Doubtify – JEE Ki Taiyaari, Simple Bhasha Mein

  ❓ Question Find the number of solutions of the equation cos ⁡ 2 θ cos ⁡ θ 2 + cos ⁡ 5 θ 2 = 2 cos ⁡ 3 5 θ 2​ in the interval [ − π / 2 ,   π / 2 ]. 🖼️ Question Image ✍️ Short Solution Step 1 — Observe structure The RHS has 2 cos ⁡ 3 5 θ 2 2\cos^3\frac{5\theta}{2} , which suggests the triple–angle identity: cos ⁡ 3 x = 4 cos ⁡ 3 x − 3 cos ⁡ x . Step 2 — Rearrange equation Move RHS to LHS: cos ⁡ 2 θ cos ⁡ θ 2 + cos ⁡ 5 θ 2 − 2 cos ⁡ 3 5 θ 2 = 0. Write 2 cos ⁡ 3 A = 1 2 ( 4 cos ⁡ 3 A ) = 1 2 ( cos ⁡ 3 A + 3 cos ⁡ A ) 2\cos^3 A = \tfrac12 (4\cos^3A) = \tfrac12(\cos3A + 3\cos A) So: cos ⁡ 2 θ cos ⁡ θ 2 + cos ⁡ 5 θ 2 − 1 2 ( cos ⁡ 15 θ 2 + 3 cos ⁡ 5 θ 2 ) = 0. Step 3 — Simplify LHS After simplification, the equation reduces to: cos ⁡ 2 θ cos ⁡ θ 2 + cos ⁡ 5 θ 2 − 1 2 cos ⁡ 15 θ 2 − 3 2 cos ⁡ 5 θ 2 = 0. Collect like terms: cos ⁡ 2 θ cos ⁡ θ 2 − 1 2 cos ⁡ 15 θ 2 − 1 2 cos ⁡ 5 θ 2 = 0. Step 4 — Use product-to-sum cos ⁡ 2 θ cos ⁡ θ 2 = 1 2 [ cos ⁡ ( 2 θ − θ 2 ) + cos ⁡ ...

  ❓ Question If the locus of z ∈ C z \in \mathbb{C} , such that Re  ⁣ ( z − 1 2 z + i ) + Re  ⁣ ( z − 1 2 z − i ) = 2 , is a circle of radius r r r and center ( a , b ) (a,b) , then find the value of 15 a b r 2 . 🖼️ Question Image

❓ Question If the orthocenter of the triangle formed by the lines y = x + 1 , y = 4 x − 8 , y = m x + c is at the point ( 3 , − 1 ) (3,-1) , then find the value of m − c m - c . 🖼️ Question Image ✍️ Short Solution Step 1 — Find intersection of the first two lines (one vertex). Solve x + 1 = 4 x − 8 x+1 = 4x - 8 . Move terms: 1 + 8 = 4 x − x 1 + 8 = 4x - x  → 9 = 3 x 9 = 3x  → x = 3 x = 3 . Then y = x + 1 = 3 + 1 = 4 y = x + 1 = 3 + 1 = 4 . So vertex A = ( 3 , 4 ) A = (3,4) . Step 2 — Use the altitude through A A . An altitude from vertex A A  passes through the orthocenter ( 3 , − 1 ) (3,-1) . The line through A ( 3 , 4 ) A(3,4)  and orthocenter ( 3 , − 1 ) (3,-1)  has the same x x -coordinate, so it is the vertical line x = 3 x = 3 . That altitude is vertical, so the side it is perpendicular to must be horizontal. Therefore the third side y = m x + c y = mx + c  must be horizontal — i.e. its slope m = 0 m = 0 . So the third line is y = c y ...

 ❓ Question Let A = { ( α ,  β )  ∈ R × R :    ∣ α  −  1 ∣  ≤  4 ,    ∣ β  −  5 ∣  ≤  6 } and B = { ( α ,  β )  ∈ R × R :    16 ( α  −  2 )²  +  9 ( β  −  6 )²  ≤  144 } . Then determine the relationship between sets A  and B  (containment, intersection, union, etc.). 🖼️ Question Image ✍️ Short Solution Step 1 — Convert descriptions into ranges/standard form For A A A : ∣ α − 1 ∣ ≤ 4 ⇒ α ∈ [ − 3 , 5 ] |\alpha-1|\le4 \Rightarrow \alpha\in[-3,5] ∣ β − 5 ∣ ≤ 6 ⇒ β ∈ [ − 1 , 11 ] |\beta-5|\le6 \Rightarrow \beta\in[-1,11] So A A  is an axis-aligned rectangle with opposite corners at ( − 3 , − 1 ) (-3,-1)  and ( 5 , 11 ) (5,11) . Center at ( 1 , 5 ) (1,5) , width = 8 =8 , height = 12 =12 . For B B : divide the inequality by 144 144 : ( α − 2 ) 2 9 + ( β − 6 ) 2 16 ≤ 1. So B B  is an ellipse centered at ( 2 , 6 ) (2,6)  with semi-axes ...

 Question: Let the length of a latus rectum of an ellipse x 2 a 2 + y 2 b 2 = 1 \dfrac{x^2}{a^2}+\dfrac{y^2}{b^2}=1  be 10 . If its eccentricity is the minimum value of the function f ( t ) = t 2 + t + 11 12 ,    t ∈ R, then  a 2 + b 2 a^2+b^2  is equal to:

  Question: Let a and b be vectors of the same magnitude such that ∣ a + b ∣ ∣ a − b ∣ = 2 + 1. Then the value of ∣ a + b ∣ + ∣ a − b ∣ ∣ a + b ∣ − ∣ a − b ∣ is: 📷 Question Image:

Question: The number of real roots of the equation x   ∣ x − 2 ∣ + 3 ∣ x − 3 ∣ + 1 = 0 is: 📷 Question Image: Short Solution (Text): Critical points for the absolute values are x = 2 x=2 and x = 3 x=3 . Solve piecewise in the three intervals. 1) Region x < 2 x<2 : ∣ x − 2 ∣ = 2 − x ,    ∣ x − 3 ∣ = 3 − x |x-2|=2-x,\; |x-3|=3-x . The equation becomes x ( 2 − x ) + 3 ( 3 − x ) + 1 = 0 ⇒ − x 2 − x + 10 = 0 ⇒ x 2 + x − 10 = 0. x(2-x) + 3(3-x) + 1 = 0 \\ \Rightarrow -x^2 - x + 10 = 0 \\ \Rightarrow x^2 + x - 10 = 0. Discriminant Δ = 1 + 40 = 41 \Delta = 1 + 40 = 41 . Roots: x = − 1 ± 41 2 . x = \frac{-1 \pm \sqrt{41}}{2}. Numeric values: − 1 + 41 2 ≈ 2.7016 \dfrac{-1+\sqrt{41}}{2}\approx 2.7016 and − 1 − 41 2 ≈ − 3.7016 \dfrac{-1-\sqrt{41}}{2}\approx -3.7016 . Only x ≈ − 3.7016 x\approx -3.7016 lies in this region ( x < 2 x<2 ) → one valid root here. 2) Region 2 ≤ x < 3 2 \le x < 3 : ∣ x − 2 ∣ = x − 2 ,    ∣ x − 3 ∣ = 3 − x |x-2|=x-2,\; |x-3|=3-x . The e...

Question: Consider the following half-cell reaction: C r 2 O 7 2 − ( a q ) + 6 e − + 14 H + ( a q ) ⟶ 2 C r 3 + ( a q ) + 7 H 2 O ( l ) The reaction was conducted with the ratio [ C r 3 + ] 2 [ C r 2 O 7 2 − ] = 10 − 6 \dfrac{[Cr^{3+}]^2}{[Cr_2O_7^{2-}]} = 10^{-6} The pH value at which the EMF of the half cell will become zero is ______. (nearest integer) Short Solution (Text): Step 1: Nernst equation (at 298 K) For the half reaction, E = E ∘ − 0.05916 n log ⁡ 10 Q where n = 6 n=6  and Q = [ C r 3 + ] 2 [ C r 2 O 7 2 − ] [ H + ] 14 Q=\dfrac{[Cr^{3+}]^2}{[Cr_2O_7^{2-}][H^+]^{14}} Set E = 0 E=0  (given) ⇒ E ∘ = 0.05916 6 log ⁡ 10 Q Step 2: insert Q Q  and given ratio Let R = [ C r 3 + ] 2 [ C r 2 O 7 2 − ] = 10 − 6 R=\dfrac{[Cr^{3+}]^2}{[Cr_2O_7^{2-}]}=10^{-6} . Then Q = R [ H + ] 14 ⇒ log ⁡ 10 Q = log ⁡ 10 R − 14 log ⁡ 10 [ H + ] . Step 3: solve for log ⁡ 10 [ H + ] \log_{10}[H^+] Using the standard potential E ∘ E^\circ  for C r 2 O 7 2 − / C r 3 + \math...

 Question: Resonance in X 2 Y X_2Y  can be represented as X = X + = Y −       ↔       : X ≡ X + − Y : The enthalpy of formation of X 2 Y X_2Y   ( X ( g ) + 1 2 Y 2 ( g ) → X 2 Y ( g ) ) \big(\mathrm{X(g)} + \tfrac{1}{2}\mathrm{Y_2(g)} \rightarrow X_2Y(g)\big) is 80 kJ mol⁻¹ . The magnitude of resonance energy of X 2 Y X_2Y  is _____ kJ mol⁻¹ (nearest integer value). 📷 Question Image: Short Solution (Text): Step 1: Identify the prototype The resonance set : X ≡ X + − Y : ↔ X = X + = Y − \mathrm{:X \equiv X^{+} - Y:} \leftrightarrow \mathrm{X = X^{+} = Y^{-}} is the classic pattern of N 2 O \mathrm{N_2O}  (take X = N X = \mathrm{N} , Y = O Y = \mathrm{O} ). Step 2: Use the given formation enthalpy For this species, the extra stabilization due to resonance (resonance energy) is taken as the magnitude corresponding to the given energetic stabilization relative to a localized single structure. With Δ H f ∘ \Delta H_f^\circ ​ provided as 80\ \mathrm{kJ\,...