2021 NIMCET (Maths)

(Total questions: 50)

Question 01: If $$\theta$$ is the acute angle between the lines $${x^2} - 7xy + 12{y^2} = 0$$, then $$\cfrac{{2\cos \theta + 3\sin \theta }}{{4\sin \theta + 5\cos \theta }}$$ is:
[a] $$\frac{{29}}{{69}}$$
[b] $$\frac{{61}}{{39}}$$
[c] $$\frac{{39}}{{61}}$$
[d] $$\frac{{69}}{{29}}$$

Choice [A]

We know that
$$\tan \theta = \cfrac{{2\sqrt {{h^2} - ab} }}{{a + b}} = \frac{{2\sqrt {\frac{{49}}{4} - 12} }}{{1 + 12}} = \frac{1}{{13}}$$
Now $$\cfrac{{2\cos \theta + 3\sin \theta }}{{4\sin \theta + 5\cos \theta }}$$= $$\frac{{2 + 3\tan \theta }}{{4\tan \theta + 5}}$$
= $$\cfrac{{2 + 3 \times \frac{1}{{13}}}}{{4 \times \frac{1}{{13}} + 5}} = \frac{{29}}{{69}}$$

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Question 02: The function $$f(x) = \cfrac{x}{{1 + x\tan x}}\,$$, $$0 \le x \le \cfrac{\pi }{2}$$ is maximum when
[a] $$x = \sec x$$
[b] $$x = \tan x$$
[c] $$x = \cos x$$
[d] None of these

Choice [C]

$$f'(x) = \cfrac{{1(1 + x\tan x) - x(x{{\sec }^2}x + \tan x)}}{{{{(1 + x\tan x)}^2}}}$$
$$= \cfrac{{1 - {x^2}{{\sec }^2}x}}{{{{(1 + x\tan x)}^2}}}$$
Here we see that when $${x^2}{\sec ^2}x < 1,\;\;f'(x) > 0$$ and when $${x^2}{\sec ^2}x > 1,\;\;f'(x) < 0$$. Hence the function is maximum at $$x\sec x = 1$$.
$$\Rightarrow \cos x = x$$

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Question 03: If $$\cfrac{{n!}}{{2!(n - 2)!}}$$ and $$\cfrac{{n!}}{{4!(n - 4)!}}$$ are in the ratio 2 : 1, then the value of $$n$$ is:
[a] 0
[b] 2
[c] 4
[d] 5

Choice [D]

The numbers are $$^n{{\rm{C}}_2}\;{\rm{and}}{\;^n}{{\rm{C}}_4}$$, clearly $$n \ge 4$$. From the options $$n$$ = 5 gives the correct ratio.

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Question 04: There are 50 questions in a paper. Find the number of ways in which a student can attempt one or more questions:
[a] $$2^{50} – 1$$
[b] $$2^{50} + 1$$
[c] $$2^{50} – 2$$
[d] $$2^{50} + 2$$

Choice [A]

Each question can be attempted or not attempted. So there are two choices for each question. So total number of ways of attempting the paper is $${2^{50}}$$, but there will be a case when no question is attempted. Except this case all other cases are to be counted. Hence the number of ways = $${2^{50}} - 1$$.

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Question 05: If $$p$$ is the probability that a man aged $$x$$ will die in year, then the probability that out of $$n$$ men A1, A2,……, An each aged $$x$$, A1 will die in a year and be first to die, is
[a] $$\frac{1}{{{n^2}}}$$
[b] 1 – (1 – p)n
[c] $$\frac{1}{{{n^2}}}\left[ {1 - {{(1 - p)}^n}} \right]$$
[d] $$\frac{1}{n}\left[ {1 - {{(1 - p)}^n}} \right]$$

Choice [D]

Probability that none of them will die is $${(1 - p)^n}$$. Hence probability that at least one of them will die is $$1 - {(1 - p)^n}$$.
Now if we consider order, probability that $${A_1}$$ is the first to die among $$n$$people is:
$$\frac{1}{n}\left( {1 - {{(1 - p)}^n}} \right)$$

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Question 06: If a variable takes values 0, 1, 2, 3, …, 50 with frequency $$1, ^{50}C_1 , ^{50}C_2 ….. {50}C_{50}$$ then A.M. is
[a] 50
[b] 25
[c] $$\frac{{{2^{50}}}}{{50}}$$
[d] 51

Choice [B]

The mean will be $$\cfrac{{\sum {fx} }}{{\sum f }} = \cfrac{{\sum\limits_{n = 0}^{50} {n{ \cdot ^{50}}{{\rm{C}}_n}} }}{{\sum\limits_{n = 0}^{50} {^{50}{{\rm{C}}_n}} }} = \cfrac{{\sum\limits_{n = 0}^{50} {n{ \cdot ^{50}}{{\rm{C}}_n}} }}{{{2^{50}}}}$$ ……(1)
We know that $${(1 + x)^{50}} = {\,^{50}}{{\rm{C}}_0} + {\,^{50}}{{\rm{C}}_1}x + {\,^{50}}{{\rm{C}}_2}{x^2} + ....... + {\,^{50}}{{\rm{C}}_n}{x^n}$$
Differentiating w.r.t $$x$$, we have
$$50{(1 + x)^{49}} = \,{\,^{50}}{{\rm{C}}_1} + \,2 \times {\,^{50}}{{\rm{C}}_2}x + ....... + \,50 \times {\,^{50}}{{\rm{C}}_{50}}{x^{49}}$$
Now put $$x = 1$$ in the above equation,
$$50 \times {2^{49}} = \,{\,^{50}}{{\rm{C}}_1} + \,2 \times {\,^{50}}{{\rm{C}}_2} + ....... + \,50 \times {\,^{50}}{{\rm{C}}_{50}}$$
Put this value in equation (1),
Mean = $$\frac{{50 \times {2^{49}}}}{{{2^{50}}}} = 25$$
The question can be solved taking smaller values, take total numbers as 5, and you will find the mean as 2.5. From the choices we can conclude that the answer is 50/2 = 25.

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Question 07: In $$\Delta ABC$$ if $${\tan ^2}\frac{A}{2} + {\tan ^2}\frac{B}{2} + {\tan ^2}\frac{C}{2} = k,$$ then $$k$$ is always
[a] > 1
[b] ≥ 1
[c] 2
[d] 1

Choice [B]

We can put values of the angles A, B and C to get an idea about the value of $$k$$, put A = B = C = 60°,
$$k = {\tan ^2}30 + {\tan ^2}30 + {\tan ^2}30 = 1$$
Put another set of values A = B = 30° and C = 120°,
$$k = {\tan ^2}15 + {\tan ^2}15 + {\tan ^2}45 > 1$$
Hence $$k \ge 1$$

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Question 08: If |k| = 5, 0° ≤ α ≤ 360°, then total number of different solutions of 3 cos α + 4 sin α = k is:
[a] 0
[b] 1
[c] 2
[d] infinite

Choice [C]

The maximum and minimum value of $$a\cos \alpha + b\sin \alpha = \pm \sqrt {{a^2} + {b^2}}$$
Hence maximum and minimum value of $$3\cos \alpha + 4\sin \alpha = \pm 5$$
Given that $$\left| k \right| = 5 \Rightarrow k = \pm 5$$
Taking $$k = 5$$, $$3\cos \alpha + 4\sin \alpha = 5$$
or $$\frac{3}{5}\cos \alpha + \frac{4}{5}\sin \alpha = 1$$, this is possible only when $$\cos \alpha = \frac{3}{5}$$ and $$\sin \alpha = \frac{4}{5}$$
Similarly when $$3\cos \alpha + 4\sin \alpha = 5$$ or $$\frac{3}{5}\cos \alpha + \frac{4}{5}\sin \alpha = - 1$$, this is possible only when $$\cos \alpha = - \frac{3}{5}$$ and $$\sin \alpha = - \frac{4}{5}$$.
Hence there are two possible values of $$\alpha$$, so there are two solutions.

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Question 09: In triangle ABC $$a\,{\cos ^2}\frac{C}{2} + c\,{\cos ^2}\frac{A}{2} = \frac{{3b}}{2}$$ then the sides of the triangle are in
[a] A.P.
[b] G.P.
[c] H.P.
[d] None

Choice [A]

Using the half angle formula,
$$a.\left( {\frac{{s(s - c)}}{{ab}}} \right) + c.\left( {\frac{{s(s - a)}}{{bc}}} \right) = \frac{{3b}}{2}$$
$$\Rightarrow s(s - c) + s(s - a) = \frac{{3{b^2}}}{2}$$
$$\Rightarrow s(2s - a - c) = \frac{{3{b^2}}}{2}$$
Or $$\frac{{b(a + b + c)}}{2} = \frac{{3b}}{2} \Rightarrow a + c = 2b$$
The sides are in AP.

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Question 10: Let $$\overrightarrow a = 2\hat i + \hat j + 2\hat k$$ and $$\overrightarrow b = \widehat i - \widehat j + 2\mathop k\limits^ \wedge$$ and $$\overrightarrow c = \widehat i + \widehat j - 2\mathop k\limits^ \wedge$$ are three vectors. Then a vector in the plane of $$\vec a$$ and $$\vec c$$ whose projection on $$\overrightarrow b$$ is of magnitude $$\frac{1}{{\sqrt 6 }}$$ is
[a] $$3\widehat i - 2\widehat j$$
[b] $$3\widehat i + 2\widehat j$$
[c] $$2\widehat i + 3\widehat j - \mathop k\limits^ \wedge$$
[d] $$3\widehat i + 2\widehat j + \mathop k\limits^ \wedge$$

Choice [B]

Any vector that is coplanar with $$\vec a$$and $$\vec c$$ is, $$\vec a + \lambda \vec c = (2 + \lambda )\hat i + (1 + \lambda )\hat j + (2 - 2\lambda )\hat k$$
Projection on $$\vec b$$is: $$(\vec a + \lambda \vec c).\hat b$$ = $$\frac{{(2 + \lambda ) - (1 + \lambda ) + 2(2 - 2\lambda )}}{{\sqrt {{1^2} + {{( - 1)}^2} + {2^2}} }} = \frac{{5 - 4\lambda }}{{\sqrt 6 }}$$
Given that projection on $$\overrightarrow b$$is of magnitude $$\frac{1}{{\sqrt 6 }}$$, hence $$5 - 4\lambda = 1 \Rightarrow \lambda = 1$$
The vector is: $$\vec a + \lambda \vec c = (2 + \lambda )\hat i + (1 + \lambda )\hat j + (2 - 2\lambda )\hat k$$ = $$3\widehat i + 2\widehat j$$

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Question 11: The standard deviation of 20 numbers is 30. If each of the numbers is increased by 4, then the new standard deviation will be
[a] 24
[b] 34
[c] 30
[d] 20

Choice [C]

Standard deviation does not change when we increase or decrease each number by a constant, hence it will remain 30.

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Question 12: If $$a\cos \theta + b\sin \theta = 2$$ and $$a\sin \theta - b\cos \theta = 3$$ then $${a^2} + {b^2}$$ is:
[a] 6
[b] 5
[c] 13
[d] 10

Choice [C]

Square and add both the equations,
$${a^2}({\sin ^2}\theta + {\cos ^2}\theta ) + {b^2}({\sin ^2}\theta + {\cos ^2}\theta ) = {2^2} + {3^2}$$
$$\Rightarrow {a^2} + {b^2} = 13$$

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Question 13: If $$\log (1 - x + {x^2}) = {a_1}x + {a_2}{x^2} + {a_3}{x^3} + ..... + {a_n}{x^n}$$, then $${a_3} + {a_6} + {a_9} + ....$$ is equal to
[a] log 2
[b] $$\frac{2}{3}\log 2$$
[c] $$\frac{1}{3}\log 2$$
[d] 2 log 2

Choice [B]

Given that, $$\log (1 - x + {x^2}) = {a_1}x + {a_2}{x^2} + {a_3}{x^3} + ..... + {a_n}{x^n}$$
Put $$x = 1,\,\,\omega ,\,\,{\omega ^2}$$ respectively,
$$\log (1) = {a_1} + {a_2} + {a_3} + ..... + {a_n}$$
$$\log (1 - \omega + {\omega ^2}) = {a_1}\omega + {a_2}{\omega ^2} + {a_3}{\omega ^3} + ..... + {a_n}{\omega ^n}$$
$$\log (1 - {\omega ^2} + {\omega ^4}) = {a_1}{\omega ^2} + {a_2}{\omega ^4} + {a_3}{\omega ^6} + .. + {a_n}{\omega ^{2n}}$$
Adding the above three equations, we have
$$\begin{array}{l}\log ( - 2\omega ) + \log ( - 2{\omega ^2}) = {a_1}(1 + \omega + {\omega ^2}) + {a_2}(1 + {\omega ^2} + {\omega ^4}) + \\\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\; + {a_3}(1 + {\omega ^3} + {\omega ^6}) + ..........\end{array}$$
Now except the terms containing $${a_3},\;{a_6},\;{a_9},...$$ all other coefficients will be zero.
$$\log ( - 2\omega \times - 2{\omega ^2}) = 3({a_3} + {a_6} + {a_9} + ...)$$
$$\Rightarrow \log 4 = 3({a_3} + {a_6} + {a_9} + ....)$$
or $${a_3} + {a_6} + {a_9} + .... = \frac{2}{3}\log 2$$

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Question 14: The area of the region bounded by the X-axis and the curve defined by $$y = \tan x,\,\frac{{ - \pi }}{3} \le x \le \frac{\pi }{3}$$ and $$y = \cot x,$$$$\frac{\pi }{6} \le x \le \frac{\pi }{2}$$ is
[a] $$- \frac{1}{2}\log 2$$
[b] $$\frac{1}{2}\log 2$$
[c] log 2
[d] None of these

Choice [C]

Both the curves will intersect at $$x = \frac{\pi }{4}$$ and the enclosed region is ACBDA.
Hence its area is:

$$\int\limits_0^{\frac{\pi }{4}} {\tan x\,dx} + \int\limits_{\frac{\pi }{4}}^{\frac{\pi }{2}} {\cot x\,dx}$$ $$= \left[ {\log \sec x} \right]_0^{\pi /4} + \left[ {\log \sin x} \right]_{\pi /4}^{\pi /2}$$
$$= \left[ {\log \sqrt 2 - \log 1} \right] + \left[ {\log 1 - \log \frac{1}{{\sqrt 2 }}} \right] = \log 2$$

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Question 15: A number $$x$$ is selected from natural numbers 1, 2….., 100. Then the probability for $$x + \frac{{100}}{x} > 29$$ is
[a] $$\frac{{37}}{{50}}$$
[b] $$\frac{{39}}{{50}}$$
[c] $$\frac{{41}}{{50}}$$
[d] $$\frac{{43}}{{50}}$$

Choice [B]

Given that $$x + \frac{{100}}{x} > 29 \Rightarrow {x^2} - 29x + 100 > 0$$
Or $$(x - 4)(x - 25) > 0$$
$$\Rightarrow x > 25\;{\rm{or}}\;x < 4$$
Thus $$x$$can take a total of 78 values. (1, 2, 3, 26, ……,100)
Required probability = $$\frac{{78}}{{100}} = \frac{{39}}{{50}}$$

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Question 16: If n is an integer between 0 to 21, then find a value of n for which the value of n!(21 – n)! is minimum
[a] 9
[b] 10
[c] 12
[d] 21

Choice [B]

We know that $$^{21}{{\rm{C}}_n} = \frac{{21!}}{{n!(21 - n)!}}$$
And $$^{21}{{\rm{C}}_n}$$ is maximum when $$n$$ is 10 or 11.
So $$n!(21 - n)!$$ is minimum when $$n$$ = 10, 11.

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Question 17: The general value of $$\theta$$, satisfying the equations, $$\tan \theta = \frac{1}{{\sqrt 3 }}$$and $$\sin \theta = \frac{{ - 1}}{2}$$ is
[a] $$\theta = n\pi + \frac{\pi }{6},n \in I$$
[b] $$n\pi + {( - 1)^n}\left( {\frac{{7\pi }}{6}} \right),n \in I$$
[c] $$\theta = 2n\pi + \frac{{7\pi }}{6},n \in I$$
[d] $$\theta = 2n\pi \pm \frac{{11\pi }}{6},n \in I$$

Choice [C]

The first value of $$\theta$$ satisfying the equations $$\tan \theta = \frac{1}{{\sqrt 3 }},\;\sin \theta = - \frac{1}{2}$$ lies in third quadrant and the value is $$\pi + \frac{\pi }{6} = \frac{{7\pi }}{6}$$. The general value of $$\theta$$ can be obtained by adding a multiple of $$2\pi$$.
General value = $$2n\pi + \frac{{7\pi }}{6}$$

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Question 18: A polygon has 44 diagonals, the number of sides are
[a] 9
[b] 10
[c] 11
[d] 12

Choice [C]

Number of diagonals in a polygon is
$$^n{{\rm{C}}_2} - n$$= 44
$$\Rightarrow \frac{{n(n - 3)}}{2} = 44$$ or $$n = 11$$

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Question 19: The probability of occurrence of two event E and F are 0.25 & 0.50 respectively. The probability of their simultaneously occurrence is 0.14 The probability that neither E nor F occurs is
[a] 0.61
[b] 0.11
[c] 0.39
[d] 0.89

Choice [C]

We know that, $$P(E \cup F) = P(E) + P(F) - P(E \cap F)$$
$$\Rightarrow P(E \cup F) = 0.25 + 0.50 - 0.14 = 0.61$$
Now probability that neither E nor F occurs is $$1 - P(E \cup F) = 1 - 0.61 = 0.39$$

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Question 20: $$\mathop {\lim }\limits_{x \to \infty } {\left( {\cfrac{{x + 7}}{{x + 2}}} \right)^{x + 5}}$$ equal to
[a] $${e^5}$$
[b] $${e^{ - 5}}$$
[c] ({e^2}\)
[d] $${e^{ - 2}}$$

Choice [A]

$$\mathop {\lim }\limits_{x \to \infty } {\left( {\frac{{x + 7}}{{x + 2}}} \right)^{x + 5}} = \mathop {\lim }\limits_{x \to \infty } {\left( {{{\left( {1 + \frac{5}{{x + 2}}} \right)}^{x + 2}}} \right)^{\frac{{x + 5}}{{x + 2}}}}$$
$$= \mathop {\lim }\limits_{x \to \infty } {\left( {{e^5}} \right)^{\frac{{x + 5}}{{x + 2}}}} = {e^5}$$

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Question 21: If A = {1, 2, 3, 4} and B = {3, 4, 5}, then the number of elements in $$(A \cup B) \times (A \cap B) \times (A\Delta B)$$ is
[a] 18
[b] 20
[c] 24
[d] 30

Choice [D]

$$A \cup B$$ = {1, 2, 3, 4, 5} $$A \cap B$$ = {3, 4} $$A\Delta B = (A - B) \cup (B - A)$$ = {1, 2, 5}. Hence number of elements in $$(A \cup B) \times (A \cap B) \times (A\Delta B)$$ is 5×2×3 = 30

