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Class 12th Maths

The number of solutions of equation e^{sin x} − 2e^{– sin x} = 2 is

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alok kumar singh

Contributor-Level 10

Take e^sinx = t (t > 0)

⇒ $t - \frac{2}{t} = 2$

⇒ $\frac{t^{2} - 2}{t} = 2$

->t² – 2t – 2 = 0

->t² – 2t + 1 = 3

⇒ (t −1)² = 3

⇒ t = 1 ± $\sqrt[]{3}$

⇒ t = 1 ± 1.73

⇒ t = 2.73 or –0.73 (rejected as t > 0)

⇒ e^{sin x} = 2.73

->log_e e^{sin x} = log_e 2.73

⇒ sin x = log_e 2.73 > 1

So no solution.

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3 months ago

Class 12th Maths

Let : f(0, R)->∞ be increasing function such that $\underset{x \to \infty}{l i m} \frac{f (7 x)}{f (x)} = 1$ then $\underset{x \to \infty}{l i m} (\frac{f (5 x)}{f (x)} - 1)$ is equal to

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alok kumar singh

Contributor-Level 10

f is increasing function

x < 5x < 7x

f (x) < f (5x) < f (7x)

$\frac{f (x)}{f (x)} < \frac{f (5 x)}{f (x)} < \frac{f (7 x)}{f (x)}$

$\underset{x \to \infty}{l i m} \frac{f (x)}{f (x)} < \underset{x \to \infty}{l i m} \frac{f (5 x)}{f (x)} < \underset{x \to \infty}{l i m} \frac{f (7 x)}{f (x)}$

-> $1 < \underset{x \to \infty}{l i m} \frac{f (5 x)}{f (x)} < 1$ $\underset{x \to \infty}{l i m} \frac{f (5 x)}{f (x)} = 1$

$\underset{x \to \infty}{l i m} (\frac{f (5 x)}{f (x)} - 1) = 0$

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3 months ago

Class 12th Matrices

If $A [\begin{matrix} 1 & 0 & 1 \end{matrix}] = 2 [\begin{matrix} 1 & 0 & 1 \end{matrix}]$ , $A [\begin{matrix} - 1 & 0 & 1 \end{matrix}] = 4 [\begin{matrix} - 1 & 0 & 1 \end{matrix}]$ and $A [\begin{matrix} 0 & 1 & 0 \end{matrix}] = 2 [\begin{matrix} 1 & 0 & 0 \end{matrix}]$ where, A is a 3 * 3 matrix and (A –3I) $[\begin{matrix} x & y & z \end{matrix}] = [\begin{matrix} - 1 & 2 & 3 \end{matrix}]$ then the value of (x, y, z) is

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alok kumar singh

Contributor-Level 10

Let $A = [\begin{matrix} x_{1} & y_{1} & z_{1} \\ x_{2} & y_{2} & z_{2} \\ x_{3} & y_{3} & z_{3} \end{matrix}]$

Given $A = [\begin{matrix} 1 & 0 & 1 \end{matrix}] = [\begin{matrix} 2 & 0 & 2 \end{matrix}]$ .(1)

$[\begin{matrix} x_{1} + z_{1} & x_{2} + z_{2} & x_{3} + z_{3} \end{matrix}] = [\begin{matrix} 2 & 0 & 2 \end{matrix}]$

∴ x₁ + z₁ = 2 … (2)

x₂ + z₂ = 0 … (3)

x₃ + z₃ = 0 … (4)

Given $A = [\begin{matrix} - 1 & 0 & 1 \end{matrix}] = [\begin{matrix} - 4 & 0 & 4 \end{matrix}]$

$[\begin{matrix} - x_{1} + z_{1} & - x_{2} + z_{2} & - x_{3} + z_{3} \end{matrix}] = [\begin{matrix} 4 & 0 & 4 \end{matrix}]$

⇒ – x₁ + z₁ = −4 … (5)

–x₂ + z₂ = 0 … (6)

