Tag: maths

Questions Related to maths

If the ellipse $\displaystyle \frac{x^{2}}{4}+\frac{y^{2}}{b^{2}}=1$ meets the ellipse $\displaystyle \frac{x^{2}}{1}+\frac{y^{2}}{a^{2}}=1$ in four distinct points and $\displaystyle a^{2} = b^{2} -4b + 8$, then $b$ lies in

graphs to solve linear and non linear equations functions and their graphs curved graphs sets, relations and functions maths

  1. $(- \infty ,0)$

  2. $(- \infty ,2)$

  3. $(2,\infty)$

  4. $[2, \infty)$

Show answer
Correct Option: B
Explanation:
Given,
$\displaystyle \dfrac{x^{2}}{1}+\dfrac{y^{2}}{a^{2}}=1$   -- (i)

$\displaystyle \dfrac{x^{2}}{4}+\dfrac{y^{2}}{b^{2}}=1$   -- (ii) 
are the two equations of the ellipses

Eliminating $y^2$ from both the equations we get, 
$ x^2 \left( \dfrac{b^2-4a^2}{4a^2b^2} \right) =  \dfrac{b^2-a^2}{a^2b^2} $

$\dfrac{x^2}{4}  = \dfrac{b^2-a^2}{b^2-4a^2} $

Substituting the value of $a^2$ we get, 
$ \dfrac{x^2}{4} = \dfrac{4b-8}{-3b^2+16b-32} $

The denominator is always negative as the discriminant of the expression is negative and the coefficient of $b^2$ is also negative. 

Hence, $4b-8 < 0$
$\Rightarrow b <2 $

Eliminating $x^2$ from the two equations we get, 

$y^2 \left ( \dfrac{4a^2-b^2}{a^2b^2} \right) = 3 $

$ \dfrac{y^2}{3} = \dfrac {a^2b^2} { 4a^2 - b^2} $

Hence, $4a^2 -b^2 >0 $
$\Rightarrow 3b^2 -16b +32 > 0 $. 

The discriminant of the expression is less than 0 and the coefficient of $b^2$ is greater than 0. Hence, the inequality holds true for all values of $b$. 

Hence the common set of the values of $b$ is $ (-\infty,  2) $. 
Hence, option B is correct

Let $A(z _a), B(z _b), C(z _c)$ are three non-collinear points where $z _a=i, z _b=\dfrac{1}{2}+2i, z _c=1+4i$ and a curve is $z=z _a\cos^4t+2z _b\cos^2t \sin^2t+z _c\sin^4t(t\in R)$
A line bisecting AB and parallel to AC intersects the given curve at

graphs to solve linear and non linear equations functions and their graphs curved graphs sets, relations and functions maths

  1. Two distinct points

  2. Two co-incident points

  3. Only one point

  4. No point

Show answer
Correct Option: C
Explanation:

${ z } _{ a }=i\Rightarrow A\left( 0,1 \right) $
${ z } _{ b }=\cfrac { 1 }{ 2 } +2i\Rightarrow B\left( \cfrac { 1 }{ 2 } ,2 \right) $
${ z } _{ c }=1+4i\Rightarrow C\left( 1,4 \right) $
Let D be the midpoint of AB
$D=\left[ \cfrac { 0+\cfrac { 1 }{ 2 }  }{ 2 } ,\cfrac { 1+2 }{ 2 }  \right] $
$D\left[ \cfrac { 1 }{ 4 } ,\cfrac { 3 }{ 2 }  \right] $
Line bisecting AB at D is parallel to AC
$\therefore $ Slope of AC $=\cfrac { 4-1 }{ 1-0 } =3$
Equation of line bisecting AB is $y-\cfrac { 3 }{ 2 } =3\left( x-\cfrac { 1 }{ 4 }  \right) $

$\Rightarrow \cfrac { 2y-3 }{ 2 } =\cfrac { 12x-3 }{ 4 } $
$\Rightarrow 4y-6=12x-3$
$\Rightarrow 12x-4y+3=0$
$\Rightarrow y=\cfrac { 12x+3 }{ 4 } $
Equation of curve is $y={ \left( x+1 \right)  }^{ 2 }$
$\cfrac { 12x+3 }{ 4 } ={ x }^{ 2 }+2x+1$
$4{ x }^{ 2 }-4x+1=0$
${ \left( 2x-1 \right)  }^{ 2 }=0$
$x=\cfrac { 1 }{ 2 } ,\quad y=\cfrac { 9 }{ 4 } $
$\left( \cfrac { 1 }{ 2 } ,\cfrac { 9 }{ 4 }  \right) \Rightarrow $ given line and curve intersect only at one point

