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Maths NCERT Exemplar Solutions Class 12th Chapter Twelve

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Maths NCERT Exemplar Solutions Class 12th Chapter Twelve Maths NCERT Exemplar Solutions Class 12th Chapter Ten

If $\vec{a} = 2 \hat{i} + \hat{j} + 2 \hat{k}$ , then the value of $| \hat{i} * (\vec{a} * \hat{i}) |^{2} + | \hat{j} * (\vec{a} * \hat{j}) |^{2} + | \hat{k} * (\vec{a} * \hat{k}) |^{2}$ is equal to

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alok kumar singh

Contributor-Level 10

Given ${(2 x^{2} + 3 x + 4)}^{10} = \sum_{r = 0}^{20} ? a_{r} x^{r}$

replace $x$ by $\frac{2}{x}$ in above identity :-

$\begin{array}{r} \frac{2^{10} {(2 x^{2} + 3 x + 4)}^{10}}{x^{20}} = \sum_{r = 0}^{20} ? ? \frac{a_{r} 2^{r}}{x^{r}} \\ \Rightarrow 2^{10} \sum_{r = 0}^{20} ? ? a_{r} x^{r} = \sum_{r = 0}^{20} ? ? a_{r} 2^{r} x^{(20 - r)} (from (i)) \end{array}$

now, comparing coefficient of $x^{7}$ from both sides

(take $r = 7$ in L.H.S. and $r = 13$ in R.H.S.)

$2^{10} a_{7} = a_{13} 2^{13} \Rightarrow \frac{a_{7}}{a_{13}} = 2^{3} = 8$

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Maths NCERT Exemplar Solutions Class 12th Chapter Twelve Maths NCERT Exemplar Solutions Class 12th Chapter Nine

Let $a_{1}, a_{2}, \dots a_{n}$ be a given A.P. whose common difference is an integer and $S_{n} = a_{1} + a_{2}$ $+ \dots + a_{n}$ . If $a_{1} = 1, a_{1} = 300$ and $15 \leq n \leq 50$ , then the ordered pair $(S_{n - 4}, a_{n - 4})$ is equal to:

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alok kumar singh

Contributor-Level 10

$x \frac{d y}{d x} - y = x^{2} (x c o s ? x + s i n ? x), x > 0$

$\frac{d y}{d x} - \frac{y}{x} = x (x c o s ? x + s i n ? x) \Rightarrow \frac{d y}{d x} + P y = Q$

So, I.F. $= e^{\int - \frac{1}{x} d x} \frac{1}{| x |} = \frac{1}{x} (x > 0)$

Thus, $\frac{y}{x} = \int \frac{1}{x} (x (x c o s ? x + s i n ? x)) d x$

$\Rightarrow \frac{y}{x} = x s i n ? x + C$

$? y (π) = π \Rightarrow C = 1$

So, $y = x^{2} s i n ? x + x \Rightarrow (y)_{π / 2} = \frac{π^{2}}{4} + \frac{π}{2}$

Also, $\frac{d y}{d x} = x^{2} c o s ? x + 2 x s i n ? x + 1$

$\Rightarrow \frac{d^{2} y}{d x^{2}} = - x^{2} s i n ? x + 4 x c o s ? x + 2 s i n ? x$

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Maths NCERT Exemplar Solutions Class 12th Chapter Twelve Maths NCERT Exemplar Solutions Class 12th Chapter Ten

If the perpendicular bisector of the line segment joining the points $P (1,4)$ and $Q (k, 3)$ has $y$ -intercept equal to -4 , then a value of $k$ is:

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alok kumar singh

Contributor-Level 10

$m a x {n (A), n (B)} \leq n (A \cup B) \leq n (U)$

$\Rightarrow 76 \leq 76 + 63 - x \leq 100$

$\Rightarrow - 63 \leq - x \leq - 39$

$\Rightarrow 63 \geq x \geq 39$

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Maths NCERT Exemplar Solutions Class 12th Chapter Twelve Maths NCERT Exemplar Solutions Class 12th Chapter One