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Question 22: Consider the following frequency distribution table

 Class Interval 10-20 20-30 30-40 40-50 50-60 60-70 70-80 Frequency 180 F1 34 180 136 F2 50
If the total frequency is 686 and the median is 42.6, then the value of F1 and F2 are
[a] 81, 25
[b] 82, 24
[c] 83, 23
[d] 84, 22

Choice [B]

Since total frequency is 686, hence F1 + F2 = 686 – sum of all other frequencies
Or F1 + F2 = 106.
Given that median is 42.6, hence the median class is 40 – 50. The formula for median is:
$${\ell _1} + \cfrac{{\,\,\cfrac{n}{2} - C\,\,}}{f}({\ell _2} - {\ell _1})$$
Where $${\ell _1},\;{\ell _2}$$ are the lower and upper limit of the class interval, $$n$$ is the total frequency, $$C$$ is the cumulative frequency of the pre median class and $$f$$ is the frequency of the median class.
$$\Rightarrow 42.6 = 40 + \cfrac{{343 - C}}{{180}} \times 10$$ or $$C$$ = 296
Hence 180 + 34 + F1 = 296 or F1 = 82. So F2 = 24.

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Question 23: For what value of $$p$$, the polynomial $$x^4-3x^3+2px^2-6$$ is exactly divisible by $$(x - 1)$$
[a] 2
[b] 4
[c] 6
[d] 8

Choice [B]

We know that when $$f(x)$$ is divided by $$(x - a)$$the remainder is $$f(a)$$. Here to be exactly divisible by $$(x - 1)$$, the value of polynomial at $$x = 1$$must be 0. Hence $${1^4} - 3({1^3}) + 2p({1^2}) - 6 = 0 \Rightarrow p = 4$$

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Question 24: The lines px + qy = 1 and qx + py – 1 = 0 are respectively the sides AB, AC of the triangle ABC and the base BC is bisected at (p, q). Equation of the median of the triangle through the vertex A is
[a] (2pq – 1) (qx + py – 1) – (p² + q² – 1) (px + qy – 1) = 0
[b] (2pq – 1) (qx + py – 1) + (p² + q² – 1) (px + qy – 1) = 0
[c] (2pq – 1) (px + qy – 1) – (p² + q² – 1) (qx + py – 1) = 0
[d] (2pq – 1) (px + qy – 1) + (p² + q² – 1) (qx + py – 1) = 0

Choice [B]

The median will pass through A and the midpoint of the opposite side i.e $$(p,\,\,q)$$.
General equation of the line that passes through the intersection point of the lines AC and AB is:
$$(px + qy - 1) + \lambda (qx + py - 1) = 0$$
As this line passes through the point $$(p,\,\,q)$$, hence
$$({p^2} + {q^2} - 1) + \lambda (2pq - 1) = 0 \Rightarrow \lambda = - \frac{{(2pq - 1)}}{{{p^2} + {q^2} - 1}}$$
Putting this value of $$\lambda$$, we get the desired equation,
$$(px + qy - 1) - \left( {\frac{{2pq - 1}}{{{p^2} + {q^2} - 1}}} \right)(qx + py - 1) = 0$$
Or $$(px + qy - 1)({p^2} + {q^2} - 1) - (2pq - 1)(qx + py - 1) = 0$$

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Question 25: If P(1, 2), Q(4, 6), R (5, 7) and S(a, b) are the vertices of a parallelogram PQRS, then
[a] a = 2, b = 3
[b] a = 3, b = 4
[c] a = 2, b = 4
[d] a = 3, b = 5

Choice [A]

The mid points of the diagonals in a parallelogram coincide, hence $$\frac{{P + R}}{2} = \frac{{Q + S}}{2}$$ $$\Rightarrow \left( {\frac{{1 + 5}}{2},\,\frac{{2 + 7}}{2}} \right) = \left( {\frac{{4 + a}}{2},\;\frac{{6 + b}}{2}} \right)$$ Or $$a = 2,\,\,b = 3$$

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Question 26: The area of the triangle formed by the vertices whose position vector are 3i + j, 5i + 2j + k, i – 2j + 3k is
[a] $$\sqrt {21}$$
[b] $$\sqrt {23}$$
[c] $$\sqrt {33}$$
[d] $$\sqrt {29}$$

Choice [D]

Suppose the three vertices are
$$A = 3i + j,{\rm{ }}B = 5i + 2j + k$$ and $$C = i - 2j + 3k$$
The side $$AB = 2i + j + k,\;AC = - 2i - 3j + 3k$$
Area of the triangle = $$\frac{1}{2}\left| {AB \times AC} \right| = \frac{1}{2}\left| {6i - 8j - 4k} \right| = \sqrt {29}$$

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Question 27: Angle between $$\overrightarrow a$$ and $$\overrightarrow b$$ is 120°. If $$|\overrightarrow b |\, = 2|\overrightarrow a |$$ and the vector $$\overrightarrow a + x\vec b$$, $$\overrightarrow a$$$$\overrightarrow b$$ are right angles, then $$x$$ is:
[a] $$\frac{1}{3}$$
[b] $$\frac{1}{5}$$
[c] $$\frac{2}{3}$$
[d] $$\frac{2}{5}$$

Choice [D]

Suppose $$\left| {\vec a} \right| = k\;{\rm{and}}\;\left| {\vec b} \right| = 2k$$
Also $${k^2} + x(2{k^2})\cos 120 - x{(2k)^2} - k.(2k)\cos 120 = 0$$
$$x = \frac{2}{5}$$.

