Tags Maths

Maths

Get insights from 6.5k questions on Maths, answered by students, alumni, and experts. You may also ask and answer any question you like about Maths

Follow Ask Question

6.5k

Questions

Discussions

Active Users

Followers

New answer posted

8 months ago

Maths Class 12th

The minimum value of the twice differentiable function f(x) = $\int_{0}^{x} e^{x - t} f' (t) d t - (x^{2} - x + 1) e^{x}, x \in R$ is:

0 Follower 6 Views

Vishal Baghel

Contributor-Level 10

$f (x) = e^{x} . \int_{0}^{x} \frac{f' (t)}{e^{t}} d t$

$f' (x) = e^{x} . \int_{0}^{x} \frac{f' (t)}{e^{t}} d t + e^{x} . \frac{f' (x)}{e^{x}} - [(2 x - 1) . e^{x} + (x^{2} - x + 1) . e^{x}]$

$\begin{array}{l} f (x) = (2 x + 1) . e^{x} - 2 e^{x} + c \\ f (x) = e^{x} (2 x - 1) c = 0 \end{array}$

$f (- \frac{1}{2}) = \frac{- 2}{\sqrt[]{e}}$

0 0

New answer posted

8 months ago

Maths Class 11th

Kindly consider the following questions

0 Follower 3 Views

Vishal Baghel

Contributor-Level 10

$a_{n + 2} a_{n + 1} - a_{n + 1} a_{n} = 2$

Series will satisfy

$a_{1} a_{2}, a_{2} a_{3}, a_{3} a_{4}, . . . . . . a_{4} a_{5}$

$\begin{array}{l} 1.2 2.2 2.3 2.4 \\ \frac{a_{n} + \frac{1}{a_{n + 1}}}{a_{n + 2}} = \frac{a_{n + 2} - \frac{1}{a_{n + 1}}}{a_{n + 2}} \end{array}$

$= 1 - \frac{1}{2 (r + 1)} = \frac{2 r + 1}{2 (r + 1)}$

Now, proof = $\frac{30}{11} \frac{(2 r + 1)}{2 (r + 1)}$ $\frac{}{} \frac{}{}$

r = 1

$= \frac{(1.3.5 . . . . . . 61)}{2^{30} (2.3 . . . . . . 31)}$

$\frac{61}{2^{60} . 31.30} = α = - 60$

0 0

New answer posted

8 months ago

Maths Functions

Let a, b and $γ$ be three positive real number. Let f(x) = ax⁵ + bx³ + $γ x, x \in R$ and g : R ® R be such that $g (f (x)) = x$ for all x $\in R .$ If $a_{1}, a_{2}, a_{3} . . . . ., a_{n}$ be in arithmetic progression with mean zero then the value of $f (g (\frac{1}{n} \sum_{i = 1}^{n} f (a_{i})))$ is equal to:

0 Follower 5 Views

Vishal Baghel

Contributor-Level 10

Case I

= = 0 then

f (x) = yx

$g (x) = \frac{x}{y}$

$\frac{1}{x} \sum_{i = 1}^{n} f (a_{i}) \Rightarrow \frac{y}{x} (a_{1} + a_{2} + . . . . + a_{x}) = 0$

$\Rightarrow f (g (0)) \Rightarrow f (0)$

0 0

New answer posted

8 months ago

Maths Sets

The number of elements in the set S = ${x \in R : 2 c o s (\frac{x^{2} + x}{6}) = 4^{x} + 4^{- x}} i s :$

0 Follower 2 Views

alok kumar singh

Contributor-Level 10

2cos $(\frac{x^{2} + x}{6}) = 4^{x} + 4^{- x}$

$- 2 \leq L H S \leq 2 L H S = 2 & R H S = 2 \Rightarrow x = 0 o n l y \to t h e n L H S = 2 a l s o$

RHS $\geq$ 2

0 0

New answer posted

8 months ago

Maths Class 12th

If the minimum value of f(x) = $\frac{5 x^{2}}{2} + \frac{α}{x^{5}}, x > 0,$ $\frac{^{}}{} \frac{}{^{}}$ is 14, then the value of is equal to:

0 Follower 7 Views

Vishal Baghel

Contributor-Level 10

$\frac{5 x^{2}}{2} + \frac{α}{2 x^{5}} + \frac{α}{2 x^{5}} \geq 7 {(\frac{α^{2}}{2^{7}})}^{\frac{1}{7}}$

$\frac{7 . {(α)}^{2 / 7}}{2} = 14$

${(α^{2})}^{\frac{1}{7}} = 2^{2} \Rightarrow α = {(2^{2})}^{\frac{7}{2}} = 2^{7}$

=128

0 0

New answer posted

8 months ago

Maths Class 11th

Let m₁, m₂ be the slopes of two adjacent sides of a square of side a such that a² + 11a + $3 (m_{1}^{2} + m_{2}^{2}) = 220 .$ If one vertex of the square is $(10 (c o s α - s i n α), 10 (s i n α + c o s α)),$ where $α \in (0, \frac{π}{2})$ and the equation of one diagonal is $(c o s α - s i n α) x + (s i n α + c o s α) y = 10,$ then $72 (s i n^{4} α + c o s^{4} α) + a^{2} - 3 a + 13$ is equal to:

0 Follower 11 Views

alok kumar singh

Contributor-Level 10

$m_{1} m_{2} = - 1,$ for square a,b,c,d let

$A (10 (c o s α - s i n α), 10 (s i n α + c o s α))$

Diagonal : (cos a - sina)x + (sina + cosa)y = 10

BD (diagonal)

Dist. Of BD from A is

$\frac{| 10 {(c o s α - s i n α)}^{2} + 10 {(s i n α + c o s α)}^{2} - 10 |}{\sqrt[]{2}} = \frac{a}{\sqrt[]{2}}$

$\frac{10}{\sqrt[]{2}} = \frac{a}{\sqrt[]{2}} \Rightarrow a = 10$

Also, a² + 11a + 3 $(m_{1}^{2} + m_{2}^{2}) = 220$

-> 210 + 3 $(c m_{1}^{2} + m_{2}^{2}) = 220$

$m_{1}^{2} + m_{2}^{2} = \frac{10}{3}$

Also, m₁ m₂ = -1

->m² + $\frac{1}{m^{2}} = \frac{10}{3}$

or $- \sqrt[]{3}, \frac{1}{\sqrt[]{3}}$

m = $\sqrt[]{3}, \frac{- 1}{\sqrt[]{3}}$

$m^{4} - \frac{10}{3} m^{2} + 1 = 0 \Rightarrow m^{2} = \frac{\frac{10}{3} \pm \sqrt[]{\frac{100}{9} - 4}}{2} - \frac{\frac{10}{3} \pm \frac{8}{3}}{2} = 3, \frac{1}{3}$

m = $\pm \sqrt[]{3}, \pm \frac{1}{\sqrt[]{3}}$

Diagonal AC:

$(s i n α + c o s α) x - (c o s α - s i n α) y$

=10 cos2a - 10cos2a = 0

Slope of AC = $\frac{s i n α + c o s α}{c o s α - s i n α} = \frac{t a n α + 1}{1 - t a n α} = t a n (α + \frac{π}{4}) α = 30 °$

...more

$m_{1} m_{2} = - 1,$ for square a,b,c,d let

$A (10 (c o s α - s i n α), 10 (s i n α + c o s α))$

Diagonal : (cos a - sina)x + (sina + cosa)y = 10

BD (diagonal)

Dist. Of BD from A is

$\frac{| 10 {(c o s α - s i n α)}^{2} + 10 {(s i n α + c o s α)}^{2} - 10 |}{\sqrt[]{2}} = \frac{a}{\sqrt[]{2}}$

