Class 12th

Which of the following oxoacids of sulphur contains “S” in two different oxidation states?

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Among the given oxides of nitrogen ; N₂O, N₂O₃, N₂O₄ and N₂O₅, the number of compound / (s) having N – N bond is:

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Among the following, basic oxide is:

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CaO – Basic Oxide

SO₃ & SiO₂ – Acidic oxides

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Hydrogen has three isotopes : protium (¹H), deuterium (²H or D) and tritium (³H or T). They have nearly same chemical properties but different physical properties. They differ in

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Isotopes are the atom of same element with different atomic mass.

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Class 12th Maths Matrices

Let $M = [\begin{matrix} 0 & - α \\ α & 0 \end{matrix}],$ where is a non-zero real number an $N = \sum_{k = 1}^{49} M^{2 k} .$ If (I – M²)N = 2I, then the positive integral value of is

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$N = M^{2} + M^{4} + . . . . . + M^{98}$

$= (- α^{2} I) + {(- α^{2} I)}^{2} + . . . . + {(- α^{2} I)}^{49}$

$= I (- α^{2} + α^{4} - α^{6} + . . . . - α^{98})$

N = $- I (α^{2} - α^{4} + α^{6} . . . . . . . + α^{98})$

$= - I \frac{α^{2} (1 - {(- α^{2})}^{49})}{1 - (- α^{2})}$

N = $\frac{- I α^{2} (1 + α^{98})}{1 + α^{2}}$

Now $(I - m^{2}) N = - 2 I$

$(I + α^{2} I) \frac{(- I α^{2} \cdot (1 + α^{98})}{1 + α^{2}} = - 2 I$

? α¹⁰⁰ + α² = 2

? α = $\pm 1$

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Let f and g be twice differentiable even functions on (2, 2) such that $f (\frac{1}{4}) = 0, f (\frac{1}{2}) = 0, f (1) = 1$ and $f (\frac{3}{4}) = 0, g (1) = 2$

Then, the minimum number of solutions of $f (x) g " (x) + f' (x) g' (x) = 0 i n (- 2, 2)$ is equal to

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f (x) is an even function

$f (- \frac{1}{4}) = f (- \frac{1}{2}) = f (\frac{1}{2}) = f (\frac{1}{4}) = 0$

So, f (x) has at least four roots in (-2, 2)

$g (\frac{- 3}{4}) = g (\frac{3}{4}) = 0$

So, g (x) has at least two roots in (2, 2)

now number of roots of f (x) $\cdot g " (x) = f' (x) \cdot g' (x) = 0$

It is same as number of roots of $\frac{d}{d x} (f (x) \cdot g' (x)) = 0$ will have atleast 4 roots in (2, 2)

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Class 12th Maths Matrices

For real numbers a, b(a > b > 0), let

$A r e a {(x, y) : x^{2} + y^{2} \leq a^{2} a n d \frac{x^{2}}{a^{2}} + \frac{y^{2}}{b^{2}} \geq 1} = 30 π a n d$

$A r e a {(x, y) : x^{2} + y^{2} \geq b^{2} a n d \frac{x^{2}}{a^{2}} + \frac{y^{2}}{b^{2}} \leq 1} = 18 π$

Then the value of (a – b)² is equal to

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Given a > b

Area common to x2 + y2 $\leq a^{2} a n d \frac{x^{2}}{a^{2}} + \frac{y^{2}}{b^{2}} \leq 1$

is $π a^{2} - π a b = 30 π . . . . . . . . . . . . . . (i)$

Similarly $π a b - π b^{2} = 18 π . . . . . . . . . . . . . . . . . (i i)$

Equation (i) and equation (ii) $\frac{a}{b} = \frac{5}{3}$

Equation (i) + equation (ii) $a^{2} - b^{2} = 48$

a²= 75, b² = 27

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Let y = y(x), x > 1, be the solution of the differential equation

$(x - 1) \frac{d y}{d x} + 2 x y = \frac{1}{x - 1}, w i t h y (2) = \frac{1 + e^{4}}{2 e^{4}} .$ If y(3) = $\frac{e^{α} + 1}{β e^{α}}$ , then the value of α+β is equal to

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$\frac{d y}{d x} + \frac{2 x}{x - 1} \cdot y = \frac{1}{{(x - 1)}^{2}}$

IF $= e^{\int \frac{2 x}{x - 1} d x}$

= $e^{2 x} \cdot {(x - 1)}^{2}$

$y \cdot e^{2 x} {(x - 1)}^{2} = {e^{2 x} \frac{{(x - 1)}^{2}}{{(x - 1)}^{2}} d x + C$

y = $\frac{e^{2 x}}{2 {(x - 1)}^{2}} + \frac{C}{{(x - 1)}^{2}}$

y (2) = $\frac{1 + e^{4}}{2 e^{4}}, \Rightarrow C = \frac{1}{2}$

y (3) = $\frac{e^{α} + 1}{β e^{α}} = \frac{e^{6} + 1}{8 e^{6}}$

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Let $\vec{a} = \hat{i} - 2 \hat{j} + 3 \hat{k}, \vec{b} = \hat{i} + \hat{j} + \hat{k} a n d \vec{c}$ be a vector such that $\vec{a} + (\vec{b} * \vec{c}) = \vec{0} a n d \vec{b} . \vec{c} = 5 .$ Then, the value of $3 (\vec{c} . \vec{a})$ is equal to

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Data contradiction.

$\vec{a} * (\vec{b} * \vec{c}) = (\vec{a} \cdot \vec{c}) \vec{b} - (\vec{a} \cdot \vec{b}) \vec{c}$

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Let A, B, C be three points whose position vectors respectively are

$\vec{a} = \hat{i} + 4 \hat{j} + 3 \hat{k}$

$\vec{b} = 2 \hat{i} + α \hat{j} + 4 \hat{k}, α \in R$

$\vec{c} = 3 \hat{i} - 2 \hat{j} + 5 \hat{k}$

If is the smallest positive integer for which $\vec{a}, \overset{\leftarrow}{b}, \vec{c}$ are noncollinear, then the length of the median, in $Δ A B C,$ through A is

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