Match the following List-I with List-II and choose the correct option: List-I (Compounds) | List-II (Shape and Hybridisation) (A) PF3_{3}3 (I) Tetrahedral and sp3^33 (B) SF6_{6}6 (III) Octahedral and sp3^33d2^22 (C) Ni(CO)4_{4}4 (I) Tetrahedral and sp3^33 (D) [PtCl4_{4}4]2−^{2-}2− (II) Square planar and dsp2^22
Let A be a 3 × 3 matrix such that det(A)=5\text{det}(A) = 5det(A)=5. If det(3 adj(2A))=2α⋅3β⋅5γ\text{det}(3 \, \text{adj}(2A)) = 2^{\alpha \cdot 3^{\beta} \cdot 5^{\gamma}}det(3adj(2A))=2α⋅3β⋅5γ, then (α+β+γ) (\alpha + \beta + \gamma) (α+β+γ) is equal to:
If dydx+2ysec2x=2sec2x+3tanx⋅sec2x \frac{dy}{dx} + 2y \sec^2 x = 2 \sec^2 x + 3 \tan x \cdot \sec^2 x dxdy+2ysec2x=2sec2x+3tanx⋅sec2x and
and f(0)=54 f(0) = \frac{5}{4} f(0)=45, then the value of 12(y(π4)−1e2) 12 \left( y \left( \frac{\pi}{4} \right) - \frac{1}{e^2} \right) 12(y(4π)−e21) equals to:
If the domain of the function f(x)=13x+10−x2+1x+∣x∣ f(x) = \frac{1}{\sqrt{3x + 10 - x^2}} + \frac{1}{\sqrt{x + |x|}} f(x)=3x+10−x21+x+∣x∣1 is (a,b) (a, b) (a,b), then (1+a)2+b2 (1 + a)^2 + b^2 (1+a)2+b2 is equal to: