Class 11th

Contributor-Level 10

Hence

CH₃ – CH₂ – CHO can't be formed.

In case of unsymmetrical alkynes, more positive charge stabilizing carbon attacked by H₂O and finally converted into carbonyl.

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The correct set from the following in which both pairs are in correct order of melting point is:

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Contributor-Level 10

MgO has more lattice enthalpy than NaCl

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If the curves x = y⁴ and xy = k cut at right angles, then (4k)⁶ equal to………….

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Payal Gupta

Contributor-Level 10

$x = y^{4}, x y = k$

$\frac{d y}{d x} = \frac{1}{4 y^{3}}, \frac{d y}{d x} = \frac{- k}{x^{2}}$

P (x₁, y₁)

where $x_{1} = y_{1}^{4} & x_{1} y_{1} = k$

$\Rightarrow y_{1} = k^{1 / 5}, x_{1} y_{1} = k$

$m_{1} m_{2} = - 1$

$\Rightarrow \frac{- 1}{4 . k^{6 / 5}} = - 1 \Rightarrow k^{6 / 5} = \frac{1}{4} \Rightarrow k^{6} = \frac{1}{4^{5}} = \frac{1}{2024}$

${(4 k)}^{6} = 2^{12} . \frac{1}{2^{10}} = 2^{2} = 4$

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Given below are two statements : one is labelled as Assertion A and the other is labelled as Reason R.

Assertion A : Hydrogen is the most abundant element in the Universe, but it is not the most abundant gas in the troposphere.

Reason R : Hydrogen is the lightest element

In the light of the above statements, choose the correct answer from the options given below:

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Contributor-Level 10

Hydrogen is most abundant element in earth crust and the lightest element. But in troposphere nitrogen gas in most abundant element

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Which one of the following compounds is non-aromatic?

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Class 11th Relations and Functions

Contributor-Level 10

Kindly consider the following figure

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4 months ago

Let f: R -> R be defined as

$f (x) = t {\begin{matrix} - 55 x, & i f x < - 5 \\ 2 x^{3} - 3 x^{2} - 120 x, & i f - 5 \leq x \leq 4 \\ 2 x^{3} - 3 x^{2} - 36 x - 336, & i f x > 4, \end{matrix}$

Let A = ${x \in R : f i s i n c r e a s i n g} .$ Then A is equal to:

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alok kumar singh

Contributor-Level 10

let f : R -> R

$f' (x) = {\begin{cases} - 55, x < - 5 \\ 6 x^{2} - 6 x - 120, - 5 < x < 4 \\ 6 x^{2} - 6 x - 36, x > 4 \end{cases}$

f' (x) increasing in x $\in (- 5, - 4) \cup (4, \infty)$

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4 months ago

Match List – I and List – II

Choose the correct answer from the options given below:

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Contributor-Level 10

Kindly consider the following figure

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4 months ago

A line is a common tangent circle and the parabola y² = 4x. If the two point of contact (a, b) and (c, d) are distinct and lie in the first quadrant, then 2(a + c) is equal to………

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Payal Gupta

Contributor-Level 10

A tangent to y² = 4x is x – ty + t² = 0

$\frac{3 + t^{2}}{\sqrt[]{1 + t^{2}}} = 3$

(3 + t²)² = 9 (1 + t²)

$9 + t^{4} + 6 t^{2} = 9 + 9 t^{2}$

Point of contact $(3, 2 \sqrt[]{3}) = (a, b)$

$x - \sqrt[]{3} y + 3 = 0$

$\frac{\sqrt[]{3} x + y - 3 \sqrt[]{3} = 0]}{4 x - 6 = 0}$

$x = \frac{3}{2}, y = \frac{\sqrt[]{3}}{2} + \sqrt[]{3}$

& $(\frac{3}{2}, \frac{\sqrt[]{3}}{2} + 2) = (c, d), 2 (a + c) = 2 (3 + \frac{3}{2}) = 9$

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4 months ago

If $\underset{x \to 0}{l i m} \frac{a x - (e^{4 x} - 1)}{a x (e^{4 x} - 1)}$ exists and is equal to b, then the value of a – 2b is…….

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Payal Gupta

Contributor-Level 10

$\underset{x \to 0}{l i m} \frac{a x - (e^{4 x} - 1)}{a x (e^{4 x} - 1)} = b,$ use of L' Hospital rule implies

$\underset{x \to 0}{l i m} \frac{a - 4 e^{4 x}}{a (e^{4 x} - 1) + a x (4 e^{4 x})}$

$= \frac{a - 4}{0} \Rightarrow a = 4$

$\Rightarrow \underset{n ? \to 0}{l i m} \frac{4 (- 4 . e^{4 x})}{4.4 e^{4 x} + 16 e^{4 x} + 16 x . 4 e^{4 x}}$

$= \frac{- 16}{16 + 16} = - \frac{1}{2} = b$

a – 2b = 4 – (1) = 5

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4 months ago