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Questions:

Multiple Choice Question

25 x 1 = 25

If m is a number such that m $\le$ 5, then the probability that quadratic equation 2x² + 2mx + m + 1 = 0 has real roots is

(a)

${1\over 5}$

(b)

${2\over 5}$

(c)

${3\over 5}$

(d)

${4\over 5}$

The probability of two events A and B are 0.3 and 0.6 respectively. The probability that both A and B occur simultaneously is 0.18. The probability that neither A nor B occurs is

(a)

0.1

(b)

0.72

(c)

0.42

(d)

0.28

A matrix is chosen at random from a set of all matrices of order 2, with elements 0 or 1 only. The probability that the determinant of the matrix chosen is non zero will be

(a)

${3\over 16}$

(b)

${3\over 8}$

(c)

${1\over 4}$

(d)

${5\over 8}$

A letter is taken at random from the letters of the word ‘ASSISTANT’ and another letter is taken at random from the letters of the word ‘STATISTICS’. The probability that the selected letters are the same is

(a)

${7\over 45}$

(b)

${17\over 90}$

(c)

${29\over 90}$

(d)

${19\over 90}$

$\int \frac{e^x\left(x^2 \tan ^{-1} x+\tan ^{-1} x+1\right)}{x^2+1} d x$ is

(a)

e^xtan^-1(x+1)+c

(b)

tan^-1(e^x)+c

(c)

$e^x{(tan^{-1}x)^2\over 2}+c$

(d)

e^x tan^-1 x+c

$\text { If } f(x)=\left\{\begin{array}{ll} a x^2-b, & -1<x<1 \\ \frac{1}{|x|}, & \text { elsewhere } \end{array} \ \text { is differentiable at } x=1\right. \text {, then }$

(a)

$a={1\over2},b={-3\over 2}$

(b)

$a={-1\over2},b={3\over 2}$

(c)

$a=-{1\over2},b=-{3\over 2}$

(d)

$a={1\over2},b={3\over 2}$

If $f(x)= \begin{cases}2 a-x, & \text { for } \quad-a<x<a \\ 3 x-2 a & \text { for } \quad x \geq a\end{cases}$, then which one of the following is true?

(a)

f(x) is not differentiable at x = a

(b)

f(x) is discontinuous at x = a

(c)

f(x) is continuous for all x in R

(d)

f(x) is differentiable for all x $\ge$ a

If f(x) = x² - 3x, then the points at which f(x) = f '(x) are

(a)

both positive integers

(b)

both negative integers

(c)

both irrational

(d)

one rational and another irrational

$lim_{x\rightarrow {\pi/2}}{2x-\pi\over cosx} $

(a)

(b)

(c)

-2

(d)

10.

If the projection of $5\hat{i}-\hat{j}-3\hat{k}$ on the vector $\hat{i}+3\hat{j}+\lambda\hat{k}$ is same as the projection of $\hat{i}+3\hat{j}+\lambda\hat{k}$ on $5\hat{i}-\hat{j}-3\hat{k}$, then $\lambda$ is equal to

(a)

$\pm 4$

(b)

$\pm 3$

(c)

$\pm 5$

(d)

$\pm 1$

11.

A vector makes equal angle with the positive direction of the coordinate axes. Then each angle is equal to

(a)

$cos^{-1}({1\over 3})$

(b)

$cos^{-1}({2\over 3})$

(c)

$cos^{-1}({1\over\sqrt 3})$

(d)

$cos^{-1}({2\over\sqrt 3})$

12.

If x₁, x₂, x₃ as well as y₁, y₂, y₃ are in geometric progression with the same common ratio, then the points (x₁, y₁ ), (x₂, y₂), (x₃, y₃ ) are

(a)

vertices of an equilateral triangle

(b)

vertices of a right angled triangle

(c)

vertices of a right angled isosceles triangle

(d)

collinear

13.

If A =$\begin{bmatrix} 1& 2 &2 \\ 2 & 1 & -2 \\ a & 2 & b \end{bmatrix}$ is a matrix satisfying the equation AA^T = 9I, where I is 3 $\times$ 3 identity matrix, then the ordered pair (a, b) is equal to

(a)

(2, - 1)

(b)

(- 2, 1)

(c)

(2, 1)

(d)

(- 2, - 1)

14.

