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Engineering
30 Views   |

Let where       Then   equals :

• Option 1)

• Option 2)

• Option 3)

• Option 4)

Option 2

Engineering
56 Views   |

If (10)9 + 2(11)1   (10)8 + 3(11)2  (10)7 +......  +10 (11)9 = k (10)9, then k is equal to :

• Option 1)

100

• Option 2)

110

• Option 3)

• Option 4)

Use

Sum of n terms of a GP -

$S_{n}= \left\{\begin{matrix} a\frac{\left ( r^{n}-1 \right )}{r-1}, &if \: r\neq 1 \\ n\, a, & if \, r= 1 \end{matrix}\right.$

- wherein

$a\rightarrow$ first term

$r\rightarrow$ common ratio

$n\rightarrow$ number of terms

and

(10)9 + 2(11)1 (10)8 + 3(11)2 (10)7 +......  +10 (11)9 = k(10)9

Take common 109

$10^{9}\left [ 1+2\times \frac{11}{10}+3\times \left ( \frac{11}{10} \right )^{2}+\cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot 10\times \left ( \frac{11}{10} \right ) ^{9}\right ]= k\left ( 10 \right )^{9}$

$\therefore k= 1+2x+3x^{2}+\cdot \cdot \cdot \cdot \cdot \cdot \cdot 10x^{9} \: where \:\:x=\frac{11}{10}$

$kx= x+2x^{2}+3x^{3}+\cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot 10x^{10}$

Subtract

$k-kx= 1+x+x^{2}+x^{3}+x^{4}\cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot (-10x^{10})$

$k(1-x)= \frac{1(x^{10}-1)}{x-1}-10x^{10}$

$k\left ( 1-\frac{11}{10} \right )= \frac{\left ( \frac{11}{10} \right )^{10}-1}{\frac{11}{10}-1}-10\times \left ( \frac{11}{10} \right )^{10}$

$-\frac{k}{10}= \frac{\left ( \frac{11}{10} \right )^{10}}{\frac{1}{10}}-10-\frac{11^{10}}{10^{9}}$

$\therefore k=100$

Option 1)

100

Option 2)

110

Option 3)

Option 4)

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Engineering
319 Views   |

If the coefficients of x3 and x4 in the expansion of (1 + ax + bx2) (1 - 2x)18 in powers of x are both zero, then (a, b) is equal to :

• Option 1)

• Option 2)

• Option 3)

• Option 4)

As we have learned Expression of Binomial Theorem -   - wherein for n  +ve integral .     coeff of   coeff of    and  and   a = 16  b = 272/3        Option 1) Option 2) Option 3) Option 4)
Engineering
282 Views   |

The sum of first 20 terms of the sequence 0.7 ,0.77,0.777,.........,is :

• Option 1)

• Option 2)

• Option 3)

• Option 4)

As we learnt

Sum of infinite terms of a GP -

$a+ar+ar^{2}+- - - - -= \frac{a}{1-r}\\here \left | r \right |<1$

- wherein

$a\rightarrow$ first term

$r\rightarrow$ common ratio

$S=0.7+0.77+0.777...upto\: \: 20\: \: terms$

$S=\frac{7}{9}(0.9+0.99+0.999...)$

$S=\frac{7}{9}(1-0.1+1-0.01+1-0.001...)$

$S=\frac{7}{9}(20-(\frac{1}{10}+\frac{1}{100}+...upto\: \: 20\: \: terms))$

$S=\frac{7}{9}(20-\frac{\frac{1}{10}(1-\frac{1}{10^{20}})}{(1-\frac{1}{10})})$

$S=\frac{7}{9}(20-\frac{1-10^{-20}}{9})$

$S=\frac{7}{81}(179+10^{-20})$

Option 1)

Option 2)

Option 3)

Option 4)

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Engineering
134 Views   |

The graph of the function  $\dpi{100} y=f\left ( x \right )$  is symmetrical about the line $\dpi{100} x=2$ ,then

• Option 1)

$f\left ( x \right )= f\left ( -x \right )$

• Option 2)