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Question 28: If $$32{\tan ^2}\theta = 2{\cos ^2}\alpha - 3\cos \alpha$$ and $$3\cos 2\theta = 1$$, then the general value of $$\alpha$$ for $$n \in Z$$ is
[a] $$n\pi \pm \frac{\pi }{3}$$
[b] $$2n\pi \pm \frac{{2\pi }}{3}$$
[c] $$2n\pi \pm \frac{\pi }{3}$$
[d] $$n\pi \pm \frac{{2\pi }}{3}$$

Choice [B]

We know that $$\cos 2\theta = \frac{{1 - {{\tan }^2}\theta }}{{1 - {{\tan }^2}\theta }}$$
Given that $$\cos 2\theta = \frac{{1 - {{\tan }^2}\theta }}{{1 - {{\tan }^2}\theta }} = \frac{1}{3} \Rightarrow {\tan ^2}\theta = \frac{1}{2}$$
Putting this value in the equation, we have,
$$32{\left( {\frac{1}{2}} \right)^4} = 2{\cos ^2}\alpha - 3\cos \alpha$$
$$\Rightarrow (2\cos \alpha + 1)(\cos \alpha - 2) = 0$$
Or $$\cos \alpha = - \frac{1}{2} \Rightarrow \alpha = 2n\pi \pm \frac{{2\pi }}{3}$$

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Question 29: If the vector $$ai + j + k,\,i + bj + k,\;i + j + ck$$ $$(a,\,b,\,c \ne 1)$$ are coplanar then $$\frac{1}{{1 - a}} + \frac{1}{{1 - b}} + \frac{1}{{1 - c}} =$$
[a] 0
[b] 1
[c] 2
[d] 3

Choice [B]

Given that the three vectors $$ai + j + k,\,i + bj + k,\;i + j + ck$$ are coplanar, hence
$$\left| {\begin{array}{*{20}{c}}a&1&1\\1&b&1\\1&1&c\end{array}} \right| = 0$$
Putting $${C_1} = {C_1} - {C_2},\;{C_2} = {C_2} - {C_3}$$,
$$\left| {\begin{array}{*{20}{c}}{a - 1}&0&1\\{1 - b}&{b - 1}&1\\1&{1 - c}&c\end{array}} \right| = 0$$
$$\Rightarrow c(a - 1)(b - 1) - (a - 1)(c - 1) + (1 - b)(1 - c) = 0$$
$$\Rightarrow \frac{c}{{1 - c}} + \frac{1}{{1 - b}} + \frac{1}{{1 - a}} = 0$$
$$\Rightarrow \frac{{c - 1 + 1}}{{1 - c}} + \frac{1}{{1 - b}} + \frac{1}{{1 - a}} = 0$$
Or $$\frac{1}{{1 - c}} + \frac{1}{{1 - b}} + \frac{1}{{1 - a}} = 1$$

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Question 30: The four geometric means between 2 and 64 are
[a] $$\frac{1}{4},\frac{1}{8},\frac{1}{{16}},\frac{1}{{32}}$$
[b] 4, 8, 16, 32
[c] $$4\sqrt 2 ,8,16\sqrt 2 ,32$$
[d] None of these

Choice [B]

Including the 4 geometric means, there are 6 terms in the GP, where the first term is 2 and the last term is 64.
$$\Rightarrow 2({r^5}) = 64$$ or $$r = 2$$
The four geometric means are 4, 8, 16, 32.

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Question 31: If $$\alpha \ne \beta$$ and $${\alpha ^2} = 5\alpha - 3\,\,\& \,{\beta ^2} = 5\beta - 3,$$ then find the equation whose roots are $$\frac{\alpha }{\beta },\frac{\beta }{\alpha }$$ is
[a] 3x2 – 25x + 3 = 0
[b] 3x2 + 5x + 3 = 0
[c] 3x2 – 5x + 3 = 0
[d] 3x2 – 19x + 3 = 0

Choice [D]

The equation that is satisfied by both $$\alpha$$ and $$\beta$$ is $${x^2} = 5x - 3$$or $${x^2} - 5x + 3 = 0$$
Now sum of the roots is:
$$\cfrac{\alpha }{\beta } + \cfrac{\beta }{\alpha } = \cfrac{{{\alpha ^2} + {\beta ^2}}}{{\alpha \beta }} = \cfrac{{{{(\alpha + \beta )}^2} - 2\alpha \beta }}{{\alpha \beta }}$$
$$= \cfrac{{25 - 2 \times 3}}{3} = \cfrac{{19}}{3}$$
Product of the roots is 1, so the equation is: $${x^2} - \frac{{19}}{3}x + 1 = 0$$ or $$3{x^2} - 19x + 3 = 0$$

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Question 32: If the system of equations 3x – y + 4z = 3, x + 2y – 3z = –2, 6x + 5y + $$\lambda$$z = –3 has at least one solution then $$\lambda$$ is:
[a] –5
[b] 3
[c] 5
[d] 6

Choice [A]

Let us eliminate $$y$$ from the first two equations, we have
$$7x + 5z = 4$$ ……..(i)
Also eliminate$$y$$ from second and third equations,
$$21x + (20 + \lambda )z = 12$$……… (ii)
There will be many solutions if equations (i) and (ii) are identical. Hence $$20 + \lambda = 5 \Rightarrow \lambda = - 5$$

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Question 33: If three throws of three dices, the probability of throwing a triplet not more than twice is
[a] $$1 - \frac{1}{{{6^2}}}$$
[b] $$1 - \frac{1}{{{6^3}}}$$
[c] $$1 - \frac{1}{{{{36}^2}}}$$
[d] $$1 - \frac{1}{{{{36}^3}}}$$

Choice [D]

Probability of throwing a triplet is $${\left( {\frac{1}{6}} \right)^3} \times 6 = \frac{1}{{36}}$$
Probability that in all the three throws a triplet is thrown = $$\frac{1}{{36}} \times \frac{1}{{36}} \times \frac{1}{{36}} = {\left( {\frac{1}{{36}}} \right)^3}$$
Probability that a triplet is not thrown more than twice = $$1 - {\left( {\frac{1}{{36}}} \right)^3}$$

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Question 34: Suppose that $${A_1},\,{A_2},\,\,{A_3},........{A_{30}}$$ are thirty sets each with five elements and $${B_1},\,{B_2},\,\,{B_3},........{B_n}$$ are n sets each with three elements such that $$\bigcup\limits_{i = 1}^{30} {{A_i} = \bigcup\limits_{i = 1}^n {{B_i} = S} }$$. If each element of $$S$$ belongs to exactly ten of the $${A_i}'s$$ and exactly 9 of the $${B_i}'s$$, then the value of $$n$$ is
[a] 15
[b] 45
[c] 75
[d] 90

Choice [B]

Since, each $${A_i}$$ has 5 elements, then $$\sum\limits_{i = 1}^{30} {{A_i}} = 30 \times 5 = 150$$
Let set $$S$$consists of $$m$$elements and given that each elements in $$S$$ belongs to exactly 10 of $${A_i}'s$$, so $$10m = 150 \Rightarrow m = 15$$
Now, since each $${B_i}'s$$ has 3 elements and each $$S$$ belongs to exactly 9 of $${B_i}'s$$
$$\Rightarrow \sum\limits_{i = 1}^n {{B_i}} = 3 \times n = 3n = 9m = 135$$
Or $$n = 45$$