The length of the perpendicular from origin to line is $\sqrt{3}x-y+24=0$ is ______________

(a)

2$\sqrt{3}$

(b)

(c)

(d)

15.

$\sqrt \frac{1-2x}{1+2x}$ is approximately equal to ______________

(a)

1- 2x-x²

(b)

1 + 2x+ x²

(c)

1+ 2x

(d)

1-2x+x²

16.

If a vertex of a square is at the origin and its one side lies along the line 4x + 3y - 20 = 0, then the area of the square is

(a)

20 sq. units

(b)

16 sq. units

(c)

25 sq. units

(d)

4 sq.units

17.

The sum up to n terms of the series $\frac { 1 }{ \sqrt { 1 } +\sqrt { 3 } } +\frac { 1 }{ \sqrt { 3 } +\sqrt { 5 } } +\frac { 1 }{ \sqrt { 5 } +\sqrt { 7 } } +$....is

(a)

$\sqrt { 2n+1 } $

(b)

$\frac { \sqrt { 2n+1 } }{ 2 } $

(c)

$\sqrt { 2n+1 } -1$

(d)

$\frac { \sqrt { 2n+1 } -1 }{ 2 } $

18.

If $\frac { kx }{ (x+2)(x-1) } =\frac { 2 }{ x+2 } +\frac { 1 }{ x-1 } $, then the value of k is

(a)

(b)

(c)

(d)

19.

Given that x, y and b are real numbers x < y, b > 0, then

(a)

xb < yb

(b)

xb > yb

(c)

xb ≤ yb

(d)

$\frac { x }{ b } \ge \frac { y }{ b } $

20.

If |x+2| $\le$ 9, then x belongs to

(a)

$(-\infty ,-7)$

(b)

[-11, 7]

(c)

$(-\infty ,-7)\cup (11,\infty)$

(d)

(-11, 7)

21.

If tan α and tan β are the roots of x² + ax + b = 0; then $\frac { sin(\alpha +\beta ) }{ sin\alpha sin\beta } $ is equal to

(a)

$\frac { b }{ a } $

(b)

$\frac { a }{ b } $

(c)

$-\frac { a }{ b } $

(d)

$-\frac { b}{ a } $

22.

Let f:R➝R be defined by f(x) = 1 - |x|. Then the range of f is

(a)

(b)

(1,∞)

(c)

(-1,∞)

(d)

(-∞,1]

23.

The inverse of f(x) = $\begin{cases} x\quad if\quad x<1 \\ { x }^{ 2 }\quad if\quad 1\le x\le 4 \\ 8\sqrt { x } \quad if\quad x>4 \end{cases}$ is

(a)

${ f }^{ -1 }(x)=\begin{cases} x\quad if\quad x<1 \\ \sqrt { x } \quad if\quad 1\le x\le 16 \\ \frac { { x }^{ 2 } }{ 64 } \quad if\quad x>16 \end{cases}$

(b)

${ f }^{ -1 }(x)=\begin{cases} -x\quad if\quad x<1 \\ \sqrt { x } \quad if\quad 1\le x\le 16 \\ \frac { { x }^{ 2 } }{ 64 } \quad if\quad x>16 \end{cases}$

(c)

${ f }^{ -1 }(x)=\begin{cases} { x }^{ 2 }\quad if\quad x<1 \\ \sqrt { x } \quad if\quad 1\le x\le 16 \\ \frac { { x }^{ 2 } }{ 64 } \quad if\quad x>16 \end{cases}$

(d)

${ f }^{ -1 }(x)=\begin{cases} { 2x }\quad if\quad x<1 \\ \sqrt { x } \quad if\quad 1\le x\le 16 \\ \frac { { x }^{ 2 } }{ 8 } \quad if\quad x>16 \end{cases}$

24.

If the function f:[-3,3]➝S defined by f(x) = x² is onto, then S is

(a)

[-9,9]

(b)

(c)

[-3,3]

(d)

[0,9]

25.

