MCQExams
0:0:1
CBSE
JEE
NTSE
NEET
Practice
Homework
×
CBSE Questions for Class 12 Commerce Applied Mathematics Quantification And Numerical Applications Quiz 10 - MCQExams.com
CBSE
Class 12 Commerce Applied Mathematics
Quantification And Numerical Applications
Quiz 10
If $$x,y,z \in +R$$ and $$x^2+y^2+z^2=27$$, then $$x^3+y^3+z^3$$ has
Report Question
0%
Minimum value of $$81$$
0%
Maximum value of $$81$$
0%
Maximum value of $$27$$
0%
Minimum value of $$27$$
Explanation
Given that $$x^2+y^2+z^2=27$$
We know that $$A.M.\geq G.M.$$
Therefore, $$\dfrac{x^2+y^2+z^2}{3}\geq \sqrt[3]{x^2y^2z^2}$$
$$\implies \dfrac{27}{3}\geq \sqrt[3]{x^2y^2z^2}$$
$$\implies 9\geq (xyz)^{\dfrac 23}$$
$$\implies (3^2)^{\dfrac 32}\geq xyz$$
$$\implies 3^3\geq xyz$$
$$\implies xyz\leq 27$$ ————-(1)
Similarly, $$\dfrac{x^3+y^3+z^3}{3}\geq \sqrt[3]{x^3y^3z^3}$$
$$\implies \dfrac{x^3+y^3+z^3}{3}\geq xyz$$
$$\implies {x^3+y^3+z^3}\geq 3xyz$$ ————-(2)
From (1) $$xyz$$ has a macimum value of 27. Substituting $$xyz=27$$ we get
$$\implies x^3+y^3+z^3\geq 3*27$$
$$\implies x^3+y^3+z^3\geq 81$$
Therefore, $$x^3+y^3+z^3$$ has minimum value of 81.
If $$0<\theta<\pi$$ then the minimum value of $$\sin ^5\theta+\mathrm{cosec} ^5\theta$$ is
Report Question
0%
$$0$$
0%
$$1$$
0%
$$2$$
0%
None of the above.
Explanation
we know that, $$A.M.\geq G.M.$$
$$\implies \dfrac{a+b}{2}\geq \sqrt{ab}$$
let $$a=sin^5\theta$$ and $$b=cosec^5\theta$$
therefore $$\dfrac{sin^5\theta+cosec^5\theta}{2}\geq \sqrt{sin^5\theta *cosec^5\theta}$$
$$\implies {sin^5\theta+cosec^5\theta}\geq2\sqrt{(sin\theta *cosec\theta)^5}$$
$$\implies {sin^5\theta+cosec^5\theta}\geq 2\sqrt{(sin\theta *cosec\theta)^5}$$
$$\implies {sin^5\theta+cosec^5\theta}\geq 2\sqrt{(sin\theta *(\dfrac{1}{sin\theta}))^5}$$
$$\implies {sin^5\theta+cosec^5\theta}\geq 2\sqrt{1^5}$$
$$\implies {sin^5\theta+cosec^5\theta}\geq 2$$
therefore the minimum value of
$$ {sin^5\theta+cosec^5\theta}$$ is $$ 2$$
If $$10-x >3$$, then $$x < 7$$.
Report Question
0%
True
0%
False
Explanation
$$10-x>3$$
$$ x<7$$
$$ TRUE$$
If $$xyz=abc$$, then the least value of $$bcx+cay+abz$$ is
Report Question
0%
$$3abc$$
0%
$$6abc$$
0%
$$abc$$
0%
$$4abc$$
Explanation
We know that $$A.M. \geq G.M.$$
$$\implies \dfrac{a+b+c}{3}\geq \sqrt[3]{abc}$$
Let $$a=bcx, b=acy, c=abz$$
Therefore,
$$\dfrac{bcx+acy+abz}{3}\geq \sqrt[3]{bcx\times acy\times abz}$$
$$\implies bcx+acy+abz\geq 3\times\sqrt[3]{xyz\times a^2b^2c^2}$$
$$\implies bcx+acy+abz\geq 3\times\sqrt[3]{(abc)^3}$$ Since, $$xyz=abc$$
$$\implies bcz+acy+abz\geq 3abc$$
Therefore the least value of $$bcx+acy+abz$$ is $$3abc$$
Roots of the equation $$f(x)=x^6-12x^5+bx^4+cx^3+dx^2+ex+64=0$$ are positive.
