16 and 8

We know the formula that \(\left( {\begin{array}{{20}{c}}m\\k\end{array}} \right) + \left( {\begin{array}{{20}{c}}m\\{k + 1}\end{array}} \right) = \left( {\begin{array}{{20}{c}}{m + 1}\\{k + 1}\end{array}} \right)\),
Here \(m = 15\) and \(k = 7\), hence the desired sum is \(\left( {\begin{array}{{20}{c}}{16}\\8\end{array}} \right)\)
Thus \(n = 16\) and \(r = 8\).

25

Suppose number of students reading Hidavat and Hindustan are \(n(A)\) and \(n(B)\), then
\(n(A \cup B) = n(A.) + n(B.) - n(A \cap B)\)
\( \Rightarrow 50 - 10 = 30 + 35 - n(A \cap B)\)
\( \Rightarrow n(A \cap B) = 25\)

A ⊂ B

Given that A = \({4^x} - 3x - 1\) and B = 9(x – 1)
By putting x = 1, 2, 3, 4… we see that,
A = {0, 9, 54, …} and B = {0, 9, 18, 27, …}
The set A is a proper subset of B or \(A \subset B\)
Alternately we can expand A as
\({(1 + 3)^x} - 3x - 1\)
\( = 1 + 3x + {\,^x}{{\rm{C}}_2} \cdot {3^2} + .... - 3x - 1\)
= a multiple of \(9\).
Here A and B are both multiple of 9, but B contains each multiple of 9 while A does not. Thus A is a subset of B.

8

A has 3 elements, hence number of subsets = \(2^3= 8\).

24

In DELHI all the letters are different. The first letter is fixed, so the remaining 4 letters can be permuted in 4! ways.
Number of such words = 4! = 24.

7

Out of the 10 friends, 3 are not attending the party, so Naresh has to select 6 friends out of 7 friends.
Number of ways = \(^7{{\rm{C}}_6} = 7\)

84

Total number of prizes are 10 and there are 10 recipients, number of ways is equal to the number of solutions of the equation \(x + y + z + w = 10\), where \(x,\;y,\;z,\,\,w \ge 1\).
Hence the number of solutions = \(^{10 - 1}{{\rm{C}}_{4 - 1}} = {\,^9}{{\rm{C}}_3} = 84\)

38.3km/hour

Suppose the distance AB = BC = CA = \(x\), then total journey = \(3x\) and total time taken in the journey = \(\frac{x}{{30}} + \frac{x}{{40}} + \frac{x}{{50}} = \frac{{47x}}{{600}}\).
Hence the average speed = \(\frac{{{\rm{Total}}\;\,{\rm{Journey}}}}{{{\rm{Total}}\;\,{\rm{time}}}}\)
\( = \frac{{3x}}{{\frac{{47x}}{{600}}}} = \frac{{1800}}{{47}} = 38.3\)

\(\frac{7}{8}\)

Probabilities that B will solve the problem = \(1 - \frac{1}{4} = \frac{3}{4}\). Let probability that C will solve the problem = \(p\), then probability that none of them is able to solve the problem = \(\frac{1}{2} \times \frac{1}{4} \times (1 - p) = \frac{{1 - p}}{8}\).
From the given information,
\(\frac{{1 - p}}{8} = \frac{1}{{64}} \Rightarrow p = \frac{7}{8}\)

\(\frac{{24}}{{25}}\)

A can select any number from the first 25 numbers with probability \(\frac{{25}}{{25}}\). Now B has to select a number different from A, this can be done with probability \(\frac{{24}}{{25}}\).
Required probability = \(\frac{{24}}{{25}}\).

Mean, Median and Mode of A are same.

Mean of A = \(\frac{{2 + 3 + 7 + 1 + 3 + 2 + 3}}{7} = 3\)
Median of A = 3 and Mode of A = 3
Choice (D) is correct.