The function f:[0,2π]➝[-1,1] defined by f(x) = sin x is

(a)

one-to-one

(b)

on to

(c)

bijection

(d)

cannot be defined

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Answers Key:

Multiple Choice Question

25 x 1 = 25

$m \leq 5$
$2 x^{2}+2 m x+m+1>0$
For real roots
$4 m^{2}-8(m+1)>0 $
$4 m^{2} - 8 m-8>0 $
$m^{2}-2 m-2>0 $
$\text {Then } m \text { can be } 5,4,3$
$n(S)=5 ,n(A)=3 ,P(A)=\frac{3}{5} $

$P(A)=0.3, \quad P(B)=0.6, \quad P(A \cap B)=0.18 $
$P(A \cup B)=P(A)+P(B)-P(A \cap B) $
$=0.3+0.6-0.18 =0.72 $
$P(\bar{A} \cup \bar{B})=1-P(A \cap B) $
$=1-0.72=0.28 $

$n(S) =2^{14}=16 $
$A =\left\{\begin{array}{ll} 1 & 0 \\ 0 & 1 \end{array}|,| \begin{array}{ll} 0 & 1 \\ 1 & 0 \end{array}|,| \begin{array}{ll} 1 & 1 \\ 0 & 1 \end{array}|,| \begin{array}{ll} 1 & 0 \\ 1 & 1 \end{array}|,| \begin{array}{ll} 0 & 1 \\ 1 & 1 \end{array}|,| \begin{array}{ll} 1 & 1 \\ 1 & 0 \end{array} \mid\right\} $
$n(A) =6 $
$P(A) =\frac{6}{16}=\frac{3}{8} $

Assistant Statistics
$\begin{array}{ll} A=2 & S=3 \\ S=3 & T=3 \\ T=2 & A=1 \\ I=1 & I=2 \\ N=1 & C=1 \end{array}$
$\mathrm{n}(\mathrm{S})=9 \mathrm{C}_{1} \times 10 \mathrm{C}_{1}=90$
One letter in I word and another letter (same) in II word the combinations are
$\begin{array}{ll} \mathrm{AA} & \mathrm{A} \\ \mathrm{SSS} & \mathrm{SSS} \\ \mathrm{TT} & \mathrm{TTT} \\ \mathrm{I} & \mathrm{II} \end{array}$
No. of doublets are
AA, AA, SS, SS, SS, SS, SS, SS, S$ SS, SS, TT TT,
TT, TT, TT, TT, II, II
n(A) = 19
$P(A) =\frac{n(A)}{n(S)} $
$=\frac{19}{90} $

$\int \frac{e^{x}\left(x^{2} \tan ^{-1} x+\tan ^{-1} x+1\right)}{x^{2}+1} d x $
$=\int e^{x}\left(\frac{\left(x^{2}+1\right) \tan ^{-1} x+1}{\left(x^{2}+1\right)}\right) d x $
$=\int e^{x}\left(\tan ^{-1} x+\frac{1}{\left(x^{2}+1\right)}\right) d x $
$=\int e^{x}\left[f(x)+f^{\prime}(x)\right] d x, f(x)=\tan ^{-1} x $
$=e^{x} f(x)+c \text { where } f(x)=\tan ^{-1} x $
$=e^{x} \tan ^{-1} x+c $

Given f is differentiable
$\therefore f^{\prime}\left(1^{-}\right)=f^{\prime}\left(1^{+}\right)=1 $
$f^{\prime}\left(1^{-}\right) =\lim _{x \rightarrow 1^{-}} \frac{f(x)-f(1)}{x+1}=\lim _{x \rightarrow 1^{-}} \frac{\left(a x^{2}-b\right)-(a-b)}{x+1} $
$=\lim _{x \rightarrow 1^{-}} \frac{a x^{2}-b-a+b}{x+1}=\lim _{x \rightarrow 1^{-}} \frac{a\left(x^{2}-1\right)}{x+1} $
$=\lim _{x \rightarrow 1^{-}}-a(x+1) $
$=a(1+1)=2 a $
$\therefore f^{\prime}\left(1^{+}\right) =\lim _{x \rightarrow 1^{+}} \frac{f(x)-f(1)}{x-1}=\lim _{x \rightarrow 1^{+}} \frac{\frac{1}{x}-1}{x-1} $
$=\lim _{x \rightarrow 1^{+}} \frac{1-x}{x(x-1)}=\lim _{x \rightarrow 1^{+}} \frac{-1}{x}=-1$
$\therefore 2 a =-1 $
$a =\frac{-1}{2} $
$\text { and } f(1)=1$
$a-b=1 $
$-1 / 2-1=b $
$b=-3 / 2 $

