Ncert Solutions Maths class 12th

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

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

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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8 months 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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8 months ago

The line of shortest distance between the lines $\frac{x - 2}{0} = \frac{y - 1}{1} = \frac{z}{1}$ and $\frac{x - 3}{2} = \frac{y - 5}{2} = \frac{z - 5}{1}$ makes an angle of $c o s^{- 1} (\sqrt[]{\frac{2}{27}})$ with the plane P : ax – y – z = 0, (a > 0). If the image of the point (1, 1, - 5) in the plane P is $(α, β, γ),$ then $α + β - γ$ is equal to……….

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Line of shortest distance will be along ${\bar{b}}_{1} * {\bar{b}}_{2}$ $_{}_{}$

where $\begin{array}{l} _{} \end{array}$ $\begin{array}{l} {\bar{b}}_{1} = \hat{j} + \hat{k} \\ a n d \\ {\bar{b}}_{2} = 2 \hat{i} + 2 \hat{j} + \hat{k} \\ {\bar{b}}_{1} * {\bar{b}}_{2} = | \begin{matrix} \hat{i} & \hat{j} & \hat{k} \\ 0 & 1 & 1 \\ 2 & 2 & 1 \end{matrix} | = - \hat{i} + 2 \hat{j} - 2 \hat{k} \end{array}$

Angle between ${\bar{b}}_{1} * {\bar{b}}_{2}$ $_{}_{}$ and plane P,

$\Rightarrow a^{2} = - \frac{25}{11} (n o t p o s s i b l e)$

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8 months ago

If $\underset{n \to \infty}{l i m} \frac{{(n + 1)}^{k - 1}}{n^{k + 1}} [(n k + 1) + (n k + 2) + . . . . + (n k + n)]$

$= 33 . \underset{n \to \infty}{l i m} \frac{1}{n^{k + 1}} .$ $[1^{k} + 2^{k} + 3^{k} + . . . + n^{k}]$ , then the integral value of k is equal to…………

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Contributor-Level 10

$l i m_{n \to \infty} {(\frac{n + 1}{n})}^{k - 1} \frac{1}{n} \sum_{r = 1}^{n} (k + \frac{r}{n})$

= $33 \int_{0}^{1} x^{k} d x$

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8 months ago

Let f(x) = ${\begin{matrix} | 4 x^{2} - 8 x + 5 |, i f 8 x^{2} - 6 x + 1 \geq 0 & [4 x^{2} - 8 x + 5], i f 8 x^{2} - 6 x + 1 < 0 \end{matrix}}$ , where [a] denotes the greatest integer less than or equal to a. Then the number of points in R where f is not differentiable is………….

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Ncert Solutions Maths class 12th Binomial Theorem

Contributor-Level 10

$f (x) = {\begin{cases} | 4 x^{2} - 8 x + 5 |, i f 8 x^{2} - 6 x + 1 \geq 0 \\ [4 x^{2} - 8 x + 5], i f 8 x^{2} - 6 x + 1 < 0 \end{cases}$

$x = \frac{1}{4}, \frac{2 - \sqrt[]{2}}{2}, \frac{1}{2}$

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8 months ago

If the maximum value of the term independent of t in the expansion of ${(t^{2} x^{\frac{1}{5}} + \frac{{(1 - x)}^{\frac{1}{10}}}{t})}^{15}$ , x $\geq$ , is K, then 8 K is equal to………….

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General term

$=^{15} C_{r} {(t^{2} x^{\frac{1}{5}})}^{15 - r} {(\frac{{(1 - x)}^{\frac{1}{10}}}{t})}^{r}$

for term independent on t

2 (15 – r) – r = 0

$\Rightarrow r = 10$

$∴ T_{11} =^{15} C_{10} * (1 - x)$

Maximum value of x (1 – x) occur at

x = $\frac{1}{2}$

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8 months ago

Let $S = {(x, y) \in N * N : 9 {(x - 3)}^{2} + 16 {(y - 4)}^{2} \leq 144}$ and $T = {(x, y) \in R * R : {(x - 7)}^{2} + {(y - 4)}^{2} \leq 36} .$ Then $n (S \cap T)$ is equal to

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Contributor-Level 10

$\frac{{(x - 3)}^{2}}{16} + \frac{{(y - 4)}^{2}}{9} \leq 1, x, y \in N, {(x - 7)}^{2} + {(y - 4)}^{2} \geq 36, x, y \in R$

Total number of common point = 1 + 5 + 7 + 5 + 5 + 3 + 1 = 27

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8 months ago

Let $\vec{a} a n d \vec{b}$ be two vectors such that $| \vec{a} + \vec{b} | = {| \vec{a} |}^{2} + 2 {| \vec{b} |}^{2}, \vec{a} . \vec{b} = 3 a n d {| \vec{a} * \vec{b} |}^{2} = 75 .$ Then ${| \overset{\leftrightarrow}{a} |}^{2}$ is equal to

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Given : ${| \vec{a} + \vec{b} |}^{2} = {| \vec{a} |}^{2} + 2 {| \vec{b} |}^{2} & \vec{a} \cdot \vec{b} = 3$

$| \vec{a} | | \vec{b} | c o s θ = 3$

$| \vec{a} | c o s θ = \frac{3}{\sqrt[]{6}} = \sqrt[]{\frac{9}{6}} = \sqrt[]{\frac{3}{2}}$

${| \vec{a} |}^{2} {| \vec{b} |}^{2} s i n^{2} θ = 75$

$| \vec{a} | s i n θ = \sqrt[]{\frac{75}{6}} = \frac{5}{\sqrt[]{2}}$

$| \vec{a} | c o s θ = \sqrt[]{\frac{3}{2}}$

${| \vec{a} |}^{2} = \frac{25}{2} + \frac{3}{2} = \frac{28}{2} = 14$

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8 months ago

If the tangent to the curve $y = x^{3} - x^{2} + x$ at the point (a, b) is also tangent to the curve y = 5x² + 2x – 25 at the point (2, 1), then $| 2 a + 9 b |$ is equal to

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Contributor-Level 10

y = 5x² + 2x – 25

P(2, -1)

T_(p) : T = 0

Þ y – 1 = 10x(2) + 2(x + 2) – 50

$\Rightarrow y = 22 x - 45$ is also tangent to y = x³ – x² + x at point (a, b)

For y = x³ -x² + x

$\frac{d y}{d x} = 3 x^{2} - 2 x + 1 > 0$

y = 22x $- \frac{1939}{27}$ which is not tangent to the curve.

$3 a^{2} - 2 a + 1 = 22$ (slope of tangent)

$\Rightarrow 3 a^{2} - 2 a - 21 = 0 \to a = \frac{2 \pm \sqrt[]{4 + 252}}{6} = \frac{2 \pm 16}{6} = 3, \frac{- 7}{3}$

b = 27 – 9 + 3 = 21

tangent : y – 21 = 22(x – 3)

⇒ y = 22x – 45

a = 3, b = 21

2a + 9b = 6 + 189 = 195

Also, $a^{3} - a^{2} + a = b$

For a = $\frac{- 7}{3}$

b = $\frac{- 343}{27} - \frac{49}{9} - \frac{7}{3}$

= $\frac{- 343 - 147 - 63}{27} = \frac{- 553}{27}$

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8 months ago

If [t] denotes the greatest integer $\leq t,$ then the number of points, at which the function $f (x) = 4 | 2 x + 3 | + 9 [x + \frac{1}{2}] - 12 [x + 20]$ is not differentiable in the open interval (20, 20), is

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