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Question 35: $$\int {{e^x}(\sinh x + \cosh x)dx}$$
[a] $${e^x}{\mathop{\rm sech}\nolimits} x$$+ C
[b] $${e^x}\cosh x$$+ C
[c] $$\sinh x$$+ C
[d] $$\cosh x$$+ C

Choice [B]

We know that
$$\sinh x = \cfrac{{{e^x} - {e^{ - x}}}}{2}$$ and $$\cosh x = \cfrac{{{e^x} + {e^{ - x}}}}{2}$$c
$$\int {{e^x}(\sinh x + \cosh x)dx} = \int {{e^x}({e^x})dx = \cfrac{{{e^{2x}}}}{2} + C}$$
The choice $${e^x}(\cosh x) = {e^x}\left( {\cfrac{{{e^x} + {e^{ - x}}}}{2}} \right) = \cfrac{{{e^{2x}}}}{2} + C$$
Hence the answer is $${e^x}\cosh x$$

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Question 36: The eccentric angle of the extremities of lauts – rectum of the ellipse $$\frac{{{x^2}}}{{{a^2}}} + \frac{{{y^2}}}{{{b^2}}} = 1$$ are given by
[a] $${\tan ^{ - 1}}\left( { \pm \frac{{ae}}{b}} \right)$$
[b] $${\tan ^{ - 1}}\left( { \pm \frac{{be}}{a}} \right)$$
[c] $${\tan ^{ - 1}}\left( { \pm \frac{b}{{ae}}} \right)$$
[d] $${\tan ^{ - 1}}\left( { \pm \frac{a}{{be}}} \right)$$

Choice [C]

We know that length of the latus rectum = $$\frac{{2{b^2}}}{a}$$and Latus Rectum is perpendicular to the major axis and passes through the focus. Hence coordinates of one of its end are $$\left( {ae,\,\,\frac{{{b^2}}}{a}} \right)$$.
If the eccentric angle is $$\theta ,$$ then coordinate of the one end of Latus Rectum are $$(a\cos \theta ,\;b\sin \theta )$$.
Hence $$a\cos \theta = ae,\;b\sin \theta = \frac{{{b^2}}}{a}$$
Or $$\tan \theta = \frac{b}{{ae}}$$.
As there are two ends of Latus Rectum, hence $$\theta = {\tan ^{ - 1}}\left( { \pm \frac{b}{{ae}}} \right)$$

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Question 37: If $$y = {\sin ^{ - 1}}\left( {\frac{{{x^2} + 1}}{{\sqrt {{x^4} + 3{x^2} + 1} }}} \right)(x > 0),$$ then $$\frac{{dy}}{{dx}} =$$
[a] $$\frac{{{x^2} - 1}}{{{x^4} + 3{x^2} + 1}}$$
[b] $$\frac{{{x^2} + 1}}{{{x^4} + 3{x^2} + 1}}$$
[c] $$\frac{{{x^2} - 1}}{{{x^4} - 3{x^2} + 1}}$$
[d] $$\frac{{{x^2} + 1}}{{{x^4} - 3{x^2} + 1}}$$

Choice [A]

$$y = {\sin ^{ - 1}}\left( {\frac{{x + \frac{1}{x}}}{{\sqrt {{x^2} + 3 + \frac{1}{{{x^2}}}} }}} \right) = {\sin ^{ - 1}}\left( {\frac{{x + \frac{1}{x}}}{{\sqrt {{{\left( {x + \frac{1}{x}} \right)}^2} + 1} }}} \right)$$
Let $$x + \frac{1}{x} = \tan \theta \Rightarrow {\sec ^2}\theta \frac{{d\theta }}{{dx}} = 1 - \frac{1}{{{x^2}}}$$
Also $$y = {\sin ^{ - 1}}\left( {\cfrac{{\tan \theta }}{{\sec \theta }}} \right) = \theta$$
So $$\cfrac{{dy}}{{dx}} = \cfrac{{dy}}{{d\theta }} \times \cfrac{{d\theta }}{{dx}} = 1 \times \cfrac{{1 - \frac{1}{{{x^2}}}}}{{{{\sec }^2}\theta }} = \cfrac{{1 - \frac{1}{{{x^2}}}}}{{{{\left( {x + \cfrac{1}{x}} \right)}^2} + 1}}$$
$$= \cfrac{{{x^2} - 1}}{{{x^4} + 3{x^2} + 1}}$$

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Question 38: Let $$\overrightarrow a = i + j$$ and $$\overrightarrow b = 2i - k.$$, then the point of intersection of the lines $$\overrightarrow r \, \times \,\overrightarrow a = \overrightarrow b \times \overrightarrow a$$ and $$\overrightarrow r \, \times \,\overrightarrow b = \overrightarrow a \times \overrightarrow b$$ is
[a] –i + j + k
[b] 3i – j + k
[c] i – j – k
[d] 3i + j – k

Choice [D]

Given that $$\overrightarrow r \, \times \,\overrightarrow a = \overrightarrow b \times \overrightarrow a$$ or $$\overrightarrow r \, \times \,\overrightarrow a - \overrightarrow b \times \overrightarrow a = 0 \Rightarrow (\vec r - \vec b) \times \vec a = 0$$
So the line passes through $$\vec b$$and parallel to $$\vec a$$
$$\Rightarrow \vec r = \vec b + \alpha \vec a$$
Again given that $$\overrightarrow r \, \times \,\overrightarrow b = \overrightarrow a \times \overrightarrow b$$
$$\Rightarrow \vec r = \vec a + \beta \vec b$$
For intersection of both the lines, $$\vec b + \alpha \vec a = \vec a + \beta \vec b$$
By comparing coefficients, $$\alpha = \beta = 1$$
Hence point of intersection is $$\vec a + \vec b =$$3i + j – k

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Question 39: If the position vector of A and B relative to O be $$\widehat i - 4\widehat j + 3\mathop k\limits^ \wedge$$ and $$- \widehat i + 2\widehat j - \mathop k\limits^ \wedge$$ respectively then find median through $$\overrightarrow O$$ of triangle ABC is:
[a] –2h + 2k
[b] –j + k
[c] –i – j + k
[d] –i – j – k

Choice [B]

The midpoint of A and B, M is $$\frac{{1 - 1}}{2},\,\frac{{ - 4 + 2}}{2},\,\frac{{3 - 1}}{2} = (0,\, - 1,\,\,1)$$
Hence the vector OM = $$- \widehat j + \mathop k\limits^ \wedge$$

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Question 40: The Number of common tangent to the circles x² + y² = 4 & x² + y² – 6x – 8y = 24
[a] 0
[b] 1
[c] 3
[d] 4

Choice [B]

Centre and radius of the first circle are (0, 0) and 2. Centre and radius of the second circle are (3, 4) and 5.
Both the circle touch internally, hence there will be only one common tangent.