$f^{\prime}\left(a^{-}\right) =\lim _{x \rightarrow a^{-}} \frac{f(x)-f(a)}{x-a}=\lim _{x \rightarrow a^{-}} \frac{(2 a-x)-(2 a-a)}{x-a} $
$=\lim _{x \rightarrow a^{-}} \frac{-x+a}{x-a}=-1 $
$f^{\prime}\left(a^{+}\right) =\lim _{x \rightarrow a^{+}} \frac{f(x)-f(a)}{x-a} $
$=\lim _{x \rightarrow a^{-}} \frac{(3 x-2 a)-(3 a-2 a)}{x-a} $
$=\lim _{x \rightarrow a^{-}} \frac{3 x-2 a-a}{x-a}=\lim _{x \rightarrow a^{-}} \frac{3 x-3 a}{x-a}=3 $
$f^{\prime}\left(a^{-}\right) \neq f^{\prime}\left(a^{+}\right) $
$\therefore \text { It is not differentiable at } x=a$
$\therefore f^{\prime}(x) \text { does not exist }$

$\text { Given } f(x)=x^{2}-3 x$
$f^{\prime}(x) =2 x-3 $
$f(x) =f^{\prime}(x) $
$x^{2}-3 x =2 x-3 $
$x^{2}-3 x-2 x+3 =0 $
$x^{2}-5 x+3 =0 $
$\text { It has irrational roots. }$

$\lim _{x \rightarrow \pi / 2} \frac{2 x-\pi}{\cos x} =\lim _{x \rightarrow \pi / 2} \frac{2 x-\pi}{\sin \left(\frac{\pi}{2}-x\right)} $
$=\lim _{\left(\frac{\pi}{2}-x\right) \rightarrow 0} \frac{-2\left(\frac{\pi}{2}-x\right)}{\sin \left(\frac{\pi}{2}-x\right)}=-2 $

10.

$\text { Projection }=\frac{5-3-3 \lambda}{\sqrt{1+9+\lambda^{2}}}=\frac{5-3-3 \lambda}{\sqrt{25+1+9}}$
$\sqrt{10+\lambda^{2}} =\sqrt{35} $
$\lambda^{2}+10 =35 $
$\lambda^{2} =25 $
$\lambda =\pm 5 $

11.

All angle are equal
$\therefore \alpha=\beta=\gamma$
$\text { W.K.T } \cos ^{2} \alpha+\cos ^{2} \beta+\cos ^{2} \gamma=1$
$\cos ^{2} \alpha+\cos ^{2} \alpha+\cos ^{2} \alpha =1 $
$3 \cos ^{2} \alpha =1 $
$\cos ^{2} \alpha =\frac{1}{3} $
$\cos \alpha =\pm \frac{1}{\sqrt{3}} $
$\alpha =\cos ^{-1}\left(\frac{1}{\sqrt{3}}\right) $

12.

(d)
collinear

13.

$A A^{T}=9 I $
${\left[\begin{array}{ccc} 1 & 2 & 2 \\ 2 & 1 & -2 \\ a & 2 & b \end{array}\right]\left[\begin{array}{ccc} 1 & 2 & a \\ 2 & 1 & 2 \\ 2 & -2 & b \end{array}\right]=\left[\begin{array}{ccc} 9 & 0 & 0 \\ 0 & 9 & 0 \\ 0 & 0 & 9 \end{array}\right]} $
${\left[\begin{array}{ccc} 1+4+4 & 2+2-4 & a+4+2 b \\ 2+2-4 & 4+1+4 & 2 a+2-2 b \\ a+4+2 b & 2 a+2-2 b & a^{2}+4+b^{2} \end{array}\right]} $
$=\left[\begin{array}{ccc} 9 & 0 & 0 \\ 0 & 9 & 0 \\ 0 & 0 & 9 \end{array}\right] $
$a+4+2 b =0 $
$a+2 b =-4 $
$2 a-2 b+2 =0 $
$2 a-2 b =-2 $
$\text { Solve (1) } \&(2) \text { we get } a=-2 b=-1$
$\therefore(a, b)=(-2,1)$

14.