If the line $y=x\sqrt{3}$ cuts the curve $x^{3}+y^{3}+3xy+5x^{2}+3y^{2}+4x+5y-1=0$ at the points $A, B$ and $C$,then $OA. OB. OC$ is equal to (where '$O$' is origin)

graphs to solve linear and non linear equations functions and their graphs curved graphs sets, relations and functions maths

  1. $\dfrac{4}{13}\left ( 3\sqrt{3}-1 \right )$

  2. $\left ( 3\sqrt{3}-1 \right )$

  3. $\dfrac{1}{\sqrt{3}}\left ( 2+7\sqrt{3} \right )$

  4. $\dfrac{4}{13}\left ( 3\sqrt{3}+1 \right )$

Show answer
Correct Option: A
Explanation:

Coordinates of a point on the line $y=x\sqrt{3}$ at a distance r from origin 
is $\left ( r\cos \theta ,r\sin \theta  \right )$
$\therefore \tan \theta =\sqrt{3}$
$\therefore \left ( \dfrac{r}{2},\dfrac{r\sqrt{3}}{2} \right )$ lies on the given curve
$\Rightarrow  \displaystyle \frac{r^{3}}{8}+\frac{r^{3}.3\sqrt{3}}{8}+3.\frac{r}{2}.\frac{r\sqrt{3}}{2}+5.\frac{r^{2}}{4}+3.\frac{r^{2}.3}{4}+4.\frac{r}{2}+5.\frac{r\sqrt{3}}{2}-1=0$

$\Rightarrow \left (\displaystyle  \frac{1+3\sqrt{3}}{8} \right )r^{3}+\dfrac{r^{2}}{4}\left ( 3\sqrt{3}+14 \right )+\dfrac{r}{2}\left ( 5\sqrt{3}+4 \right )-1=0$

$\Rightarrow r _{1}.r _{2}.r _{3}=\dfrac{8}{3\sqrt{3}+1}$

$=\dfrac{8}{27-1}\times \left ( 3\sqrt{3}-1 \right )$

$=\dfrac{4}{13} \left ( 3\sqrt{3}-1 \right )$

The pair of lines $6{ x }^{ 2 }+7xy+\lambda { y }^{ 2 }=0\left( \lambda \neq -6 \right) $ forms a right angled triangle with $x+3y+4=0$ then $\lambda=$

graphs to solve linear and non linear equations functions and their graphs curved graphs sets, relations and functions maths

  1. $3$

  2. $-3$

  3. $1/3$

  4. $-1/3$

Show answer
Correct Option: A
Explanation:

Given line is $L: x+3y+4=0$


$\implies  y=-\dfrac{1}{3}(x+4)$

Slope of this line is $m=\dfrac{-1}{3}$

Now, $6x^2+7xy+\lambda y^2=0$ $(\lambda\neq 6)$

$x^2+\dfrac{7}{6}xy+\dfrac{\lambda}{6}y^2=0$

$\implies (x+ay)(x+by)=0$

$\implies x+ay=0$ and $x+by=0$ are the two equations with 

$a+b=\dfrac{7}{6}$    and $ab=\dfrac{\lambda}{6}$

Slope of these lines are $m _1=\dfrac{-1}{a}$ and $m _2=\dfrac{-1}{b}$

Now, $m _1m _2=\dfrac{1}{ab}=\dfrac{\lambda}{6}\neq -1$  since $\lambda\neq -6$

Hence the lines $x+ay=0$ and $x+by=0$ are not prependicular.

From these two only one is normal to $L$.

Let $x+ay$ is normal to $L$.