1.607

The formula for standard deviation is
\(\sqrt {\frac{{\sum {f{x^2}} }}{n} - {{\left( {\bar x} \right)}^2}} \), where \(\bar x\) is the mean of the data and \(\bar x = \frac{{\sum {fx} }}{n}\)
Hence \(\bar x = \frac{{3 \cdot 6 + 6 \cdot 7 + 9 \cdot 8 + .... + 4 \cdot 12}}{{3 + 6 + 9 + ... + 4}} = 9\)
\(\sum {f{x^2}} = 4012\)
Hence standard deviation = \(\sqrt {\frac{{4012}}{{48}} - {9^2}} = 1.607\)

\(\left[ {\begin{array}{left{20}{c}}{\cos n\alpha }&{\sin n\alpha }\\{ - \sin n\alpha }&{\cos n\alpha }\end{array}} \right]\)

Given that \(A = \left[ {\begin{array}{{20}{c}}{\cos \alpha }&{\sin \alpha }\\{ - \sin \alpha }&{\cos \alpha }\end{array}} \right]\), then
\({A^2} = \left[ {\begin{array}{{20}{c}}{\cos \alpha }&{\sin \alpha }\\{ - \sin \alpha }&{\cos \alpha }\end{array}} \right]\left[ {\begin{array}{{20}{c}}{\cos \alpha }&{\sin \alpha }\\{ - \sin \alpha }&{\cos \alpha }\end{array}} \right]\)
\( = \left[ {\begin{array}{{20}{c}}{{{\cos }^2}\alpha - {{\sin }^2}\alpha }&{2\sin \alpha \cos \alpha }\\{ - 2\sin \alpha \cos \alpha }&{{{\cos }^2}\alpha - {{\sin }^2}\alpha }\end{array}} \right]\)
\( = \left[ {\begin{array}{{20}{c}}{\cos 2\alpha }&{\sin 2\alpha }\\{ - \sin 2\alpha }&{\cos 2\alpha }\end{array}} \right]\)
Similarly, it can be proved that
\({A^n} = \left[ {\begin{array}{{20}{c}}{\cos n\alpha }&{\sin n\alpha }\\{ - \sin n\alpha }&{\cos n\alpha }\end{array}} \right]\).

\(\frac{b}{c}\)

Roots are \(m\) and \(n\), thus
\(m + n = \frac{{2b}}{a},\;mn = \frac{c}{a}\)
Now \(\frac{b}{{a{n^2} + c}} + \frac{b}{{a{m^2} + c}} = \frac{b}{a}\left( {\frac{1}{{{n^2} + c/a}} + \frac{1}{{{m^2} + c/a}}} \right)\)
= \(\frac{{m + n}}{2}\left( {\frac{1}{{{n^2} + mn}} + \frac{1}{{{m^2} + mn}}} \right)\)
= \(\frac{{m + n}}{2}\left( {\frac{{m + n}}{{mn(m + n)}}} \right) = \frac{{m + n}}{{2mn}} = \frac{b}{c}\).

2

To possess a non-zero solution,
\(\left| {\begin{array}{*{20}{c}}4&k&2\\k&4&1\\2&2&1\end{array}} \right| = 0\)
\( \Rightarrow 4(2) - k(k - 2) + 2(2k - 8) = 0\)
\( \Rightarrow {k^2} - 6k + 8 = 0\)
Henec \(k = 2,\;4\). There are two possible values

\(\det (cA ) = c\det (A )\)

Only third choice is wrong, the correct statement should be \(\det (cA) = {c^n}\det (A)\)where \(n\) is the order of the determinant.

√3 x – y + 7 = 0

The equation of the ellipse is \(\frac{{{x^2}}}{{16}} + \frac{{{y^2}}}{1} = 1\) and the slope of the tangent = \(\tan 60 = \sqrt 3 \).
If the equation of the tangent is \(y = mx + c\), then \({c^2} = {a^2}{m^2} + {b^2}\)
\( \Rightarrow {c^2} = 16 \times 3 + 1 = 49\)
Or \(c = \pm 7\)
Equation of the tangent is \(y = \sqrt 3 x \pm 7\)

2

The centre of the ellipse is \((0,\,1)\) and its major axis lies on y axis. The circle passes through one of the vertex \((0,\, - 2)\) of the ellipse.

Hence there will be three intersection points

3/4

Your solution here...

3

\(\frac{{{a^2}}}{{bc}} + \frac{{{b^2}}}{{ca}} + \frac{{{c^2}}}{{ab}}\; = \frac{{{a^3} + {b^3} + {c^3}}}{{abc}}\;\)
\( = \frac{{{a^3} + {b^3} + {c^3} - 3abc + 3abc}}{{abc}}\;\)
\(\frac{{ = (a + b + c)({a^2} + {b^2} + {c^2} - ab - bc - ca)}}{{abc}} + 3\)
= 3.

