Doubtify – JEE Ki Taiyaari, Simple Bhasha Mein

 ❓ Question If the sum of the 2nd, 4th and 6th terms of a G.P. of positive terms is 21 and the sum of its 8th, 10th and 12th terms is 15309 , then find the sum of its first nine terms . 🖼️ Question Image ✍️ Short Solution Let the G.P. be a ,   a r ,   a r 2 , … a,\,ar,\,ar^{2},\dots  with a > 0 ,    r > 0 a>0,\; r>0 Write the given sums: Sum of 2nd, 4th, 6th terms: S 1 = a r + a r 3 + a r 5 = a r ( 1 + r 2 + r 4 ) = 21. Sum of 8th, 10th, 12th terms: S 2 = a r 7 + a r 9 + a r 11 = a r 7 ( 1 + r 2 + r 4 ) = 15309. Divide S 2 S_2  by S 1 S_1  to eliminate the common factor ( 1 + r 2 + r 4 ) and  a r a r : S 2 S 1 = a r 7 ( 1 + r 2 + r 4 ) a r ( 1 + r 2 + r 4 ) = r 6 = 15309 21 . Compute the right-hand side: 15309 21 = 729. So r 6 = 729 = 3 6 r^{6} = 729 = 3^{6} . Since r > 0 r>0 , we get r = 3. Now substitute r = 3 r=3  back into S 1 S_1 : a r ( 1 + r 2 + r 4 ) = a r ( 1 + 9 + 81 ) = a r ⋅ 91 = 21. Hence a r = 21 91...

  Question A bag contains 19 unbiased coins and one coin with heads on both sides . One coin is drawn at random and tossed; a head turns up. If the probability that the drawn coin was unbiased is m n \dfrac{m}{n} n m ​ where gcd ⁡ ( m , n ) = 1 \gcd(m,n)=1 , then find: n 2 − m 2 Question Image Short Solution Define events: U U : chosen coin is unbiased D D : chosen coin is double-headed H H : head turns up Use Bayes’ theorem: P ( U ∣ H ) = P ( U ) P ( H ∣ U ) P ( U ) P ( H ∣ U ) + P ( D ) P ( H ∣ D )​ Compute each term: P ( U ) = 19 20 P(U)=\dfrac{19}{20} ​ P ( D ) = 1 20 P(D)=\dfrac{1}{20} ​ P ( H ∣ U ) = 1 2 P(H|U)=\dfrac{1}{2} fair coin) P ( H ∣ D ) = 1 P(H|D)=1 (double-headed always gives head) Plug values, simplify to find m m  and n n . Compute n 2 − m 2 n^{2}-m^{2} Image Solution Conclusion Step 1: Apply Bayes P ( U ∣ H ) = 19 20 ⋅ 1 2 19 20 ⋅ 1 2 + 1 20 ⋅ 1 ​ ​ Step 2: Simplify Numerator: 19 20 ⋅ 1 2 = 19...

  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...