Which of the following has the greatest absolute value?
Report Question
0%
$$b$$
0%
$$c$$
0%
$$d$$
0%
$$e$$
Explanation
Given
$$ { x }^{ 6 }-12{ x }^{ 5 }+b{ x }^{ 5 }+c{ x }^{ 3 }+d{ x }^{ 2 }+ex+64=0$$
Since $$ { \alpha }_{ 1, }{ \alpha }_{ 2 },{ \alpha }_{ 3 },{ \alpha }_{ 4 },{ \alpha }_{ 5 },{ \alpha }_{ 6 }$$ are the roots of the equation
Let us consider the sum of the roots
$$ { \alpha }_{ 1 },{ \alpha }_{ 1 },+.......+{ \alpha }_{ 6 }=12$$ ..... $$(1)$$
The product of roots would be
$${ \alpha }_{ 1 },{ \alpha }_{ 2 },.......{ \alpha }_{ 6 }=64$$
..... $$(2)$$
If we apply arithmetic and geometric mean to equation roots
$$A.M >G.M$$
$$ { \alpha }_{ 1 }+{ \alpha }_{ 2 }+.......+{ \alpha }_{ 6 }\ge \quad ({\alpha}_{1},{ \alpha }_{ 2 }.......+{ \alpha }_{ 6 }{ ) }^{ { 1/6 }_{ }^{ } }$$
$$ \dfrac { 12 }{ 6 } \ge ({ \alpha }_{ 1 },{ \alpha }_{ 2 },.......{ \alpha ) }^{ 1/6 }$$
$${ 2 }^{ 6 }\ge \dfrac { 12 }{ 6 } \ge ({ \alpha }_{ 1 },{ \alpha }_{ 2 },.......{ { \alpha }_{ 6 }) }$$
$${ \alpha }_{ 1 },{ \alpha }_{ 2 },.......{ { \alpha }_{ 6 }) }\le 64$$
$$({ \alpha }_{ 1 },{ \alpha }_{ 2 },.......{ { \alpha }_{ 6 }) }=64$$
The roots product to equality if all the roots are equal
$$\Rightarrow ({ \alpha }_{ 1 },{ \alpha }_{ 2 },.......{ { \alpha }_{ 6 }) }=2$$
Hence the polynomial
$${ x }^{ 6 }-12{ x }^{ 5 }+b{ x }^{ 4 }+c{ x }^{ 3 }+d{ x }^{ 2 }+e{ x }+64=0$$
Since all the root are $$2$$
Hence the maximum of the $$d$$ is greatest absolute value to cancel out the polynomial to zero.
For $$0<x<\dfrac {\pi}2$$, $$(1+4\mathrm{cosec} x)(1+8\sec x)$$, is
Report Question
0%
$$\geq 81$$
0%
$$>81$$
0%
$$\geq 83$$
0%
$$>83$$
Explanation
$$(1+4\text{cosec }x)(1+8\sec x)=\left(1+\dfrac { 4 }{\sin x} \right)\left(1+\dfrac { 8 }{ \cos x } \right)\\ \dfrac { 2(\sin x+4)(\cos x+8) }{ 2\sin x\cos x } =\dfrac { \sin 2x+16\sin x+8\cos x+64 }{\sin 2x } $$
The minimum value of above equation occurs when denominator have maximum value and maximum value of denominator will be at $$x=\dfrac { \pi }{ 4 } $$
Putting this value in numerator, we get
$$1+\dfrac { 16 }{ \sqrt { 2 } } +\dfrac { 8 }{ \sqrt { 2 } } +64=81.970\approx 82$$
Hence, the answer is $$>81$$.
A motorboat covers a given distance in $$6$$ hours moving downstream on a river. It covers the same distance in $$10$$ hours moving upstream. The time it takes to cover the same distance in still water is:
Report Question
0%
$$6.5$$ hours
0%
$$8$$ hours
0%
$$9$$ hours
0%
$$7.5$$ hours
Explanation
Let velocity in still water be $$S$$
Velocity of water be $$W$$
distance to be covered be $$D$$
$$D=6(S+W)$$
$$D=10(S-W)$$
which gives $$S=4W$$
So, $$D=7.5hrs$$.
