Doubtify – JEE Ki Taiyaari, Simple Bhasha Mein

❓ Question: If the function f ( x ) = tan ⁡ ( tan ⁡ x ) − sin ⁡ ( sin ⁡ x ) tan ⁡ x − sin ⁡ x​ is continuous at x = 0 x=0 , then find the value of f ( 0 ) f(0) . 🖼️ Question Image ✍️ Short Solution At x = 0 x=0 : tan ⁡ ( tan ⁡ 0 ) = tan ⁡ 0 = 0 , sin ⁡ ( sin ⁡ 0 ) = sin ⁡ 0 = 0. \tan(\tan 0) = \tan 0 = 0,\quad \sin(\sin 0)=\sin 0=0. So numerator = 0 − 0 = 0 0-0=0 Denominator = tan ⁡ 0 − sin ⁡ 0 = 0 − 0 = 0. \tan 0 - \sin 0 = 0-0=0. ⇒ Indeterminate form 0 / 0 0/0 . We must evaluate the limit as x → 0 x\to0 Apply series expansion around x = 0 x=0 : tan ⁡ x = x + x 3 3 + O ( x 5 ) . \tan x = x + \tfrac{x^3}{3} + O(x^5). sin ⁡ x = x − x 3 6 + O ( x 5 ) . \sin x = x - \tfrac{x^3}{6} + O(x^5). Hence, tan ⁡ x − sin ⁡ x = ( x + x 3 3 ) − ( x − x 3 6 ) + O ( x 5 ) = x 3 2 + O ( x 5 ) . \tan x - \sin x = \Big(x+\tfrac{x^3}{3}\Big) - \Big(x-\tfrac{x^3}{6}\Big) + O(x^5) = \tfrac{x^3}{2} + O(x^5). Expand numerator: tan ⁡ ( tan ⁡ x ) ≈ tan ⁡  ⁣ ( x + x 3 3 ) . \ta...

 ❓ 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...

❓ Question Let a n a_n a n ​ be the n th n^{\text{th}}  term of an A.P. If S n = a 1 + a 2 + ⋯ + a n = 700 ,    a 6 = 7 S_n=a_1+a_2+\cdots+a_n=700,\; a_6=7  and S 7 = 7 S_7=7 , then find a n a_n ​ . 🖼️ Question Image ✍️ Short Solution Let the A.P. have first term a 1 a_1 ​ and common difference d d . Then: a 6 = a 1 + 5 d = 7. Sum of first 7 terms: S 7 = 7 2 ( 2 a 1 + 6 d ) = 7. S_7 = \frac{7}{2}\big(2a_1 + 6d\big) = 7. Divide both sides by 7: 1 2 ( 2 a 1 + 6 d ) = 1 ⇒ a 1 + 3 d = 1. \frac{1}{2}\big(2a_1 + 6d\big) = 1 \quad\Rightarrow\quad a_1 + 3d = 1. Now subtract the two linear equations: ( a 1 + 5 d ) − ( a 1 + 3 d ) = 7 − 1    ⇒    2 d = 6    ⇒    d = 3. (a_1+5d) - (a_1+3d) = 7 - 1 \;\Rightarrow\; 2d = 6 \;\Rightarrow\; d = 3. Then a 1 + 3 d = 1 ⇒ a 1 + 9 = 1 ⇒ a 1 = − 8. a_1 + 3d = 1 \Rightarrow a_1 + 9 = 1 \Rightarrow a_1 = -8. So the A.P. is − 8 ,    − 5 ,    − 2 ,    1 ,    4 ,    7 ,    10 , … Next, use the condition S n = 700 S_n = 700 . Sum ...

❓ Question Let a random variable X X  take values 0 , 1 , 2 , 3 0,1,2,3  with P ( X = 0 ) = p , P ( X = 1 ) = p , P ( X = 2 ) = P ( X = 3 ) and it is given that    E ( X 2 ) = 2 E ( X ) . \;E(X^2)=2E(X). Find the value of 8 p − 1. 🖼️ Question Image ✍️ Short Solution Let q = P ( X = 2 ) = P ( X = 3 ) q = P(X=2)=P(X=3) . Since total probability is 1: 2 p + 2 q = 1 ⇒ p + q = 1 2 ⇒ q = 1 2 − p . Compute the expectations: E ( X ) = 0 ⋅ p + 1 ⋅ p + 2 ⋅ q + 3 ⋅ q = p + 5 q . E(X) = 0\cdot p + 1\cdot p + 2\cdot q + 3\cdot q = p + 5q. E ( X 2 ) = 0 2 ⋅ p + 1 2 ⋅ p + 2 2 ⋅ q + 3 2 ⋅ q = p + 13 q . E(X^2) = 0^2\cdot p + 1^2\cdot p + 2^2\cdot q + 3^2\cdot q = p + 13q. Use the given relation E ( X 2 ) = 2 E ( X ) E(X^2)=2E(X) : p + 13 q = 2 ( p + 5 q ) . Simplify: p + 13 q = 2 p + 10 q ⇒ − p + 3 q = 0 ⇒ 3 q = p . Substitute q = 1 2 − p q = \tfrac12 - p  into 3 q = p 3q = p : 3 ( 1 2 − p ) = p ⇒ 3 2 − 3 p = p 3\left(\tfrac12 - p\right) = p \quad\Rightarr...

  Question Let f : R → R f : \mathbb{R} \to \mathbb{R}  be a polynomial function of degree 4 having extreme values at x = 4 x = 4  and x = 5 x = 5 . If lim ⁡ x → 0 f ( x ) x 2 = 5 , then find f ( 2 ) f(2) . Question Image Short Solution Express derivative using extreme points: Since f ( x ) f(x)  is degree 4 and has extreme values at x = 4 , 5 x=4,5 , the derivative f ′ ( x ) f'(x)  must vanish there: f ′ ( x ) = k ( x − 4 ) ( x − 5 ) ( x − α ) for some real α \alpha  and constant k k . Degree of f ′ ( x ) f'(x)  is 3, consistent with degree 4 for f ( x ) f(x) . Integrate to get f ( x ) f(x) : Let’s assume a convenient factorization for simplicity: f ′ ( x ) = A ( x − 4 ) ( x − 5 ) ( x − r ) Integrate to get f ( x ) = ∫ f ′ ( x ) d x = quartic in  x + C. Use the limit condition: lim ⁡ x → 0 f ( x ) x 2 = 5    ⟹    f ( x ) ∼ 5 x 2  as  x → 0 So the constant term in f ( x ) f(x)  is 0, and the coefficient...