1 Quadratic Equations
Let α1?,α2? and β1?, β2? be the roots of ax2+bx+c=0 and px2+qx+r=0 respectively. If the system of equations α1?y+α2?z=0 and β1?y+β2?z=0 has a non-trivial solution, then
1) 1b2pr=q2ac
2) bpr2=qac2
3) bp2r=qa2c
4) None of these
Given system has non-trivial solution, then
⇒b2p2ar=a2q2cp
⇒b2pr=q2ac
2 Trigonometric Equations
If 4nα = π, then the value of tanα. tan2α. tan3α. tan4α. ….. tan(2n – 2)α tan (2n – 1)α is 1) 0 2) 1 3)-1 4) 1/2
Solution
2nα = π/2
(tanα tan(2n – 1)α) (tan 2α tan(2n – 2)α) ….. (tan (n – 1)α tan (n + 1)α).tan nα
(tanα tan(π/2 - α))(tan2αtan(π/2-2α))......tanπ/4 =1
3 Limits Continuity and Differentiability
A function is defined as follows: The function is
1) Discontinuous at x = 1 2) Differentiable at x = 1 3) Continuous but not differentiable at x = 1 4) None
4 Matrices
If $$A = \begin{bmatrix}1 & -2 & 3 \\ -4 & 2 & 5\end{bmatrix}$$ and $$B = \begin{bmatrix}1 & 3 \\ -1 & 0 \\ 2 & 4\end{bmatrix}$$. Show that $$(AB)' = ?$$
1) B'A' 2) A'B' 3) AB' 4) A'B
We have, $$\quad A' = \begin{bmatrix}1 & -4 \\ -2 & 2 \\ 3 & 5\end{bmatrix}$$ and $$\quad B' = \begin{bmatrix}1& -1 & 2 \\ 3 & 0 & 4 \end{bmatrix}$$ $$\therefore\quad AB = \begin{bmatrix}1 & -2 & 3 \\ -4 & 2 & 5\end{bmatrix}\times\begin{bmatrix}1 & 3 \\ -1 & 0 \\ 2 & 4\end{bmatrix}$$ $$\quad = \begin{bmatrix}1+2+6 & 3+0+12 \\ -4-2+10 & -12+0+20\end{bmatrix} = \begin{bmatrix}9& 15 \\ 4 & 8 \end{bmatrix}$$ $$\therefore\quad (AB)' = \begin{bmatrix}9& 4 \\ 15 & 8 \end{bmatrix}\quad ...(i)$$ and $$B'A' = \begin{bmatrix}1& -1 & 2 \\ 3 & 0 & 4 \end{bmatrix}\times\begin{bmatrix}1 & -4 \\ -2 & 2 \\ 3 & 5\end{bmatrix}$$ $$\quad = \begin{bmatrix}1+2+6 & -4-2+10 \\ 3+0+12 & -12+0+20\end{bmatrix}$$ $$\quad = \begin{bmatrix}9 & 4 \\ 15 & 8\end{bmatrix} \quad ...(ii)$$ From $$(i)$$ and $$(ii)$$, we get $$\quad (AB)' = B'A'$$
5 Fundamentals of Mathematics
p: She is beautiful. q:She is happy. Then (∼p)⇒(∼q), means
1) If she is beautiful, then she is not happy.
2) If she is not beautiful, then she is happy.
3) If she is not beautiful, then she is not happy.
4) If she is not happy, then she is not beautiful.
It is expanded to she is not beautiful implies she is not happy.
6 Sequences and Series
Let S k be the sum of an infinite GP series whose first term k is k and common ratio is k/(k + 1) (k> 0). Then, the value of
1) loge 4 2) loge 2 – 1 3) 1 - loge 2 4) 1 - loge 4
SOLUTION
7 Trigonometric Equations
If sin-1 a + sin-1 b + sin-1 c = π, then the value of a√(1- a2) + b √(1 - b2) + c√(1 - c2) will be
1) 2abc 2) abc 3) (1/2)abc 4) (1/3) abc
8 Fundamentals of Mathematics
The area of the region described by $$\left\{ \left( x,y \right) /{ x }^{ 2 }+{ y }^{ 2 }\le and\quad { y }^{ 2 }\le 1-x \right\} $$ is
1) $$\frac { \pi }{ 2 } -\frac { 2 }{ 3 } $$
2) $$\frac { \pi }{ 2 }+\frac { 2 }{ 3 } $$
3) $$\frac { \pi }{ 2 } +\frac { 4 }{ 3 } $$
4) $$\frac { \pi }{ 2 } -\frac { 4 }{ 3 } $$
$${ x }^{ 2 }+{ y }^{ 2 }=1$$ $${ y }^{ 2 }=1-x$$ $$Area = \frac { \pi }{ 2 } +\frac { 4 }{ 3 } $$
9 Sets relations and functions
Find the range of the function f (x) = sin -1 + cos -1 , where [ . ] is the greatest integer function.
\( <p>f (x) = sin <sup>-1</sup> \(\left[ {{x}^{2}}+\cfrac{1}{2} \right]+\)cos <sup>-1</sup>\(\left[ {{x}^{2}}-\cfrac{1}{2} \right]\)<br /> </p> <p>= sin <sup>-1</sup> \(\left[ {{x}^{2}}+\cfrac{1}{2} \right]+\) cos <sup>-1</sup> \(\left[ {{x}^{2}}+\cfrac{1}{2}-1 \right]\)<br /> </p> <p>= sin <sup>-1</sup> \(\left[ {{x}^{2}}+\cfrac{1}{2} \right]+\) cos <sup>-1</sup> \(\left( \left[ {{x}^{2}}+\cfrac{1}{2} \right]-1 \right)\)<br /> </p> <p>Since \({{x}^{2}}+\cfrac{1}{2}\) \(\ge \cfrac{1}{2}\), \(\left[ {{x}^{2}}+\cfrac{1}{2} \right]\) = 0 or 1 as<br /> Sin <sup>-1</sup>\(\left[ {{x}^{2}}+\cfrac{1}{2} \right]\) is defined only for these two values<br /> </p> <p>Hence \(\left[ {{x}^{2}}+\cfrac{1}{2} \right]=\) 0<br /> </p> <p>⇒f (x) = sin <sup>-1</sup> 0 + cos <sup>-1</sup> ( - 1) = \(\pi \left[ {{x}^{2}}+\cfrac{1}{2} \right]=\) 1<br /> ⇒ f (x) = sin -1 (1) + cos -10 = π</p> <p>Therefore range of f (x) = {\(\pi \)}</p>\)
10 Differentiation
The value of 'a' in order f(x)= √ 3?sinx−cosx−2ax+b decrease for all real values of x, is given by
1) a > 1 2) a ≥ 1 3) a ≥ 2 4) a < 2
Given, f(x)= √ 3?sinx−cosx−2ax+b
f′(x)=√3?cosx+sinx−2a
f′(x)=2((√3 /2)cosx + (1 /2)sin x) - 2a
f′(x)=2(sin(π / 3 ? + x )−a)
if a>1, then for all real values of x ,f′(x)<0.
So, a>1
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