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Engineering
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Let f_{k}\left ( x \right )= \frac{1}{k}\left ( \sin ^{k}x+\cos ^{k}x \right )where x\epsilon R\: \: and\: \: k\geqslant 1      Then f_{4}(x)-f_{6}(x)  equals :

  • Option 1)

    \frac{1}{4}

  • Option 2)

    \frac{1}{12}

  • Option 3)

    \frac{1}{6}

  • Option 4)

    \frac{1}{3}

 

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)

    \frac{121}{10}

  • Option 4)

    \frac{441}{100}

 

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)

\frac{121}{10}

Option 4)

\frac{441}{100}

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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)

    \left ( 14,\; \frac{272}{3} \right )

  • Option 2)

    \left ( 16,\; \frac{272}{3} \right )

  • Option 3)

    \left ( 16,\; \frac{251}{3} \right )

  • Option 4)

    \left ( 14,\; \frac{251}{3} \right )

 
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)

    \frac{7}{9}\left ( 99+10^{-20} \right )

  • Option 2)

    \frac{7}{81}\left ( 179-10^{-20} \right )

  • Option 3)

    \frac{7}{9}\left ( 99-10^{-20} \right )

  • Option 4)

    \frac{7}{81}\left ( 179+10^{-20} \right )

 

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)

\frac{7}{9}\left ( 99+10^{-20} \right )

Option 2)

\frac{7}{81}\left ( 179-10^{-20} \right )

Option 3)

\frac{7}{9}\left ( 99-10^{-20} \right )

Option 4)

\frac{7}{81}\left ( 179+10^{-20} \right )

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Engineering
134 Views   |  

The graph of the function  y=f\left ( x \right )  is symmetrical about the line 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

Symmetric about Y - axis

 

 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   f(x)+2f\left ( \frac{1}{x} \right )=3x,x\neq 0,  and  S=\left \{ x\, \epsilon \, R : f(x)=f(-x)\right \};then S:

  • 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 k for which the equation, 2x^2+3x+k = 0  has two distinct real roots in \left [ 0,1 \right ]

 

 

  • 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 \alpha and \beta be the roots of equation px^{2}+qx+r=0,p\neq 0.\; if\; p,q,r are in A.P. and \frac{1}{\alpha }+\frac{1}{\beta }=4,   then the value of \left | \alpha -\beta \right | is ?

  • Option 1)

    \frac{\sqrt{34}}{9}

  • Option 2)

    \frac{2\sqrt{13}}{9}

  • Option 3)

    \frac{\sqrt{61}}{9}

  • Option 4)

    \frac{2\sqrt{17}}{9}

 

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)

\frac{\sqrt{34}}{9}

Option 2)

\frac{2\sqrt{13}}{9}

Option 3)

\frac{\sqrt{61}}{9}

Option 4)

\frac{2\sqrt{17}}{9}

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Engineering
133 Views   |  

If z is a complex number of unit modulus and argument \theta ,then arg  (\frac{1+z}{1+\bar{z}})  equals:

  • Option 1)

    \pi - \theta

  • Option 2)

    -\theta

  • Option 3)

    \dfrac{\pi}{2}-\theta

  • Option 4)

    \theta

 

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)

\pi - \theta

Option 2)

-\theta

Option 3)

\dfrac{\pi}{2}-\theta

Option 4)

\theta

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Engineering
135 Views   |  

If the equations x^{2}+2x+3=0\; and\; ax^{2}+bx+c=0,a,b,c\; \epsilon R, 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 \left | A \right | = 4,then \alpha 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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