Let $⋃_{i = 1}^{50} X_{i} = ⋃_{i = 1}^{n} Y_{i} = T$ , where each $X_{i}$ contains 10 elements and each $Y i$ contains 5 elements. If each element of the set $T$ is an element of exactly 20 of sets $X_{i}$ 's and exactly 6 of sets $Y_{i}$ 's then $n$ is equal to:

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alok kumar singh

Contributor-Level 10

$f (x) = \int_{1}^{3} ? \frac{\sqrt[]{x} d x}{(1 + x)^{2}} = \int_{1}^{\sqrt[]{3}} ? \frac{t . 2 t d t}{{(1 + t^{2})}^{2}}$ (put $\sqrt[]{x} = t$ )

$= {(- \frac{t}{1 + t^{2}})}_{1}^{\sqrt[]{3}} + {({t a n}^{- 1} ? t)}_{1}^{\sqrt[]{3}}$ [Applying by parts]

$\begin{array}{r} = - (\frac{\sqrt[]{3}}{4} - \frac{1}{2}) + \frac{π}{3} - \frac{π}{4} \\ = \frac{1}{2} - \frac{\sqrt[]{3}}{4} + \frac{π}{12} \end{array}$

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Maths NCERT Exemplar Solutions Class 12th Chapter Twelve Maths NCERT Exemplar Solutions Class 12th Chapter Eleven

Let $x = 4$ be a directrix to an ellipse whose centre is at the origin and its eccentricity is $\frac{1}{2}$ . If $P (1, β), β > 0$ is a point on this ellipse, then the equation of the normal to it at $P$ is:

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alok kumar singh

Contributor-Level 10

$[x]^{2} + 2 [x + 2] - 7 = 0$

$\begin{array}{r} \Rightarrow [x]^{2} + 2 [x] + 4 - 7 = 0 \\ \Rightarrow [x] = 1, - 3 \\ \Rightarrow x \in [1,2) \cup [- 3, - 2) \end{array}$

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Maths NCERT Exemplar Solutions Class 12th Chapter Twelve Maths NCERT Exemplar Solutions Class 12th Chapter Three

The angle of elevation of a cloud $C$ from a point $P, 200 m$ above a still take is $30^{\circ}$ . If the angle of depression of the image of $C$ in the lake from the point $P$ is $60^{\circ}$ , then $P C$ (in $m$ ) is equal to

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alok kumar singh

Contributor-Level 10

$\sum_{r = 0}^{20} ?^{50 - r} C_{6} =^{50} C_{6} +^{49} C_{6} +^{48} C_{6} + \dots . +^{30} C_{6}$

$\begin{array}{r} =^{50} C_{6} +^{49} C_{6} +^{48} C_{6} + \dots . + (^{30} C_{6} +^{30} C_{7}) -^{30} C_{7} \\ =^{50} C_{6} +^{49} C_{6} +^{48} C_{6} + \dots . + (^{31} C_{6} +^{31} C_{7}) -^{30} C_{7} \\ =^{50} C_{6} +^{50} C_{7} -^{30} C_{7} \\ =^{51} C_{7} -^{30} C_{7} \end{array}$

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Maths NCERT Exemplar Solutions Class 12th Chapter Twelve Maths NCERT Exemplar Solutions Class 12th Chapter Thirteen

In a game two players $A$ and $B$ take turns in throwing a pair of fair dice starting with player $A$ and total of scores on the two dice, in each throw is noted. A wins the game if he throws a total of 6 before $B$ throws a total of 7 and $B$ wins the game if he throws a total of 7 before $A$ throws a total of six. The game stops as soon as either of the players wins. The probability of $A$ winning the game is:

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alok kumar singh

Contributor-Level 10

$A^{2} = (\begin{array}{l} c o s ? 2 θ & i s i n ? 2 θ \\ i s i n ? 2 θ & c o s ? 2 θ \end{array})$