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Question 41: If X and Y are two sets, then $$X \cap Y' \cap (X \cup Y)'$$ is
[a] $$X'$$
[b] $$Y'$$
[c] $$\phi$$
[d] None of these

Choice [C]

We know that $$(X \cup Y)' = X' \cap Y'$$
So $$X \cap Y' \cap (X \cup Y)' = X \cap Y' \cap X' \cap Y' = \phi$$

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Question 42: If e1 = (1, 1, 1) & e2 = (1, 1, –1) and a and b are two vectors such that $${e_1} = 2\overrightarrow a + \overrightarrow b \,\,\,\& \,\,{e_2} = \overrightarrow a + 2\overrightarrow b$$. Then the angle between $$\overrightarrow a$$ & $$\overrightarrow b$$ is
[a] $${\cos ^{ - 1}}\left( {\cfrac{{ - 7}}{{11}}} \right)$$
[b] $${\cos ^{ - 1}}\left( {\cfrac{7}{{11}}} \right)$$
[c] $${\cos ^{ - 1}}\left( {\cfrac{7}{9}} \right)$$
[d] $${\cos ^{ - 1}}\left( {\cfrac{{6\sqrt 2 }}{{11}}} \right)$$

Choice [A]

Given that $${e_1} = 2\overrightarrow a + \overrightarrow b \,\,\,\& \,\,{e_2} = \overrightarrow a + 2\overrightarrow b$$, hence $$b = \cfrac{{2{e_2} - {e_1}}}{3} = \cfrac{{i + j - 3k}}{3}$$
$$a = \cfrac{{2{e_1} - {e_2}}}{3} = \cfrac{{i + j + 3k}}{3}$$
Now $$a \cdot b = \left| a \right|\,\left| b \right|\cos \theta$$ or $$\cfrac{{1 + 1 - 9}}{9} = \cfrac{{\sqrt {11} }}{3} \times \cfrac{{\sqrt {11} }}{3} \times \cos \theta$$
$$\Rightarrow \cos \theta = \cfrac{{ - 7}}{{11}}$$.

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Question 43: The locus of the point of intersection of tangents to the ellipse $$\frac{{{x^2}}}{{{a^2}}} + \frac{{{y^2}}}{{{b^2}}} = 1$$ which meet at right angle is
[a] a Circle
[b] a Parabola
[c] an ellipse
[d] a Hyperbola

Choice [A]

This is a standard result. The locus will be a circle. This is known as director circle

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Question 44: $${H_1},\,{H_2},\,{H_3},\,......{H_n}$$ be $$n$$ Harmonic mean between $$a,\;b$$ then $$\frac{{{H_1} + a}}{{{H_1} - a}} + \frac{{{H_n} + b}}{{{H_n} - b}} =$$
[a] 2n
[b] n + 1
[c] n – 1
[d] 2n + 1

Choice [A]

We can write, $$\cfrac{1}{a} + d = \cfrac{1}{{{H_1}}}$$ and $$\cfrac{1}{b} - d = \cfrac{1}{{{H_n}}}$$
Now $$\cfrac{{{H_1} + a}}{{{H_1} - a}} + \cfrac{{{H_n} + b}}{{{H_n} - b}} = \cfrac{{\cfrac{1}{a} + \cfrac{1}{{{H_1}}}}}{{\cfrac{1}{a} - \cfrac{1}{{{H_1}}}}} + \cfrac{{\cfrac{1}{b} + \cfrac{1}{{{H_n}}}}}{{\cfrac{1}{b} - \cfrac{1}{{{H_n}}}}}$$
$$= \cfrac{{\cfrac{1}{a} + \cfrac{1}{a} + d}}{{ - d}} + \cfrac{{\cfrac{1}{b} + \cfrac{1}{b} - d}}{d} = \cfrac{{2\left( {\cfrac{1}{b} - \cfrac{1}{a}} \right) - 2d}}{d} = \cfrac{{2(n + 1)d - 2d}}{d} = 2n$$

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Question 45: The value of tan 9° – tan 27° – tan 63° + tan 81° is equal to
[a] 5
[b] 3
[c] 4
[d] 6

Choice [C]

tan 9° – tan 27° – tan 63° + tan 81°
= $$\cfrac{{\sin 9^\circ }}{{\cos 9^\circ }} - \cfrac{{\sin 27^\circ }}{{\cos 27^\circ }} - \left[ {\cfrac{{\sin {{63}^ \circ }}}{{\cos {{63}^ \circ }}} - \cfrac{{\sin 81^\circ }}{{\cos 81^\circ }}} \right]$$ = A – B
Where A = $$\left( {\cfrac{{\sin {9^ \circ }\cos {{27}^ \circ } - \cos {9^ \circ }\sin {{27}^ \circ }}}{{\cos {9^ \circ }\cos {{27}^ \circ }}}} \right) = \cfrac{{ - \sin {{18}^ \circ }}}{{\cos {9^ \circ }\cos {{27}^ \circ }}}$$$$= \cfrac{{ - 2\sin {9^ \circ }}}{{\cos {{27}^ \circ }}}$$
B = $$\cfrac{{\sin {{63}^ \circ }\cos {{81}^ \circ } - \sin {{81}^ \circ }\cos {{63}^ \circ }}}{{\cos {{63}^ \circ }\cos {{81}^ \circ }}}$$$$= \cfrac{{ - \sin {{18}^ \circ }}}{{\sin {{27}^ \circ }\sin {9^ \circ }}} = \cfrac{{ - 2\cos {9^ \circ }}}{{\sin {{27}^ \circ }}}$$
Hence A – B = $$\cfrac{{2\cos {9^ \circ }}}{{\sin {{27}^ \circ }}} - \cfrac{{2\sin {9^ \circ }}}{{\cos {{27}^ \circ }}}$$$$= \cfrac{{2\left( {\cos {9^ \circ }\cos {{27}^ \circ } - \sin {9^ \circ }\sin {{27}^ \circ }} \right)}}{{\sin {{27}^ \circ }\cos {{27}^ \circ }}}$$
$$= \cfrac{{2\cos {{36}^ \circ }}}{{\left( {\cfrac{{\sin {{54}^ \circ }}}{2}} \right)}} = \cfrac{{4\cos {{36}^ \circ }}}{{\sin {{54}^ \circ }}}$$ = 4.

