Tag: maths

Questions Related to maths

Number obtained by incrementing an integral part of $27.25$ by $1$ is?

  1. $35.26$

  2. $28.25$

  3. $26.25$

  4. $16.25$


Correct Option: B
Explanation:

Integral part of a decimal number is the part to the left of the decimal separator. 


$\Rightarrow$ Integral part $=27$
By incrementing integral part by $1$ we get $28$

$\therefore$ New number obtained $= 28.25$

The Number obtained by interchanging the integral and fractional part of $50.23$ is?

  1. $23.05$

  2. $23.5$

  3. $32.05$

  4. $23.4$


Correct Option: B
Explanation:

Integral part of a decimal number is the part to the left of the decimal separator. The part from the decimal separator to the right is the fractional part.


$\Rightarrow$ Integral part $=50$ Fractional part $=0.23$

$\therefore$ New number obtained by interchanging $= 23.50 = 23.5$

The number obtained on interchanging the integral and fractional part of $26.081$ is

  1. $18.026$

  2. $260.81$

  3. $81.26$

  4. $81.62$


Correct Option: C
Explanation:

Integral part of a decimal number is the part to the left of the decimal separator. The part from the decimal separator to the right is the fractional part.


$\Rightarrow$ Integral part $= 26$ Fractional part $=0.081$

New number obtained $= 81.26$

The point $\left( \sin { \theta  } ,\cos { \theta  }  \right) ,\theta $ being any real number, lie inside the circle ${ x }^{ 2 }+{ y }^{ 2 }-2x-2y+\lambda =0$, if

  1. $\lambda <1+2\sqrt { 2 } $

  2. $\lambda >2\sqrt { 2 } -1$

  3. $\lambda <-1-2\sqrt { 2 } $

  4. $\lambda >1+2\sqrt { 2 } $


Correct Option: C
Explanation:
${ x }^{ 2 }+{ y }^{ 2 }-2x-2y+\lambda =0$
Radius of circle$=\sqrt { 1+1-\lambda  } $
$=\sqrt { 2-\lambda  } $
Maximum distance of $\left( \sin { \theta  } ,\cos { \theta  }  \right) $
From center of above circle is when $\theta =\cfrac { 5\pi  }{ 4 } $
Thus distance$=\sqrt { { \left( 1+\cfrac { 1 }{ \sqrt { 2 }  }  \right)  }^{ 2 }+{ \left( 1+\cfrac { 1 }{ \sqrt { 2 }  }  \right)  }^{ 2 } } $\
$=\left( \sqrt { 2 } +1 \right) $
$\therefore \sqrt { 2-\lambda  } >\sqrt { 2+1 } $ [For all points to lie in center]
$\therefore 2-\lambda >2+1+2\sqrt { 2 } $
$\lambda <2-3-2\sqrt { 2 } $
$\lambda <-1-2\sqrt { 2 } $

Number of solutions of the equation $[2x]-3{2x}=1$ is?

(where $[\cdot]$ and ${\cdot }$ denote greatest integer an fractional part function respectively).

  1. $1$

  2. $2$

  3. $3$

  4. $0$


Correct Option: C

Integral part of $05.89$ is?

  1. $5$

  2. $89$

  3. $2$

  4. $3$


Correct Option: A
Explanation:

The integer part, or integral part of a decimal number is the part to the left of the decimal separator. 

So. Integral Part of given number 05.89 = 5.89 is 5

If a and b are any two such real numbers that ab $ = 0 $ , then

  1. $a = 0, b \leq 0$

  2. $b = 0, a \leq 0$

  3. a = 0 or b = 0 or both

  4. $a = b$ and $b = 0$


Correct Option: C
Explanation:

if both number are real the either a or b or both should be zero.
then only ab will be 0.
if any real number is multiplied by 0 then result will be zero.
So, answer is
C
 
a = 0 or b = 0 or both

If $f(x)-2f(1-x) = x^2+2$, then what is $f(x)$?

  1. $f(x)=-x^2+\dfrac{4}{3}x-\dfrac{3}{8}$

  2. $f(x)=−x^2+\dfrac{4}{3}x−\dfrac{8}{3}$

  3. $f(x)=−x^2+\dfrac{8}{3}x−\dfrac{4}{3}$

  4. $f(x)=−x^2+\dfrac{3}{8}x−\dfrac{3}{4}$


Correct Option: B
Explanation:
$f\left(x\right)-2f\left(1-x\right)={x}^{2}+2$      .......$(1)$

Setting $x=1-x$ then we get

$f\left(1-x\right)-2f\left(1-1+x\right)={\left(1-x\right)}^{2}+2$ 

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

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

Adding $(1)$ and $(2)$ we get

$-3f\left(x\right)=3{x}^{2}-4x+8$ 

$\therefore f\left(x\right)=-{x}^{2}+\dfrac{4}{3}x-\dfrac{8}{3}$ 

If $A\subset B$, then $n[P(A)]$ ______ $n[P(B)]$

  1. $=$

  2. $<$

  3. $\leq $

  4. $>$


Correct Option: B
Explanation:

Assume $ A \subset B$ is true. 

Then, every element of $A$ i.e. $a _1,a _2, ... , a _n$ in A are also in B.

So, number of elements in $B$ will always be greater than no. of elements in $A$

And $P(A)$ will contain less number of subsets than $P(B)$

Hence, $n[P(A)] <  n[P(B)]$

If $A $and $B$ are not disjoint, then $\displaystyle n\left( A \cup  B \right) $ is equal to

  1. $\displaystyle n\left( A \right) +n\left( B \right) $

  2. $\displaystyle n\left( A \right) +n\left( B \right) -n\left( A \cap B \right) $

  3. $\displaystyle n\left( A \right) +n\left( B \right) +n\left( A \cap B \right) $

  4. $\displaystyle n\left( A \right) .n\left( B \right) $


Correct Option: B
Explanation:

$\displaystyle n\left( A\quad \cup \quad B \right) =n\left( A \right) +n\left( B \right) -n\left( A\quad \cap \quad B \right) $