Solve $$4(x - 1) \leq 8$$
Report Question
0%
$$(-\infty,3)$$
0%
$$(-\infty,-3)$$
0%
$$(-\infty,2)$$
0%
None of these
Explanation
$$x-1\leq 2$$
$$x-1+2\leq 2+2$$
$$x\leq 3$$
$$x\epsilon \left ( -\infty ,3 \right )$$
Solve the following inequation
$$2(x - 2) < 3$$
Report Question
0%
$$x < 3.5$$
0%
$$x > 3.5$$
0%
$$x < 2.5$$
0%
None of these
Explanation
$$2(x-2) < 3 $$
$$\Rightarrow (x-2) < \dfrac32$$
$$\Rightarrow x < 2+\dfrac{3}{2}$$
$$\Rightarrow x < 3.5$$
Solve the following inequation
$$3x + 14 \geq 8$$
Report Question
0%
$$x \geq - 2$$
0%
$$x \leq - 2$$
0%
$$x \geq 2$$
0%
None of these
Explanation
$$3x+14\geq 8$$
$$3x\geq -6$$
$$x\geq -2$$
Solve the following inequation
$$2(x + 7) \leq 9$$
Report Question
0%
$$x \leq - 2.5$$
0%
$$x \leq 2.5$$
0%
$$x \geq - 2.5$$
0%
None of these
Explanation
$$2(x+7)\leq 9$$
$$x+7\leq 4.5$$
$$x \leq -2.5$$
Solve the following inequations.
$$2x + 7 > 15$$
Report Question
0%
$$x > 4$$
0%
$$x > 7$$
0%
$$x < 4$$
0%
None of these
Explanation
subtract 7 both sides
$$2x > 8 $$
Divide by 2
$$x > 4$$
The least integer satisfying $$\cfrac { 396 }{ 10 } -\cfrac { 19-x }{ 10 } <\cfrac { 376 }{ 10 } -\cfrac { 19-9x }{ 10 } $$ is
Report Question
0%
$$1$$
0%
$$2$$
0%
$$3$$
0%
$$4$$
0%
$$5$$
Explanation
In equation is
$$\cfrac { 396 }{ 10 } -\cfrac { 19-x }{ 10 } <\cfrac { 376 }{ 10 } -\cfrac { 19-9x }{ 10 } $$
$$\Rightarrow 396-(19-x)<376-(19-9x)\quad $$
$$\Rightarrow x>\cfrac { 20 }{ 8 } \Rightarrow x>2.5$$
Less integer is $$x=3$$
The set of points $$(x, y)$$ satisfying the inequalities $$x + y \leq 1, -x - y \leq 1$$ lie in the region bounded by the two straight lines passing through the respective pair of points.
Report Question
0%
$$\left \{(1, 0), (0, 1)\right \}$$ and $$\left \{(-1, 0), (0, -1)\right \}$$
0%
$$\left \{(1, 0), (1, 1)\right \}$$ and $$\left \{(-1, 0), (0, -1)\right \}$$
0%
$$\left \{(-1, 0), (0, -1)\right \}$$ and $$\left \{(1, 0), (-1, 1)\right \}$$
0%
$$\left \{(1, 0), (0, -1)\right \}$$ and $$\left \{(-1, 0), (0, 1)\right \}$$
0%
$$\left \{(1, 0), (1, 1)\right \}$$ and $$\left \{(-1, 0), (-1, -1)\right \}$$
Explanation
Given constraints are
$$x + y \leq 1$$
and $$-x - y\leq 1$$
Assume these inequalities as equality and draw these lines on graph paper, we get
and $$\left.\begin{matrix}x + y = 1\\ \ x + y = -1
\end{matrix}\right\}$$
So, pair of points $$\left \{(1, 0), (0, 1)\right \}$$ and $$\left \{(-1, 0), (0, -1)\right \}$$ are satisfying the inequalities
$$x + y \leq 1$$ and $$-x - y \leq 1$$.