–x₃ + z₃ = 4

...more

Let $A = [\begin{matrix} x_{1} & y_{1} & z_{1} \\ x_{2} & y_{2} & z_{2} \\ x_{3} & y_{3} & z_{3} \end{matrix}]$

Given $A = [\begin{matrix} 1 & 0 & 1 \end{matrix}] = [\begin{matrix} 2 & 0 & 2 \end{matrix}]$ .(1)

$[\begin{matrix} x_{1} + z_{1} & x_{2} + z_{2} & x_{3} + z_{3} \end{matrix}] = [\begin{matrix} 2 & 0 & 2 \end{matrix}]$

∴ x₁ + z₁ = 2 … (2)

x₂ + z₂ = 0 … (3)

x₃ + z₃ = 0 … (4)

Given $A = [\begin{matrix} - 1 & 0 & 1 \end{matrix}] = [\begin{matrix} - 4 & 0 & 4 \end{matrix}]$

$[\begin{matrix} - x_{1} + z_{1} & - x_{2} + z_{2} & - x_{3} + z_{3} \end{matrix}] = [\begin{matrix} 4 & 0 & 4 \end{matrix}]$

⇒ – x₁ + z₁ = −4 … (5)

–x₂ + z₂ = 0 … (6)

–x₃ + z₃ = 4 … (7)

Given $A = [\begin{matrix} 0 & 1 & 0 \end{matrix}] = [\begin{matrix} 0 & 2 & 0 \end{matrix}]$

$[\begin{matrix} y_{1} & y_{2} & y_{3} \end{matrix}] = [\begin{matrix} 0 & 2 & 0 \end{matrix}]$

∴ y₁ = 0, y₂ = 2, y₃ = 0

∴ from (2), (3), (4), (5), (6) and (7)

x₁ = 3, x₂ = 0, x₃ = –1

y₁ = 0, y₂ = 2, y₃ = 0

z₁ = –1, z₂ = 0, z₃ = 3

$A = [\begin{matrix} 3 & 0 & - 1 \\ 0 & 2 & 0 \\ - 1 & 0 & 3 \end{matrix}]$

Now (A – 3I) $[\begin{matrix} x & y & z \end{matrix}]$ = $[\begin{matrix} - 1 & 2 & 3 \end{matrix}]$

$[\begin{matrix} 0 & 0 & - 1 \\ 0 & - 1 & 0 \\ - 1 & 0 & 0 \end{matrix}] [\begin{matrix} x & y & z \end{matrix}] = [\begin{matrix} - 1 & 2 & 3 \end{matrix}]$

$[\begin{matrix} - z & - y & - x \end{matrix}]$ = $[\begin{matrix} - 1 & 2 & 3 \end{matrix}]$

[z = 1], [y = –2], [x = –3]

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3 months ago

Class 12th Maths Application of Integrals

The area of the region enclosed by the parabolas y = 4 – x² and 3y = (x – 4)² is in (sq. unit)?

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alok kumar singh

Contributor-Level 10

Area = $| \int_{1}^{4} [{(4 - x)}^{2} - \frac{{(x - 4)}^{2}}{3}] | d x$

Area = $= | (16 - \frac{64}{3}) - (4 - \frac{1}{3} + \frac{27}{9}) |$

$= | (16 - \frac{64}{3}) - (4 - \frac{1}{3} + \frac{27}{9}) |$

$= | 16 - \frac{64}{3} - 4 + \frac{1}{3} + 3 |$

$= | 15 - 2 | = 6$

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3 months ago

Class 12th Maths

A coin is biased such that head has two chances than tails, what is the probability of getting 2 heads and 1 tail?

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alok kumar singh

Contributor-Level 10

Let probability of tail is $\frac{1}{3}$

⇒ Probability of getting head = $\frac{2}{3}$

∴ Probability of getting 2 heads and 1 tail

$= (\frac{2}{3} * \frac{2}{3} * \frac{1}{3}) * 3$

$= \frac{4}{27} * 3$

$= \frac{4}{9}$

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3 months ago

Class 12th Chemistry

The potassium chloride is heated with potassium dichromate and conc. sulphuric acid to give products. The oxidation state of chromium in product is (+)_______.

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