None of these

We know that \( - 1 \le \sin \left( {\frac{1}{x}} \right) \le 1\), so
\({e^{ - 1}} \le {e^{\sin \left( {\frac{1}{x}} \right)}} \le {e^1}\)
\( \Rightarrow \mathop {\lim }\limits_{x \to 0} \,{x^2}{e^{\sin \left( {\frac{1}{x}} \right)}} = \mathop {\lim }\limits_{x \to 0} \,{x^2}\left( \lambda \right)\), where \(\lambda \)is a finite number that lies between \({e^{ - 1}}\) and \(e\).
Therefore \(\mathop {\lim }\limits_{x \to 0} \,{x^2}{e^{\sin \left( {\frac{1}{x}} \right)}} = 0\).

Point of minima

The graph of the function is given here. The point at \(x = 0\) is the point of minima.

None of these

The value of the function\(g(x)\) when \(x \ne 0\) is \(\frac{x}{2} - \frac{1}{2}\).
When \(x = 0 + h\) or \(x = 0 - h\), then \(g(x) = - \frac{1}{2}\).
As the function is continuous, so \(k = - \frac{1}{2}\).

(– 3, ∞) only

The graph is given by \(y = 2{x^3}{e^x}\)
\(\frac{{dy}}{{dx}} = 6{x^2}{e^x} + 2{x^3}{e^x} = 2{e^x}(3 + x)\)
For increasing, \(\frac{{dy}}{{dx}} > 0\)
\( \Rightarrow 2{e^x}(3 + x) > 0\) or \(x > - 3\)

1

\(\int {{{\sec }^2}x\,\,{\rm{cose}}{{\rm{c}}^4}x} \,dx = \int {\frac{1}{{{{\cos }^2}x{{\sin }^4}x}}} \,dx\)
\( = \int {\frac{1}{{{{\cot }^2}x{{\sin }^6}x}}dx = \int {\frac{{{\rm{cose}}{{\rm{c}}^6}x}}{{{{\cot }^2}x}}} \;dx} \)
\( = \int {\frac{{{{(1 + {{\cot }^2}x)}^2}{\rm{cose}}{{\rm{c}}^2}x}}{{{{\cot }^2}x}}dx} \)
Now put \(\cot x = t\), then \( - {\rm{cose}}{{\rm{c}}^2}x\,dx = dt\)
\( \Rightarrow - \int {\frac{{{{(1 + {t^2})}^2}}}{{{t^2}}}\,dt} = - \frac{{{t^3}}}{3} - 2t + \frac{1}{t}\)
= \( - \frac{1}{3}{\cot ^3}x - 2\cot x + \tan x\)
Hence \(k = 1\)

\({e^x}\tan x + c\)

\(\int {{e^x}} \left( {\frac{{1 + \sin x\cos x}}{{{{\cos }^2}x}}} \right)dx = \int {{e^x}} (\tan x + {\sec ^2}x)\,dx\)
Using the result \(\int {{e^x}[f(x) + f'(x)]} dx = {e^x}f(x)\), the answer is :
\({e^x}\tan x + c\).

\({I_n} = \frac{{2n^2{a^2}}}{{2n+ 1}}{I_{n - 1}}\)

\({I_n} = \int\limits_0^a {{{({x^2} - {a^2})}^n}dx} \)
Integrating by parts,
\( = \left. {x{{({x^2} - {a^2})}^n}} \right|_0^a - \int\limits_0^a {n{{({x^2} - {a^2})}^{n - 1}}2{x^2}dx} \)
\( \Rightarrow {I_n} = 0 - 2n\int\limits_0^a {{x^2}{{({x^2} - {a^2})}^{n - 1}}dx} \)
\( \Rightarrow {I_n} = - 2n\int\limits_0^a {({x^2} - {a^2} + {a^2}){{({x^2} - {a^2})}^{n - 1}}dx} \)
\( - 2n\int\limits_0^a {{{({x^2} - {a^2})}^n}dx + 2n{a^2}\int\limits_0^a {{{({x^2} - {a^2})}^{n - 1}}dx} } \)
\( \Rightarrow {I_n} = - 2n{I_n} + 2n{a^2}{I_{n - 1}}\)
Hence \({I_n} = \frac{{2n{a^2}}}{{2n + 1}}\)

Value of \(c\)

\(\int\limits_{ - 2}^2 {(a{x^5} + b{x^3} + c)dx} \) = \(\left( {\frac{{a{x^6}}}{6} + \frac{{b{x^4}}}{4} + cx} \right)_{ - 2}^2\)
= \(4c\). Therefore, the answer depends upon the value of \(c\).