A boat can go across a lake and return in time $$T_{0}$$ at a speed $$v$$. On a rough day there is a uniform current at speed $$v_{1}$$ to help the onward journey and impede the return journey. If the time taken to go across and return on the same day be $$T$$, then $$T/T_{0}$$ will be
Report Question
0%
$$\dfrac {1}{(1 - v_{1}^{2}/ v^{2})}$$
0%
$$\dfrac {1}{(1 + v_{1}^{2}/ v^{2})}$$
0%
$$(1 - v_{1}^{2} / v^{2})$$
0%
$$\left (1 + \dfrac {v_{1}^{2}}{v_{2}}\right )$$
Explanation
Let $$d$$ be the width of lake to be crossed, then
$$T_{0} = \dfrac {2d}{v}$$
and $$T = \dfrac {d}{(v + v_{1})} - \dfrac {d}{(v - v_{1})} = \dfrac {2dv}{v^{2} - v_{1}^{2}}$$
$$= \dfrac {2dv}{v^{2}(1 - v_{1}^{2}/ v^{2})} = \dfrac {2d}{v(1 - v_{1}^{2}/ v^{2})}$$
So, $$\dfrac {T}{T_{0}} = \dfrac {1}{(1 - v_{1}^{2}/ v^{2})}$$.
Solution of a for the inequality $$\left ( \frac{1}{2} \right )^{log_3\left ( 2^{2a}-1 \right )} > \left ( \frac{1}{2} \right )^{log_3\left ( 2^{2a}+1 \right )} $$
Report Question
0%
$$(1, 2)$$
0%
$$(0, 1)$$
0%
$$(-1, 1)$$
0%
$$(-1, 0)$$
Explanation
Since,$$\cfrac { 1 }{ 2 } <1$$
$$\log _{ 3 }{ ({ 2 }^{ 2a } } -1)<\log _{ 3 }{ ({ 2 }^{ 2a } } +1)$$
$$\log _{ 3 }{ \left[\dfrac{ { 2 }^{ 2a }-1}{{ 2 }^{ 2a }+1} \right] <0 } $$
$$\dfrac{{2 }^{ 2a -1}}{{ 2 }^{ 2a+1 }}<1$$
Now, $$({ 2 }^{ 2a }+1)>0$$
$${ 2 }^{ 2a }-1<{ 2 }^{ 2a }+1$$
which is true for each case.
Now for $$\log _{ 3 }{ ({ 2 }^{ 2a } } -1)$$ to exist
$${ 2 }^{ 2a }-1>0$$
$${ 2 }^{ 2a }>1$$
$$({ 2 }^{ a })^{ 2 }>1$$
$$ -1>{ 2 }^{ 2a }>1$$
$${ 2 }^{ 2a }>0$$
$$ a>0$$
The number of positive integral solutions $$x^2 + 9 < (x + 3)^2 < 8x + 25 $$, is
Report Question
0%
$$2$$
0%
$$3$$
0%
$$4$$
0%
$$5$$
Explanation
$${ x }^{ 2 }+9<{ (x+3 })^{ 2 }<8x+25\\ \Rightarrow { x }^{ 2 }+9<{ (x+3 })^{ 2 }<9\\ \Rightarrow { x }^{ 2 }+9<{ x }^{ 2 }+9+6x\\ \Rightarrow 6x>0\\ \Rightarrow x>0$$
Also $${ (x+3 })^{ 2 }<8x+25$$
$$\Rightarrow { x }^{ 2 }-2x-16<0\\ \Rightarrow 1-\sqrt { 17 } <x<1+\sqrt { 17 } $$
But $$x>0$$
$$\Rightarrow 0<x<1+\sqrt { 17 } \\ \Rightarrow x\in (0,1+\sqrt { 17 } )$$
So the integer values in our domain are $$1,2,3,4,5$$
So, option D is correct.