(d)
12

15.

(d)
1-2x+x²

16.

Perpendicular distance from origin to the line is
4x + 3y - 20 = 0 is
$\left(\frac{-20}{\sqrt{16+9}}\right)=\frac{20}{\sqrt{25}}=\frac{20}{5}=4 \text { units }$
Area of the square = 4 $\times$ 4 = 16 sq. units.

17.

$\frac{1}{\sqrt{1}+\sqrt{3}} =\frac{1}{\sqrt{3}+\sqrt{1}} \times \frac{\sqrt{3}-1}{\sqrt{3}-1}=\frac{\sqrt{3}-1}{2} $
$\frac{1}{\sqrt{3}+\sqrt{5}} =\frac{1}{\sqrt{5}+\sqrt{3}} \times \frac{\sqrt{5}-\sqrt{3}}{\sqrt{5}-\sqrt{3}} $
$=\frac{\sqrt{5}-\sqrt{3}}{2} $
$\text { Sum to } \mathrm{n} \text { terms }=\frac{(\sqrt{3}-1)}{2}+\frac{(\sqrt{5}-\sqrt{3})}{2}+\ldots . .\left(\frac{\sqrt{2 n+1}-\sqrt{2 n-1}}{2}\right)$
$=\frac{\sqrt{2 n+1}-1}{2}$

18.

$\frac{2}{x+2}+\frac{1}{x-1}=\frac{2 x-2+x+2}{(x+2)(x-1)}=\frac{3 x}{(x+2)(x-1)}= k = 3$

19.

(a)
xb < yb

20.

$|x+2| \leq 9 $
$ \Rightarrow-9 \leq x+2 \leq 9 $
$ \Rightarrow-11 \leq x \leq 7 $
$x \text { belongs to }[-11,7]$

21.

$x^{2}+a x+b=0$
$\text { Let } \tan \alpha \text { and } \tan \beta \text { be the roots. }$
$\tan \alpha+\tan \beta =-a $
$\tan \alpha \tan \beta =b $
$\frac{\sin (\alpha+\beta)}{\sin \alpha \sin \beta} =\frac{\sin \alpha \cos \beta+\cos \alpha \sin \beta}{\sin \alpha \sin \beta} $
$=\cot \beta+\cot \alpha $
$=\frac{1}{\tan \alpha}+\frac{1}{\tan \beta} $
$=\frac{\tan \alpha+\tan \beta}{\tan \alpha \tan \beta}=\frac{-a}{b} $

22.

$\mathrm{f}: \mathbb{R} \rightarrow \mathbb{R} \text { is defined by }$
$\mathrm{f}(x)=1-|x|$
$\text { The range is }(-\infty, 1] \text { as } f(-\infty)=-\infty$
$f(0)=1$
$f(\infty) =-\infty$

23.

$\text { i) Let } y=x \text { then } x=y \Rightarrow f^{-1}(x)=x$
$\text { ii) Let } y=x^{2} \text { then } y=\sqrt{x} \Rightarrow f^{-1}(x)=\sqrt{x}$
$\text { iii) Let } y=8 \sqrt{x} \text { then } \frac{y^{2}}{64}=x \Rightarrow \mathrm{f}^{-1}(x)=\frac{x^{2}}{64}$
$\therefore \mathrm{f}^{-1}(x)= \begin{cases}x & x<1 \\ \sqrt{x} & 1 \leq x \leq 16 \\ \frac{x^{2}}{64} & x>16\end{cases}$

24.

f(0) = 0, f(-3) = 9 and f(3) = 9
.'. S is [0,9]

25.

It is onto not one-one

$\text { Since } \sin 30^{\circ}=\frac{1}{2}$
$\sin 150^{\circ}=\frac{1}{2}$