$\implies m _1m=-1$

$\implies \dfrac{1}{3a}=-1$

$\implies a=\dfrac{-1}{3}$

Now, $a+b=\dfrac{7}{6}\implies b=\dfrac{7}{6}-\dfrac{-1}{3}$

$\implies b=\dfrac{3}{2}$

Now, $ab=\dfrac{\lambda}{6}$

$\implies \lambda=6ab=6\times \dfrac{-1}{3}\times \dfrac{3}{2}$

$\implies \lambda=-3$

Answer-(B)

Let $y=f(x)$ and $y=g(x)$ be the pair of curves such that
(i) The tangents at point with equal abscissae intersect on y-axis.
(ii) The normal drawn at points with equal abscissae intersect on x-axis and
(iii) curve f(x) passes through $(1, 1)$ and $g(x)$ passes through $(2, 3)$ then: The curve g(x) is given by.

graphs to solve linear and non linear equations functions and their graphs curved graphs sets, relations and functions maths

  1. $x-\displaystyle\frac{1}{x}$

  2. $x+\displaystyle\frac{2}{x}$

  3. $x^2-\displaystyle\frac{1}{x^2}$

  4. $(x+\displaystyle\frac{1}{x})$$(x+\displaystyle\frac{2}{x})$

Show answer
Correct Option: A
Explanation:
$y=f(x)$ and $y=g(x)$
(i)
$\dfrac{dy}{dy}=c\Rightarrow c _{1}=1$
for second curve 
$\dfrac{dy}{dy}=c _{2}\Rightarrow c _{2}=1$

(ii)
$\dfrac{dy}{dx}=d _{1}---(1)$
for second curve 
$\dfrac{dy}{dx}=d _{2}----(2)$

(iii)
From eq (1) and (2)
$d _{1}=2=d _{2}$

$\int _{1}^{2} (g(x)-f(x))d(x)$

$\int _{1}^{2} (4-2)d(x)$

$2\int _{1}^{2} d(x)$

$2(2-1)=2$

SO $g(x)=x-\dfrac{1}{x}$

What is the value of $7.854 \times 10$ ?

maths decimal fraction multiplying decimals multiplication of decimals multiplication of division of decimal fraction by whole number and decimal fraction

  1. $785.4$

  2. $78.54$

  3. $7854$

  4. $78540$

Show answer
Correct Option: B
Explanation:

$7.854$

  $\times 10$
$\overline{78.540}$

Hence, $7.854 \times 10 = 78.54$

Solve the following:

$0.05 \times 0.09 \times 5 $ 

maths decimal fraction multiplying decimals multiplication of decimals multiplication of division of decimal fraction by whole number and decimal fraction

  1. $0.025$

  2. $0.225$

  3. $0.005$

  4. $0.0225$

Show answer
Correct Option: D
Explanation:

$0.05 \times 0.09 \times 5$

$=0.05 \times 0.45$
$=0.0225$

$58.326 \times 463.9 \times 0.0081$ is same as 

maths decimal fraction multiplying decimals multiplication of decimals multiplication of division of decimal fraction by whole number and decimal fraction

  1. $5.8326 \times 4.639 \times 8.1$

  2. $5.8326 \times 4.639 \times 0.81$

  3. $58326 \times 4639 \times 0.0000081$

  4. None of these

Show answer
Correct Option: A
Explanation:

In option A,

$58.326\div10 = 5.8326$
$463.9\div100 = 4.639$
$0.0081\times1000 = 8.1$

Multiplication by $1000$ and Division by $10\times100 = 1000$ gives the same result.

Answer : $5.8326 \times 4.639 \times 8.1$

$5.\overline { 07 } \quad \times 10=y$, then the value of $y$ is

maths decimal fraction multiplying decimals multiplication of decimals multiplication of division of decimal fraction by whole number and decimal fraction

  1. $50.\overline { 07 } $

  2. $50.0\overline { 7 } $

  3. $50.7\overline { 07 } $

  4. $50.70$

Show answer
Correct Option: C
Explanation:

$5.\overline { 07 } \quad \times 10=50.7070707=50.7\overline { 07 } $

If we write November 8, 1988, as 8.11.88, we see $8\times11=88$. How many such days are in 1972?

maths decimal fraction multiplying decimals multiplication of decimals multiplication of division of decimal fraction by whole number and decimal fraction

  1. 6

  2. 4

  3. 3

  4. 2

Show answer
Correct Option: A
Explanation:

Multiples of 72 = 2,4,6,8,9,12,18,24,36,72
But in a year we have only 12 months and in a month we have only 30 days
So We have
$24.3.72 = 24\times 3 = 72$
$18.4.72 = 18\times 4 = 72$
$12.6.72= 12\times6 = 72$
$9.8.72 = 9\times8 = 72$
$8.9.72 = 8\times 9 = 72$
$6.12.72 = 6\times 12 = 72$