If $$\left| x-1 \right| +\left| x-3 \right| \le 8$$, then the values of $$x$$ lie in the interval
Report Question
0%
$$\left( -\infty ,-2 \right) $$
0%
$$\left[ -2,6 \right] $$
0%
$$(-3,7)$$
0%
$$\left( -2,\infty \right) $$
0%
$$[6,\infty )$$
Explanation
Given inequality is
$$\left| x-1 \right| +\left| x-3 \right| \le 8$$
Case I $$x< 1$$
$$\left| x-1 \right| +\left| x-3 \right| \le 8$$
$$\Rightarrow -(x-1)-(x-3)\le 8$$
$$\Rightarrow -2x+4\le 8$$
$$\Rightarrow -2x\le 4\Rightarrow x\ge -2$$
Interval $$[-2,1)$$
Case II $$1\le x<3$$
$$\Rightarrow x-1-x+3\le 8\Rightarrow 2\le 8$$
Interval $$[1,3]$$
Case III $$x\ge 3$$
$$\Rightarrow x-1+x-3\le 8\Rightarrow x\le 6\quad $$
Interval $$[3,6]$$
From all the interval
values of $$x$$ lie in the interval $$[-2,6]$$
Hence (b) is the correct option
If the ratio of the sum and the difference of two numbers is 7 : 2, then the ratio of these two numbers is
Report Question
0%
7 : 5
0%
9 : 5
0%
9 : 7
0%
7 : 4
Explanation
Let $$x$$ and $$y$$ be the numbers then
$$\dfrac{x+y}{x-y}=\dfrac{7}{2}$$(given)
$$\Rightarrow 2x+2y=7x-7y$$
$$\Rightarrow 2x-7x=-7y-2y$$
$$\Rightarrow -5x==9y$$
$$\Rightarrow \dfrac{x}{y}=\dfrac{9}{5}$$
or $$x:y=9:5$$
The house on one side of a road are numbered using consecutive even numbers. The sum of the numbers of all the house in that row is 170.If there are least 6 house in that row and a is the number of the sixth house,then
Report Question
0%
$$2\le a \le 6$$
0%
$$8 \le a \le 12$$
0%
$$14 \le a \le 20$$
0%
$$22 \le a \le 30$$
Explanation
Let house no are $$\alpha ,\alpha ,+2,\alpha +4,\alpha +6,\alpha +8,\alpha +10,........\alpha + =a \Rightarrow =\alpha = a-10$$
House no.will be (+)
$$\Rightarrow \alpha =a-10>0$$
$$\Rightarrow \alpha >10$$
$$ \Rightarrow \alpha \ge 12$$ as a is each too
Now $$s_n =\dfrac {n}{2} [2\alpha +(n-1)d]$$
$$170=\dfrac{n}{2}[2\alpha +(n-1)(2)]$$
$$=n(\alpha +(n-1))$$
$$=n(a -10+n-1)$$
$$=n(a-11+n)$$
$$\Rightarrow n^2 =n(a-11)-170=0$$
$$n=\dfrac {(11-a) \pm {\sqrt {(a-11)^2 + 680}}}{2} \ge 6$$
$$\Rightarrow a \le \dfrac {800}{24}$$
From (2) and (4) $$ Rightarrow 12 \le a \le 32$$
now checking through (3) for a =12,14,......
we have a =18, n=10 and $$S_n=170$$
Hence option C is the answer
The corner points of the feasible region determined by the following system of linear inequalities :
$$2x + y \leq 10, x + 3y \leq 15, x, y \geq 0$$ are $$(0, 0), (5, 0), (3, 4)$$ and $$(0, 5)$$. Let $$Z = px + qy$$, where $$p, q > 0$$. Condition on p and q so that the maximum of $$Z$$ occurs at both $$(3, 4)$$ and $$(0, 5)$$ is
Report Question
0%
$$p = q$$
0%
$$p = 2q$$
0%
$$p = 3q$$
0%
$$q = 3p$$
Explanation
Since, maximum of $$Z=px+qy$$ occurs at both points $$(3,4)$$ and $$(0,5)$$
Hence value of $$Z=px+qy$$ must be equal at both the points.
$$\therefore p\times 3+q\times 4=p\times 0+q\times 5$$
$$\Rightarrow 3p+4q=5q$$
$$\Rightarrow 3p=q$$
So, $$q=3p$$ is the correct answer.
Choose the correct answer from the alternatives given :
If $$x:v:: 1:3 :5$$ then value of $$\dfrac{\sqrt {x^2 \, + \, 7y^2 \, + \, 9z^2}}{x}$$ is
Report Question
0%
$$7$$
0%
$$17$$
0%
$$13$$
0%
$$1$$
Explanation
Let $$x, y\ and\ z\ $$be $$k, 3k\ and\ 5k$$ respectively.