None of these

The bounded region will be OAB, where OA = 4, OB = 4.

So the area = \(\frac{{4 \times 4}}{2} = 8\).

\(-\left[ {\vec a\;\vec b\;\vec c} \right]\)

\((\vec a + \vec b + \vec c).\left[ {\left( {\vec a + \vec b} \right) \times \left( {\vec a + \vec c} \right)} \right]\) can be expanded as:
= \((\vec a + \vec b + \vec c).\left[ {\vec a \times \vec a + \vec a \times \vec c + \vec b \times \vec a + \vec b \times \vec c} \right]\)
\(\vec a \cdot (\vec b \times \vec c) + \vec b \cdot (\vec a \times \vec c) + \vec c \cdot (\vec b \times \vec a)\)
= \(\left[ {\vec a\;\vec b\;\vec c} \right] + \left[ {\vec b\;\vec a\;\;\vec c} \right] + \left[ {\vec c\;\vec b\;\vec a\;} \right] = - \left[ {\vec a\;\vec b\;\vec c} \right]\)

\(\theta = 90\)

Resultant force of two vectors \({F_1},\;{F_2}\) is
\(\sqrt {F_{_1}^2 + F_2^2 + 2{F_1}{F_2}\cos \theta } \)
When \(\theta = 90,\) then resultant is \(\sqrt {\left\{ {F_1^2 + F_2^2} \right\}} \). Thus \(\theta = 90\).

\(\vec 0\)

Given that \(\vec a + \vec b + \vec c\;\;\)is collinear with \(\vec d\), so
\(\vec a + \vec b + \vec c\;\; = \lambda \vec d\) …………. (1)
Also \(\vec b + \vec c + \;\vec d\;\) is collinear with \(\vec a,\) so
\(\vec b + \vec c + \;\vec d\; = \mu \vec a\) …………….(2)
From equation (1) – equation (2),
\(\vec a - \vec d = \lambda \vec d - \mu \vec a\)
By comparison, \(\lambda = - 1,\;\,\mu = - 1\;\)
Putting the value of \(\lambda \) in equation (1), we have,
\(\vec a + \vec b + \vec c = - \vec d\)
\( \Rightarrow \vec a + \vec b + \vec c + \vec d = 0\)

33 units

Three forces are:
\({f_1} = 5\left( {\frac{{6i + 2j + 3k}}{{\sqrt {{6^2} + {2^2} + {3^2}} }}} \right) = \frac{5}{7}\left( {6i + 2j + 3k} \right)\)
\({f_2} = 3\left( {\frac{{3i - 2j + 6k}}{{\sqrt {{3^2} + {{( - 2)}^2} + {6^2}} }}} \right) = \frac{3}{7}\left( {3i - 2j + 6k} \right)\)
\({f_3} = \left( {\frac{{2i - 3j - 6k}}{{\sqrt {{2^2} + {{( - 3)}^2} + {{( - 6)}^2}} }}} \right) = \frac{1}{7}\left( {2i - 3j - 6k} \right)\)
Net force \(f = \left( {\frac{{41i + i + 27k}}{7}} \right)\)
Displacement = \(\vec d = (3i + 4k)\)
Work done = \(f \cdot d = \frac{{123 + 108}}{7} = 33\)

\(\frac{{n\vec a + m\vec b}}{{m + n}}\)

This is a direct formula. Position vector of the point \(p\) is \(\frac{{n\vec a + m\vec b}}{{m + n}}\).