Similarly, $A^{5} = (\begin{matrix} c o s ? 5 θ & i s i n ? 5 θ \\ i s i n ? 5 θ & c o s ? 5 θ \end{matrix}) = (\begin{array}{l} a & b \\ c & d \end{array})$

(1) $a^{2} + b^{2} = {c o s}^{2} ? 5 θ - {s i n}^{2} ? 5 θ = c o s ? 10 θ = c o s ? 75^{?}$

(2) $a^{2} - d^{2} = {c o s}^{2} ? 5 θ - {c o s}^{2} ? 5 θ = 0$

(3) $a^{2} - b^{2} = {c o s}^{2} ? 5 θ + {s i n}^{2} ? 5 θ = 1$

(4) $a^{2} - c^{2} = {c o s}^{2} ? 5 θ + {s i n}^{2} ? 5 θ = 1$

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Maths NCERT Exemplar Solutions Class 12th Chapter Twelve Maths NCERT Exemplar Solutions Class 12th Chapter Eleven

The distance of the point $(1, - 2,3)$ from the plane $x - y + z = 5$ measured parallel to the line $\frac{x}{2} = \frac{y}{3} = \frac{z}{- 6}$ is:

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alok kumar singh

Contributor-Level 10

$\int {(\frac{x}{x s i n ? x + c o s ? x})}^{2} d x = \int (\frac{x}{c o s ? x}) \cdot \frac{x c o s ? x d x}{(x s i n ? x + c o s ? x)^{2}}$

$\begin{array}{r} = \frac{x}{c o s ? x} (- \frac{1}{x s i n ? x + c o s ? x}) + \int (\frac{c o s ? x + x s i n ? x}{{c o s}^{2} ? x}) (\frac{1}{x s i n ? x + c o s ? x}) d x \\ = - \frac{x s e c ? x}{x s i n ? x + c o s ? x} + \int {s e c}^{2} ? x d x \\ = - \frac{x s e c ? x}{x s i n ? x + c o s ? x} + t a n ? x + C \end{array}$

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Maths NCERT Exemplar Solutions Class 12th Chapter Twelve Maths NCERT Exemplar Solutions Class 12th Chapter Five

Let $f : (0, \infty) \to (0, \infty)$ be a differentiable function such that $f (1) = e$ and ${l i m}_{t \to x} \frac{t^{2} f^{2} (x) - x^{2} f^{2} (t)}{t - x} = 0$ . If $f (x) = 1$ , then $x$ is equal to:

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alok kumar singh

Contributor-Level 10

Since $(3,3)$ lies on $\frac{x^{2}}{a^{2}} - \frac{y^{2}}{b^{2}} = 1$

$\frac{9}{a^{2}} - \frac{9}{b^{2}} = 1$

Now, normal at $(3,3)$ is $y - 3 = - \frac{a^{2}}{b^{2}} (x - 3)$ ,

which passes through $(9,0) \Rightarrow b^{2} = 2 a^{2}$

So, $e^{2} = 1 + \frac{b^{2}}{a^{2}} = 3$

Also, $a^{2} = \frac{9}{2}$

(From (i) and (ii)

Thus, $(a^{2}, e^{2}) = (\frac{9}{2}, 3)$

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Maths NCERT Exemplar Solutions Class 12th Chapter Twelve Maths NCERT Exemplar Solutions Class 12th Chapter Three

The minimum value of $2^{s i n x} + 2^{c o s x}$ is:

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alok kumar singh

Contributor-Level 10

$(\frac{K - 2}{h - 1}) (\frac{1 - 2}{3 - 1}) = - 1 \Rightarrow K = 2 h$

$? [? A B C] = 5 \sqrt[]{5}$

$\Rightarrow \frac{1}{2} (\sqrt[]{5}) \sqrt[]{(h - 1)^{2} + (K - 2)^{2}} = 5 \sqrt[]{5}$

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