$f\left (2+ x \right )= f\left ( 2-x \right )$

• Option 3)

$f\left ( x +2\right )= f\left ( x-2 \right )$

• Option 4)

$f\left ( x \right )=- f\left ( -x \right )$

As we learnt in

Even Function -

$f(-x)= f(x)$

- wherein

Since a graph symmetric about y-axis

means  x = 0 then it is even function and f(-x) = f(x)

$\therefore$    f(0 - x) = f(0 + x)     (b < z it is symmetric about v = 0 )

But in question it is symmetric about x = 2

then f(x - 2) = f(x + 2)

Correct option is 3.

Option 1)

$f\left ( x \right )= f\left ( -x \right )$

Option 2)

$f\left (2+ x \right )= f\left ( 2-x \right )$

Option 3)

$f\left ( x +2\right )= f\left ( x-2 \right )$

Option 4)

$f\left ( x \right )=- f\left ( -x \right )$

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Engineering
133 Views   |

if     and

• Option 1)

is an empty set.

• Option 2)

contains exactly one element.

• Option 3)

contains exactly two elements.

• Option 4)

contains more than two elements.

As we learnt in

FUNCTIONS -

A relation f from a set A to a set B is said to be a function if every element of set A has one and only one image in set B.

-

$f(x)+2f \left(\frac{1}{x} \right )=3x$

Put $\frac{1}{x}$ at the place of

$f\left(\frac{1}{x} \right )+2f(x)=\frac{3}{x}$                                                    (i)

$2f\left(\frac{1}{x} \right )+f(x)=3x$                                                (ii)

Multiplying (i) by 2

$2f\left(\frac{1}{x} \right )+4f(x)=\frac{6}{x}$

$\underline{2f\left(\frac{1}{x} \right )+f(x)=3x}$

$3f(x)=\frac{6}{x}-3x$

$f(x)=\frac{2}{x}-3x$

and             $f(-x)=\frac{2}{-x}+x$

$\therefore\ \; \frac{2}{x}-x=-\frac{2}{x}+x$

$\Rightarrow\ \; \frac{4}{x}-2x=0$

$\Rightarrow\ \; \frac{4-2x^{2}}{x}=0$

$\Rightarrow\ \; 4=2x^{2}$

$\Rightarrow\ \; x^{2}=2$

$x=\pm \sqrt{2}, \; x \neq 0$

Correct option is 3.

Option 1)

is an empty set.

Option 2)

contains exactly one element.

Option 3)

contains exactly two elements.

Option 4)

contains more than two elements.

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Engineering
124 Views   |

The real number for which the equation, $2x^2+3x+k = 0$  has two distinct real roots in

• Option 1)

does not exist.

• Option 2)

lies between 1 and 2 .

• Option 3)

lies between 2 and 3 .

• Option 4)

lies between -1 and 0 .

As we have learned

Quadratic Expression Graph when a> 0 & D > 0 -

Real and distinct roots of

$f\left ( x \right )= ax^{2}+bx+c$

& $D= b^{2}-4ac$

- wherein

$\frac{-b}{2a}=-3/4$    is the abscissa of vertex

and , it should lie in(0,1 ) but it's not true

S, no value of 'k' exists

Option 1)

does not exist.

Option 2)

lies between 1 and 2 .

Option 3)

lies between 2 and 3 .

Option 4)

lies between -1 and 0 .

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Engineering
124 Views   |

Let and be the roots of equation are in A.P. and   then the value of is ?