$\frac{10}{\sqrt[]{2}} = \frac{a}{\sqrt[]{2}} \Rightarrow a = 10$

Also, a² + 11a + 3 $(m_{1}^{2} + m_{2}^{2}) = 220$

-> 210 + 3 $(c m_{1}^{2} + m_{2}^{2}) = 220$

$m_{1}^{2} + m_{2}^{2} = \frac{10}{3}$

Also, m₁ m₂ = -1

->m² + $\frac{1}{m^{2}} = \frac{10}{3}$

or $- \sqrt[]{3}, \frac{1}{\sqrt[]{3}}$

m = $\sqrt[]{3}, \frac{- 1}{\sqrt[]{3}}$

$m^{4} - \frac{10}{3} m^{2} + 1 = 0 \Rightarrow m^{2} = \frac{\frac{10}{3} \pm \sqrt[]{\frac{100}{9} - 4}}{2} - \frac{\frac{10}{3} \pm \frac{8}{3}}{2} = 3, \frac{1}{3}$

m = $\pm \sqrt[]{3}, \pm \frac{1}{\sqrt[]{3}}$

Diagonal AC:

$(s i n α + c o s α) x - (c o s α - s i n α) y$

=10 cos2a - 10cos2a = 0

Slope of AC = $\frac{s i n α + c o s α}{c o s α - s i n α} = \frac{t a n α + 1}{1 - t a n α} = t a n (α + \frac{π}{4}) α = 30 °$

FIGURE

? = $72 (\frac{1}{16} + \frac{9}{16}) + 100 - 30 + 13 = \frac{720}{16} + 83 = 128$

less

0 0

New answer posted

8 months ago

Maths Class 12th

The foot of the perpendicular from a point on the circle x² + y² = 1, z = 0 to the plane 2x + 3y + z = 6 lies on which one of the following curves?

0 Follower 7 Views

Vishal Baghel

Contributor-Level 10

$\frac{h - c o s θ}{2} = \frac{k - s i n θ}{3} = \frac{w - 0}{1}$

$= \frac{- 1 (2 c o s θ + 3 s i n θ - 6)}{14} \Rightarrow h = \frac{c o s - 2 (2 c o s θ + 3 s i n θ - 6)}{14}$

$k = \frac{5 s i n θ - 6 c o s θ + 18}{14}$

${(5 h + 6 k - 12)}^{2} + 4 {(3 h + 5 k - 9)}^{2} = 1$

0 0

New answer posted

8 months ago

Maths Class 12th

Let $S_{1} = {z_{1} \in C : | z_{1} - 3 | = \frac{1}{2}}$ and $S_{2} = {z_{2} \in C : | z_{2} - | z_{2} + 1 | | = | z_{2} + | z_{2} - 1 | |} .$ Then, for $z_{1} \in S_{1}$ and $s_{2} \in S_{2},$ the least value of $| z_{2} - z_{1} | i s :$

0 Follower 12 Views

Vishal Baghel

Contributor-Level 10

$S_{1} = {z, \in c : ⌊ z_{1} - 3 ⌋ = \frac{1}{2}} a n d S_{2} = {z_{2} \in c : | z_{2} - | z_{2} + 1 | | = | z_{2} + | z_{2} - 1 | |}$

Then, for $z_{1} \in S_{1}$ and $z_{2} \in S_{2},$ the least value of |z₂ – z₁|

${| z_{2} + | z_{2} - 1 | |}^{2} = {| z_{2} - | z_{2} + 1 | |}^{2}$

$\Rightarrow (z_{2} + {\bar{z}}_{2}) (| z_{2} - 1 | + | z_{2} + 1 | - 2) = 0$

$∴ z_{2} + {\bar{z}}_{2} = 0 o r | z_{2} - 1 | + | z_{2} + 1 | - 2 = 0$

$∴$ z₂ lies on imaginary axis or on real axis with in [-1, 1]

also $| z_{1} - 3 | = \frac{1}{2}$ lie on circle having centre 3 and radius $\frac{1}{2}$

Clearly $| z_{1} - z_{2} | m i n = \frac{5}{2} - 1 = \frac{3}{2}$

0 0

New answer posted

8 months ago

Maths Matrices

Let the matrix $A = [\begin{matrix} 0 & 1 & 0 \\ 0 & 0 & 1 \\ 1 & 0 & 0 \end{matrix}]$ and the matrix B₀ = A²⁹ + 2A⁹⁸. If B_n = Adj(B_n=1) for all $n \geq 1,$ then det