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Question 46: If $$y = {\tan ^{ - 1}}\left( {\cfrac{{3x - {x^3}}}{{1 - 3{x^3}}}} \right)\cfrac{{ - 1}}{{\sqrt 3 }} \le x \le \cfrac{1}{{\sqrt 3 }}$$then $$\cfrac{{dy}}{{dx}}$$ is
[a] $$- \cfrac{1}{{1 + {x^2}}}$$
[b] $$\cfrac{3}{{1 + {x^2}}}$$
[c] $$\cfrac{3}{{\sqrt {1 + {x^2}} }}$$
[d] $$\cfrac{1}{{\sqrt {1 + {x^2}} }}$$

Choice [B]

Put $$\tan \theta = x$$
$$y = {\tan ^{ - 1}}\left( {\cfrac{{3\tan \theta - {{\tan }^3}\theta }}{{1 - 3{{\tan }^2}\theta }}} \right) = {\tan ^{ - 1}}(\tan 3\theta ) = 3\theta$$
$$\cfrac{{dy}}{{dx}} = \cfrac{{dy}}{{d\theta }} \times \cfrac{{d\theta }}{{dx}} = 3 \times \cfrac{1}{{{{\sec }^2}\theta }} = \cfrac{3}{{1 + {x^2}}}$$

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Question 47: $$\int {{3^{{3^{{3^x}}}}}{3^{{3^x}}}{3^x}dx}$$
[a] $$\cfrac{{{3^{{3^x}}}{3^x}}}{{{{(\log 3)}^3}}} + C$$
[b] $$\cfrac{{{3^x}}}{{{{(\log 3)}^3}}} + C$$
[c] $$\cfrac{{{3^3}^{^x}}}{{{{(\log 3)}^3}}} + C$$
[d] $$\cfrac{{{3^3}^{^{{3^x}}}}}{{{{(\log 3)}^3}}} + C$$

Choice [D]

Let $${3^{{3^{{3^x}}}}} = t$$
$$\Rightarrow {3^{{3^{{3^x}}}}}\log {3.3^{{3^x}}}\log {3.3^x}\log 3dx = dt$$
$$\Rightarrow {3^{{3^{{3^x}}}}}dx = \cfrac{{dt}}{{{{(\log 3)}^3}}}$$
Hence the integration becomes,
$$\int {\cfrac{{dt}}{{{{\left( {\log 3} \right)}^3}}}} = \cfrac{t}{{{{\left( {\log 3} \right)}^3}}} = \cfrac{{{3^{{3^{{3^x}}}}}}}{{{{\left( {\log 3} \right)}^3}}}$$

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Question 48: If $$f(\theta ) = \left[ {\begin{array}{*{20}{c}}{\cos \theta }&{ - \sin \theta }&0\\{\sin \theta }&{\cos \theta }&0\\0&0&1\end{array}} \right]$$ then find $$f(\theta ) \cdot f(\alpha )$$ is equal to:
[a] $$f(\theta \alpha )$$
[b] $$f\left( {\frac{\theta }{\alpha }} \right)$$
[c] $$f(\theta + \alpha )$$
[d] $$f(\theta - \alpha )$$

Choice [C]

Let us multiply the two matrices
$$f(\theta )f(\alpha ) = \left[ {\begin{array}{*{20}{c}}{\cos \theta }&{ - \sin \theta }&0\\{\sin \theta }&{\cos \theta }&0\\0&0&1\end{array}} \right]\left[ {\begin{array}{*{20}{c}}{\cos \alpha }&{ - \sin \alpha }&0\\{\sin \alpha }&{\cos \alpha }&0\\0&0&1\end{array}} \right]$$
= $$\left[ {\begin{array}{*{20}{c}}{\cos (\theta + \alpha )}&{ - \sin (\theta + \alpha )}&0\\{\sin (\theta + \alpha )}&{\cos (\theta + \alpha )}&0\\0&0&1\end{array}} \right]$$ = $$f(\theta + \alpha )$$

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Question 49: If $$f:R \to R$$ is defined by $\left\{ {\begin{array}{*{20}{c}}{\frac{{x + 2}}{{{x^2} + 3x + 2}}}&{{\rm{if}}\,x \in \,R - \{ - 1 - 2\} }\\\begin{array}{l} - 1\\0\end{array}&\begin{array}{l}{\rm{if}}\,x = - 2\\{\rm{if}}\,x\, = - 1\end{array}\end{array}} \right.$ Then $$f(x)$$is continuous on the set
[a] R
[b] R – {–2}
[c] R – {–1}
[d] R – {–1, –2}

Choice [C]

$$\mathop {\lim }\limits_{x \to - 2} \frac{{x + 2}}{{{x^2} + 3x + 2}} = \frac{{(x + 2)}}{{(x + 2)(x + 1)}} = \frac{1}{{(x + 1)}} = - 1$$
Also $$f( - 2) = - 1$$
So the function is continuous at $$x = - 2$$
When $$x \to - 1$$, the limit of the function does not exist. Hence the function is not continuous at $$x = - 1$$. Correct answer is R – {–1}

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Question 50: In a triangle, if the sum of two sides is x and their product is y such that (x + z) (x – z) = y, where z is the third side of the triangle, then the triangle is
[a] Equilateral
[b] Right angled
[c] Isosceles
[d] Obtuse angled

Choice [D]

Let the three sides are $$a,\,b,\,\,c$$, then
$$x = a + b,\,\,y = ab$$
Given that $$(a + b + c)(a + b - c) = ab$$
Using $$2s = a + b + c$$, we have $$2s(2s - 2c) = ab$$
$$\Rightarrow \left( {\cfrac{{s(s - c)}}{{ab}}} \right) = \cfrac{1}{4}$$
Or $${\cos ^2}\cfrac{C}{2} = \cfrac{1}{4} \Rightarrow \cos \cfrac{C}{2} = \cfrac{1}{2}$$
Thus $$\angle C = 120$$ and the triangle is an obtuse angled triangle.

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