Class 12th

Let the area enclosed by the x-axis, and the tangent and normal drawn to the curve 4x³ -3xy² + 6x² -5xy -8y² + 9x + 14 = 0 at the point (-2, 3) be A. Then 8A is equal to

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Contributor-Level 10

$4 x^{3} - 3 x y^{2} + 6 x^{2} - 5 x y - 8 y^{2} + 9 x + 14 = 0$ differentiating both sides we get

$12 x^{2} - 3 y^{2} - 6 x y y' + 12 x - 5 y - 5 x y = 16 y y' + 9 = 0$

At the point (2, 3)

48 – 27 + 36y' – 24 – 15 + 10y' – 48y' + 9 = 0

$∴ A r e a = \frac{1}{2} * B a s e * H e i g h t$

$A = \frac{1}{2} * (\frac{- 4}{3} + \frac{31}{2}) (3) = \frac{1}{2} (\frac{85}{0}) . 3 = \frac{85}{4} = 8 A = 170$

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New answer posted

a year ago

Let f be a twice differentiable function on R. If $f' (0) = 4$ and $f (x) + \int_{0}^{x} (x - t) f' (t) d t = (e^{2 x} + e^{- 2 x}) c o s 2 x + \frac{2}{a} x,$ then (2a + 1)⁵a² is equal to

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Contributor-Level 10

$f (x) + \int_{0}^{x} (x - t) f' (t) d t = (e^{2 x} + e^{- 2 x}) c o s 2 x + \frac{2 x}{a} . . . . . . . (i)$

Here f (0) = 2 ………. (ii)

On differentiating equation (i) w.r.t. x we get :

$f' (x) + f \int_{0}^{x} f' (t) d t + x f' (x) - x f' (x)$

$4 = \frac{2}{a} \Rightarrow a = \frac{1}{2}$

$∴ {(2 a + 1)}^{5} . a^{2} = 2^{5} . \frac{1}{2^{2}} = 2^{3} = 8$

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New answer posted

a year ago

Let y = y(x) be the solution of the differential equation $\frac{d y}{d x} = \frac{4 y^{3} + 2 y x^{2}}{3 x y^{2} + x^{3}}, y (1) = 1 .$ .If for some $n \in N, y (2) \in [n - 1, n],$ then n is equal to

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Contributor-Level 10

$\frac{d y}{d x} = \frac{y}{x} \frac{(4 y^{2} + 2 x^{2})}{(3 y^{2} + x^{2})}$

Put y = vx

$\Rightarrow \frac{d y}{d x} = v + x \frac{d v}{d x}$

$\Rightarrow v + x \frac{d v}{d x} = \frac{v (4 v^{2} + 2)}{(3 v^{2} + 1)}$

$l n (\frac{y^{2}}{8} + \frac{y}{2}) = 2 l n 2 \Rightarrow \frac{y^{3}}{8} + \frac{y}{2} = 4$

$\Rightarrow [y (2)] = 2$

$\Rightarrow n = 3$

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New answer posted

a year ago

The sum of the maximum and minimum values of the function f(x) = $| 5 x - 7 | + [x^{2} + 2 x]$ in the interval $[\frac{5}{4}, 2],$ where [t] is the greatest integer $\leq t$ is

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Contributor-Level 10

$f (x) = | 5 x - 7 | + [x^{2} + 2 x]$

$= | 5 x - 7 | + [{(x + 1)}^{2}] - 1$

f $(\frac{7}{4}) = 0 + 4 = 4$

as both |5x – 7| and x² + 2x are increasing in nature after x = 7/5

f (2) = 3 + 8 = 11

$f {(\frac{7}{5})}_{m i n} - 4 a n d f {(2)}_{m a x} = 11$

Sum is 4 + 11 = 15

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New answer posted

a year ago

Class 12th Maths Matrices

Let A = $| \begin{matrix} 1 & a & a \\ 0 & 1 & b \\ 0 & 0 & 1 \end{matrix} |$ , a, b $\in$ R. If for some n $\in$ N, Aⁿ = $| \begin{matrix} 1 & 48 & 2160 \\ 0 & 1 & 96 \\ 0 & 0 & 1 \end{matrix} |$ then n + a + b is equal to