0

Given that \(\vec a + 2\vec b\;\;\)is collinear with \(\vec c\)
\( \Rightarrow \vec a + 2\vec b = \lambda \vec c\) …….. (1)
Also \(\vec b + \overrightarrow {3c} \;\) is collinear with \(\vec a,\)
\( \Rightarrow \vec b + 3\vec c = \mu \vec a\) …….. (2)
Equating the value of \(2\vec b\) from the above two equations, we have
\(\lambda \vec c - \vec a = 2\mu \vec a - 6\vec c\)
By comparison, we have \(\lambda = - 6\). Putting this value in the first equation, we have
\(\vec a + 2\vec b = - 6\vec c\)
Or \(\left| {\vec a + 2\vec b + 6\vec c} \right| = 0\)

Right angle triangle

Suppose the vertices are A, B and C, then
AB = \(i - 3j - 3k\)
BC = \(2i + j\)
CA = \(i - 2j + 3k\)
Here \({\rm{BC}} \cdot {\rm{AC}} = 2 - 2 = 0\)
Hence the triangle is right angled.

9/2

Volume = \(\left| {\begin{array}{*{20}{c}}2&3&4\\1&\alpha &2\\1&2&\alpha \end{array}} \right| = 15\)
\( \Rightarrow 2({\alpha ^2} - 4) - 3(\alpha - 2) + 4(2 - \alpha ) = 15\)
\( \Rightarrow 2{\alpha ^2} - 7\alpha - 9 = 0\)
\( \Rightarrow (\alpha + 1)(2\alpha - 9) = 0\)
Hence \(\alpha = - 1,\;\frac{9}{2}\)

None of these

Given that \({\sin ^2}x - \sin x - 2 = 0\)
\( \Rightarrow (\sin x - 2)(\sin x + 1) = 0\)
\( \Rightarrow \sin x = - 1\)
Hence \(x = \frac{{3\pi }}{2}\)

38/3

Let \(\tan x = 2k,\;\tan y = 3k,\;\tan z = 5k\)
We know that when \(x + y + z = \pi \),
\(\tan x + \tan y + \tan z = \tan x\tan y\tan z\)
\( \Rightarrow 2k + 3k + 5k = (2k)(3k)(5k)\)
\( \Rightarrow 10k = 30{k^3}\) or \({k^2} = \frac{1}{3}\)
Hence \({\tan ^2}x + {\tan ^2}y + {\tan ^2}z = 4{k^2} + 9{k^2} + 25{k^2}\)
\( = 38{k^2} = \frac{{38}}{3}\)

1/8

Multiply and divide by \(\sin 72\), we get
\(\frac{{\sin 12\sin 48\sin 72\sin 54}}{{\sin 72}}\)
\( = \frac{{\sin 36\sin 54}}{{4\sin 72}}\)
(using \(\sin \theta \sin (60 - \theta )\sin (60 + \theta ) = \frac{{\sin 3\theta }}{4}\))
\(\)\( = \frac{{2\sin 36\sin 54}}{{8\sin 72}} = \frac{{2\sin 36\cos 36}}{{8\sin 72}} = \frac{1}{8}\)

\(\frac{{\sqrt 5 - 1}}{2}\)

Eliminating \(y\) from the given relations,
\(\cos x = \tan y = \frac{1}{{\cot y}} = \frac{1}{{\tan z}} = \cot z = \tan x\)
\( \Rightarrow \cos x = \tan x\)
Or \(\sin x = {\cos ^2}x = 1 - {\sin ^2}x\)
\(\Rightarrow \sin x = \frac{{ - 1 + \sqrt 5 }}{2}\)

tan θ + sec θ

\(\tan \left( {45 + \frac{\theta }{2}} \right) = \frac{{1 + \tan \frac{\theta }{2}}}{{1 - \tan \frac{\theta }{2}}}\)
\( = \frac{{\sin \frac{\theta }{2} + \cos \frac{\theta }{2}}}{{\cos \frac{\theta }{2} - \sin \frac{\theta }{2}}} = \frac{{{{\left( {\sin \frac{\theta }{2} + \cos \frac{\theta }{2}} \right)}^2}}}{{{{\cos }^2}\frac{\theta }{2} - {{\sin }^2}\frac{\theta }{2}}}\)
\( = \frac{{1 + \sin \theta }}{{\cos \theta }} = \sec \theta  + \tan \theta \)

\(\frac{1}{8}\)

