the point condition and given equation to find a , b , c , d a, b, c, d . Finally compute a + b + c + d a + b + c + d . Image Solution

  ❓ Question Let e 1 e_{1} ​ and e 2 e_{2} ​ be the eccentricities of Ellipse:  x 2 b 2 + y 2 25 = 1 and Hyperbola:  x 2 16 − y 2 b 2 = 1 respectively. If b < 5 b<5  and e 1 e 2 = 1 e_{1}e_{2}=1 , then find the eccentricity of the ellipse whose axes are along the coordinate axes and which passes through all four foci (two of the given ellipse and two of the hyperbola). 🖼️ Question Image ✍️ Short Solution Step 1 — Find e 1 e_1 e 1 ​ and e 2 e_2 e 2 ​ relations For ellipse: x 2 b 2 + y 2 25 = 1 \dfrac{x^2}{b^2} + \dfrac{y^2}{25} = 1 Major axis = along y y y -axis (since 25 > b 2 25 > b^2 ). Eccentricity: e 1 = 1 − b 2 25 . e_1 = \sqrt{1 - \frac{b^2}{25}}. . For hyperbola: x 2 16 − y 2 b 2 = 1 Eccentricity: e 2 = 1 + b 2 16 . Given e 1 e 2 = 1 e_1 e_2 = 1 : 1 − b 2 25    1 + b 2 16 = 1. Step 2 — Solve for b b b Square both sides: ( 1 − b 2 25 ) ( 1 + b 2 16 ) = 1. Expand: 1 + b 2 16 − b 2 25 − b 4 400 = 1. Simplify: b 2 16 − b ...