0 Follower 4 Views

Vishal Baghel

Contributor-Level 10

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

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

$a \leftrightarrow$ R2

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

$R_{2} \leftrightarrow R_{3}$

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

$\begin{array}{l} B_{0} = A^{49} + 2 A^{98} = A + 2 I \\ B_{n} = A d j (B_{n} - 1) \\ B_{4} = A d j (A d j (A d j (A d j B_{0}))) \end{array}$

$= {| B_{0} |}^{{(n - 1)}^{4}} = {| B_{0} |}^{16}$

$B_{0} = [\begin{matrix} 0 & 1 & 0 \\ 0 & 0 & 1 \\ 1 & 0 & 0 \end{matrix}] + [\begin{matrix} 2 & 0 & 0 \\ 0 & 2 & 0 \\ 0 & 0 & 2 \end{matrix}] = [\begin{matrix} 2 & 1 & 0 \\ 0 & 2 & 1 \\ 1 & 0 & 2 \end{matrix}]$

$= 2 (4 - 0) - 1 (0 - 1) = 9$

$B_{4} {(9)}^{16} = 3^{32}$

0 0

New answer posted

8 months ago

Maths Class 12th

Let y = y(x) be the solution curve of the differential equation $\frac{d y}{d x} + (\frac{2 x^{2} + 11 x + 13}{x^{3} + 6 x^{2} + 11 x + 6}) y = \frac{(x + 3)}{x + 1}, x > - 1,$ which passes through the point (0, 1). Then y(1) is equal to:

0 Follower 2 Views

alok kumar singh

Contributor-Level 10

$\frac{d y}{d x} + (\frac{2 x^{2} + 11 x + 13}{x^{3} + 6 x^{2} + 11 x + 6}) y = \frac{x + 3}{x + 1}, x > - 1$

IF = $e^{\int p d x} = \frac{{(x + 1)}^{2} (x + 2)}{x + 3}$

$\int P d x = \int \frac{2 x^{2} + 11 x + 13}{x^{3} + 6 x^{2} + 11 x + 6} d n = \int (\frac{2}{x + 1} + \frac{1}{x + 2} - \frac{1}{x + 3}) d x$

$= l n ({(x + 1)}^{2} \cdot (x + 2) / (x + 3))$

$\frac{2 x^{2} + 11 x + 13}{(x + 1) (x + 2) (x + 3)} = \frac{A}{x + 1} + \frac{B}{x + 2} + \frac{C}{x + 3}$

$2 x^{2} + 11 x + 13 = A (x + 2) (x + 3) + B (x + 1) (x + 3) + C (x + 1) (x + 2)$

x = -1

->4 = 2A Þ A = 2

x = -2

-> -1 = -B Þ B = 1

x₂ – 3 Þ -2 = 2c

c = -1

$y \cdot \frac{{(x + 1)}^{2} (x + 2)}{x + 3} = \int \frac{x + 3}{x + 1} \cdot \frac{{(x + 1)}^{2} (x + 2)}{x + 3} d x$

= $\int (x + 1) (x + 2) d x$

= $\frac{x^{3}}{3} + \frac{3 x^{2}}{2} + 2 x + c$

$(0, 1) \Rightarrow 1 \cdot \frac{2}{3} = c$

x = 1 $y \cdot (3) = \frac{1}{3} + \frac{3}{2} + 2 + \frac{2}{3} = \frac{3}{2} + 3 = \frac{9}{2}$

y = $\frac{3}{2}$

0 0

Get authentic answers from experts, students and alumni that you won't find anywhere else

On Shiksha, get access to

66k Colleges
1.2k Exams
687k Reviews
1800k Answers

Community GuidelinesUser Points System

QuestionsDiscussionsTags

Need guidance on career and education? Ask our experts

Your Question

Maths

Questions

Discussions

Active Users

Followers

All

Discussions

Unanswered Questions

Share Your College Life Experience