• Option 1)

• Option 2)

• Option 3)

• Option 4)

As we have learned

Sum of Roots in Quadratic Equation -

$\alpha +\beta = \frac{-b}{a}$

- wherein

$\alpha \: and\beta$ are root of quadratic equation

$ax^{2}+bx+c=0$

$a,b,c\in C$

Product of Roots in Quadratic Equation -

$\alpha \beta = \frac{c}{a}$

- wherein

$\alpha \: and\ \beta$ are roots of quadratic equation:

$ax^{2}+bx+c=0$

$a,b,c\in C$

@1449

$|\alpha -\beta | = \left | \frac{\sqrt{q^2}-4pr}{p} \right |$

$\left ( \because \left | \frac{\sqrt{D}}a{} \right | \right )$

Also $\frac{\alpha +\beta }{\alpha \beta }= 4$

$\Rightarrow \frac{-q}{r}= 4$

$\Rightarrow q = -4r ....(1)$

$= \sqrt{16(\frac{r}{p})^2-(4\frac{r}{p})}$

Also p+r =2q

$\Rightarrow p+r = -8r \Rightarrow r/p = -1/9$

$\therefore \frac{\left | \alpha -\beta \right |}{16\times 1/81+4/9}= \sqrt{\frac{52}{81}}=\frac{2\sqrt{13}}{9}$

Option 1)

Option 2)

Option 3)

Option 4)

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Engineering
133 Views   |

If z is a complex number of unit modulus and argument  ,then arg    equals:

• Option 1)

• Option 2)

• Option 3)

• Option 4)

As we have learned

Euler's Form of a Complex number -

$z=re^{i\theta}$

- wherein

r denotes modulus of z and $\theta$ denotes argument of z.

Polar Form of a Complex Number -

$z=r(cos\theta+isin\theta)$

- wherein

r= modulus of z and $\theta$ is the argument of z

$|z| = 1$

Arg (z)= $\theta$

$\Rightarrow z = e^{i\theta }= \cos \theta + i \sin \theta$

So, $\frac{1+z}{1+z}= \frac{1+\cos \theta +i \sin \theta }{1+\cos \theta -i\sin \theta }$

$\frac{2 \cos^2\theta h+2 i\sin \theta h\cos \theta /2}{2\cos ^{2}\theta h-2i\sin \theta h\cos \theta }$

$=\frac{\cos \theta h+i\sin \theta h}{\cos \theta h-i\sin \theta h}$

$=\frac{e^{i\theta h}}{e^{-i\theta h}}= e^{i\theta }$

$\left ( \frac{1+z}{1+\bar{z}} \right )= \theta$

Option 1)

Option 2)

Option 3)

Option 4)

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Engineering
135 Views   |

If the equations have a common root, then a : b : c is :

• Option 1)

3 : 1 : 2

• Option 2)

1 : 2 : 3

• Option 3)

3 : 2 : 1

• Option 4)

1 : 3 : 2

As we have learned

Quadratic Expression Graph when a > 0 & D < 0 -

No Real and Equal root of

$f\left ( x \right )= ax^{2}+bx+c$

& $D= b^{2}-4ac$

- wherein

Condition for both roots common -

$\frac{a}{{a}'}=\frac{b}{{b}'}=\frac{c}{{c}'}$

- wherein

$ax^{2}+bx+c=0$ &

$a'x^{2}+b'x+c'=0$

are the 2 equations

For $x^2+2x+3=0$

Discriminant = 4-12 = -8 < 0

Both the roots are common as complex roots occur in conjugate $\therefore a:b:c= 1:2:3$

Option 1)

3 : 1 : 2

Option 2)

1 : 2 : 3

Option 3)

3 : 2 : 1

Option 4)

1 : 3 : 2

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Engineering
105 Views   |

If    is the adjoint of a 3 x 3  matrix A and = 4,then is equal to :

• Option 1)

0

• Option 2)

4

• Option 3)

11

• Option 4)

5

As we have learned

Property of adjoint of A -

$\left | adj A \right |=\left | A \right |^{n-1}$

- wherein

$adj A$ denotes adjoint of $A$ and  $\left |A \right |$  denotes determinant  of $A$ and $n$ is the order of the matrix

$|adj \; \; A| = |A|^{3-1}$

$\Rightarrow |adj \; \; A| =4^2=16$

$\Rightarrow 0-\alpha (-2)+3(-2)=16$

$\Rightarrow 2\alpha -6 = 16$

$\Rightarrow \alpha = 11$

Option 1)

0

Option 2)

4

Option 3)

11

Option 4)

5

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