Putting the value in the equation,
$$\Rightarrow \dfrac {\sqrt{k^2 + 7 \times (3k)^2 + 9(5k)^2}}{k}$$
$$\Rightarrow \dfrac {\sqrt{k^2 + 63k^2 + 225k^2}}{k}$$
$$\Rightarrow \dfrac {\sqrt{289k^2}}{k}$$
$$\dfrac {17k}{k}$$
$$= 17$$
Choose the correct answer from the alternatives given:
The ratio of copper and zinc in the mixture is 5 :If 1.25 kg of zinc is added to 17.5 kg of mixture. Find the ratio of copper and Zinc in the new mixture.
Report Question
0%
1:2
0%
2:1
0%
2:3
0%
3:2
Explanation
Copper : Zinc
$$= 5 : 2$$
C
opper in alloy =
$$\displaystyle \frac{17.5}{7} \, \times \, 5 = 12.5\ kg$$
Zinc in alloy =
$$\displaystyle \frac{17.5}{7} \, \times \, 5 = 5 \ kg$$
After mixing zinc = $$1 .25 + 5 = 6.25$$ kg
New ratio of $$\displaystyle \frac{Copper}{Ziinc} \, = \, \frac {12.5}{6.25} = 2 : 1.$$
Three pipes A, B,C can fill a cistern in $$20$$ minutes, $$15$$ minutes and $$12$$ minutes respectively. The time in minutes that three pipes together will take to fill the cistern, is
Report Question
0%
$$5$$ minutes
0%
$$10$$ minutes
0%
$$12$$ minutes
0%
$$15 \frac{2}{3}$$ minutes
Explanation
Let the capacity of cistern be $$60$$ units $$[L.C.M\ of\ 20, 15\ and\ 12]$$
A does = $$\dfrac{60}{20} = 3\ units\ / \ minute$$
B does =
$$\dfrac{60}{12} = 5 \ units\ /\ minute$$
(A + B + C)'s per minute work = $$(3 + 4 +5) units\ /\ minute = 12 units\ /\ minute$$
Together they will fill the tank is $$\dfrac{60}{12} = 5$$ minutes
The solution of the inequation $$\dfrac{{\left(x-2\right)}^{10000}{\left(x+1\right)}^{253}{\left(x-\frac{1}{2}\right)}^{971}{\left(x+8\right)}^{4}}{{x}^{500}{\left(x-3\right)}^{75}{\left(x+2\right)}^{93}}\ge0$$ is
Report Question
0%
$$\left(-\infty,-2\right)\cup\left[-1,0\right)\cup\left(0,\frac{1}{2}\right]\cup\left(3,\infty\right)$$
0%
$$\left(-\infty,-2\right]\cup\left[-1,0\right)\cup\left[0,\frac{1}{2}\right]\cup\left[3,\infty\right)$$
0%
$$\left(-\infty,-2\right]\cup\left[-1,0\right]\cup\left(0,\frac{1}{2}\right]\cup\left[3,\infty\right)$$
0%
$$\left(-\infty,-2\right)\cup\left(-1,0\right)\cup\left(0,\frac{1}{2}\right)\cup\left(3,\infty\right)$$
Explanation
$$\dfrac{{\left(x-2\right)}^{10000}{\left(x+1\right)}^{253}{\left(x-\frac{1}{2}\right)}^{971}{\left(x+8\right)}^{4}}{{x}^{500}{\left(x-3\right)}^{75}{\left(x+2\right)}^{93}}\ge0$$
The critical points are $$\left(-8\right),\left(-2\right),\left(-1\right),0,\dfrac{1}{2},2,3$$
$$\left[\because x\neq-2,0,3\right]$$
$$\therefore,x\in\left(-\infty,-2\right]\cup\left[-1,0\right)\cup\left(0,\dfrac{1}{2}\right]\cup\left(3,\infty\right)$$
Thus,$$\left(-\infty,-2\right)\cup\left[-1,0\right)\cup\left(0,\dfrac{1}{2}\right]\cup\left(3,\infty\right)$$
A river flows with a speed more than the maximum speed with which a person can swim in still water. He intends to cross the river by the shortest possible path (i.e., he wants to reach the point on the opposite bank which directly opposite to the starting point). Which of the following is correct?