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Contributor-Level 10

$A = [\begin{matrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{matrix}] + [\begin{matrix} 0 & a & a \\ 0 & 0 & b \\ 0 & 0 & 0 \end{matrix}] = l + B$

$B^{2} = [\begin{matrix} 0 & a & a \\ 0 & 0 & b \\ 0 & 0 & 0 \end{matrix}] * [\begin{matrix} 0 & a & a \\ 0 & 0 & b \\ 0 & 0 & 0 \end{matrix}] = [\begin{matrix} 0 & 0 & a b \\ 0 & 0 & 0 \\ 0 & 0 & 0 \end{matrix}]$

B³ = 0

$∴ A^{n} = {(1 + B)}^{n} =^{n} C_{0} l +^{n} C_{1} B +^{n} C_{2} B^{2} +^{n} C_{3} B^{3} + . . . .$

On comparing we get na = 48, nb = 96 and na + $\frac{n (n - 1)}{2} a b = 2160$

$\Rightarrow a = 4, n = 12 a n d b = 8$

$\Rightarrow n + a + b = 24$

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New answer posted

a year ago

Let f(x) be a quadratic polynomial with leading coefficient 1 such that f(0) = p, p $\neq 0$ , and f(1) = $\frac{1}{3} .$ If the equations f(x) = 0 and fo fo fo f(x) = 0 have a common real root , then f(3) is equal to

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Contributor-Level 10

Let f (x) = (x -α) (x - β)

It is given that f (0) = p = p

and

$f (1) = \frac{1}{3} \Rightarrow (1 - α) (1 - β) = \frac{1}{3}$

Now, let us assume that is the common root of f (x) = 0 and fofofof (x) = 0

fofofof (x) = 0

$f (- 3) = (- 3 - \frac{7}{6}) (3 - 3) = 25$

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New answer posted

a year ago

The total number of monobromo derivatives formed by the alkanes with molecular formula C₅H₁₂ is (excluding stereo isomers)___________.

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Contributor-Level 10

Consider the image below

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New answer posted

a year ago

The separation of two coloured substances was done by paper chromatography. The distance travelled by solvent front, substance A and substance B from the base line are 3.25cm, 2.08cmand 1.05cm, respectively. The ratio of R_f values of A to B is ________.

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Contributor-Level 10

$R_{f}^{0} = distance travelled by the solute$ $/$ $/ distance travelled by the solvent$

${(R_{f})}_{A} = \frac{2.08}{3.25} {(R_{f})}_{B} = \frac{1.05}{3.25}$

$\frac{{(R_{f})}_{A}}{{(R_{f})}_{B}} = \frac{2}{1}$

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New answer posted

a year ago

A sample of 4.5mg of an unknown monohydric alcohol, R-OH was added to methylmagnesium iodide. A gas is evolved and is collected and its volume measured to be 3.1 mL. The molecular weight of the unknown alcohol is ____________ g/mol.

[Nearest integer]

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Contributor-Level 10

(Calculation is done considering STP condition)

$R O H + C H_{3} M g l \to C H_{4} + R O M g l$

No. of moles ROH = no. of moles of CH4

$\frac{4.5 * 1 0^{- 3}}{M} = \frac{31}{22400} \Rightarrow 32.52 \approx 33 g m / m o l e$

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New answer posted

a year ago

The spin-only magnetic moment value of M³+ ion (in gaseous state) from the pairs Cr³⁺/Cr²⁺, Mn³⁺/Mn²⁺, Fe³⁺/Fe²⁺ and Co³⁺/Co²⁺ that has negative standard electrode potential, is________ B.M [Nearest integer]

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