The value of \(\sin 10\sin 50\sin 70\) \( = \frac{{\sin (3 \times 10)}}{4} = \frac{1}{8}\)
(using \(\sin \theta \sin (60 - \theta )\sin (60 + \theta ) = \frac{{\sin 3\theta }}{4}\))

sec A cosec A + 1

The given expression is:
\(\frac{{\frac{{\sin x}}{{\cos x}}}}{{\,1 - \frac{{\cos x}}{{\sin x}}\,}} + \frac{{\frac{{\cos x}}{{\sin x}}}}{{\,1 - \frac{{\sin x}}{{\cos x}}\,}}\)
\( = \frac{{{{\sin }^2}x}}{{\cos x\,(\sin x - \cos x)}} + \frac{{{{\cos }^2}x}}{{\sin x\,(\cos x - \sin x)}}\)
\( = \frac{{{{\sin }^3}x - {{\cos }^3}x}}{{\sin x\cos x(\sin x - \cos x)}}\)
\( = \frac{{(\sin x - \cos x)({{\sin }^2}x + {{\cos }^2}x + \sin x\cos x)}}{{\sin x\cos x(\sin x - \cos x)}}\)
\( = 1 + \sec x{\rm{cosec}}x\)

\(\sqrt 3 :1\)

Let the tower is PQ, and its height is \(h\), then

\(AQ = h\sqrt 3 \)
\(BQ = h\)
\(CQ = \frac{h}{{\sqrt 3 }}\)
\( \Rightarrow \frac{{AB}}{{BC}} = \frac{{h\sqrt 3  - h}}{{h - \frac{h}{{\sqrt 3 }}}} = \frac{{3 - \sqrt 3 }}{{\sqrt 3  - 1}} = \sqrt 3 \)

1/6 sq. unit

The graphs will intersect at the points (1, 1) and (0, 0)

Shaded area = \(\int\limits_0^1 {\left( {\sqrt x - x} \right)} \,dx = \left[ {\frac{{{x^{2/3}}}}{{2/3}} - \frac{{{x^2}}}{2}} \right]_0^1\)
\( = \frac{2}{3} - \frac{1}{2} = \frac{1}{6}\)

Discontinuous at \(x = 2\)

The function is discontinuous at \(x = 2\)

tan⁻¹x

The given expression can be written as
\(2{\tan ^{ - 1}}\left[ {{\rm{cosec}}\left( {{\rm{cose}}{{\rm{c}}^{ - 1}}\left( {\frac{{\sqrt {1 + {x^2}} }}{x}} \right)} \right)} \right]\) \( - \tan \left( {{{\tan }^{ - 1}}\left( {\frac{1}{x}} \right)} \right)\)
\( = 2{\tan ^{ - 1}}\left[ {\frac{{\sqrt {1 + {x^2}} }}{x} - \frac{1}{x}} \right] = 2{\tan ^{ - 1}}\left[ {\frac{{\sqrt {1 + {x^2}} - 1}}{x}} \right]\)
Put \(x = \tan \theta \),
\( = 2{\tan ^{ - 1}}\left[ {\frac{{\sec \theta - 1}}{{\tan \theta }}} \right] = 2{\tan ^{ - 1}}\left( {\frac{{1 - \cos \theta }}{{\sin \theta }}} \right)\)
\( = 2{\tan ^{ - 1}}\left( {\frac{{2{{\sin }^2}\frac{\theta }{2}}}{{2\sin \frac{\theta }{2}\cos \frac{\theta }{2}}}} \right) = 2{\tan ^{ - 1}}\left( {\tan \frac{\theta }{2}} \right)\)
\( = \theta = {\tan ^{ - 1}}x\)

1

Given that \(\frac{3}{5}\sin x + \frac{4}{5}\cos x = 1\)
Since \({\sin ^2}x + {\cos ^2}x = 1\), so \(\sin x = \frac{3}{4},\;\cos x = \frac{4}{5}\)
\( \Rightarrow \frac{{1 - {{\tan }^2}\frac{x}{2}}}{{1 - {{\tan }^2}\frac{x}{2}}} = \frac{4}{5}\) or \(\tan \frac{x}{2} = \frac{1}{3}\)
\( \Rightarrow 6\tan \frac{x}{2} - 9{\tan ^2}\frac{x}{2} = \frac{6}{3} - \frac{9}{9} = 1\).

Cardinality of A

Given that B and C are both subsets of \(A\) , hence \({\rm{A}} \cup {\rm{B}} \cup {\rm{C}} = A\). So, cardinality will be same as that of \(A\).