Report Question
0%
He should start normal to the river bank
0%
He should start in such a way that he moves normal to the bank, relative to the bank
0%
He should start in a particular (calculated) direction making an obtuse angle with the direction of water current
0%
The man cannot cross the river in that way
Explanation
We know that to cross the river by the shortest path,
$$\sin \alpha = \dfrac {u}{v}$$
But $$u > v \Rightarrow \sin \alpha > 1$$, which is not possible.
The solution set of $$x$$ for the inequations $$2x+3\ge 8$$ and $$3x+1\le 12$$ is
Report Question
0%
$$\dfrac{5}{2}< x\le \dfrac{11}{3}$$
0%
$$\dfrac{5}{2}< x<\dfrac{11}{3}$$
0%
$$\dfrac{5}{2}\le x\le \dfrac{11}{3}$$
0%
$$\dfrac{5}{2}\ge x\ge \dfrac{11}{3}$$
A boat os mass 40kg is at rest. A dog of mass 4kg moves in the boat with a velocity of 10m/s. What is the velocity of boat(nearly)?
Report Question
0%
4 m/s
0%
-1 m/s
0%
8 m/s
0%
7 m/s
Explanation
Initial momentum $$=0$$
so, final momentum $$=0$$
$$\begin{array}{l} 40\times v+4\times 10=0 \\ v=-1\, m/s \end{array}$$
Hence,Option $$B$$ is correct.
Find the set of all $$x$$ for which$$\dfrac{2x}{2x^2+5x+2}>\dfrac{1}{x+1}$$
Report Question
0%
$$x\epsilon(-2,-1)\cup(\dfrac{-2}{3},\dfrac{-1}{2})$$
0%
$$-2\geq x\geq-1$$
0%
$$-2\geq{x}<-1 $$
0%
$$-2<{x}\leq-1 $$
Explanation
$$\cfrac { 2x }{ { x }^{ 2 }+5x+2 } >\cfrac { 1 }{ x+1 } $$
$$ \cfrac { 2x }{ { 2x }^{ 2 }+4x+x+2 } >\cfrac { 1 }{ x+1 } $$
$$ \cfrac { 2x }{ (2x+1)(x+2) } -\cfrac { 1 }{ x+1 } >0$$
$$ \cfrac { { 2x }^{ 2 }+2x-{ 2x }^{ 2 }-5x-2 }{ (2x+1)(x+2)(x+1) } >0$$
$$ \cfrac { -3x-2 }{ (2x+1)(x+2)(x+1) } >0$$
$$ \cfrac { 3x+2 }{ (2x+1)(x+2)(x+1) } <0$$
$$ x\quad \varepsilon (-2,-1)\cup (\cfrac { -2 }{ 3 } ,\cfrac { -1 }{ 2 } )$$
A river is flowing from west to east at a speed of $$5\ m/min$$. A man on the south bank of the river, capable of swimming at $$10\ m/min$$ in still water, wants to swim across the river in the shortest time. Finally he will move in a direction.
Report Question
0%
$$\tan^{-1} (2) E$$ of $$N$$
0%
$$tan^{-2} (2) N$$ of $$E$$
0%
$$30^{\circ}E$$ of $$N$$
0%
$$60^{\circ}E$$ of $$N$$
Explanation
Finally, he will swim along $$B$$ $$\tan \theta =\dfrac{v}{u}=\dfrac{10}{5}=2$$
$$\Rightarrow \theta =tan^{-1}(2) N of E$$
0:0:1
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
27
28
29
30
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
27
28
29
30
0
Answered
0
Not Answered
0
Not Visited
Correct : 0
Incorrect : 0
Report Question
×
What's an issue?
Question is wrong
Answer is wrong
Other Reason
Want to elaborate a bit more? (optional)
Practice Class 12 Commerce Applied Mathematics Quiz Questions and Answers
<
>
Support mcqexams.com by disabling your adblocker.
×
Please disable the adBlock and continue.
Thank you.
Reload page
