MCQ Questions for Class 12 Commerce Applied Mathematics Quantification And Numerical Applications Quiz 10

$x,y,z \in +R$ and

$x^2+y^2+z^2=27$ , then

$x^3+y^3+z^3$ has

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.

$0<\theta<\pi$ then the minimum value of

$\sin ^5\theta+\mathrm{cosec} ^5\theta$ is

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

$2$

$10-x >3$ , then

$x < 7$ .

0%
True
0%
False

Explanation

$10-x>3$

$x<7$

$TRUE$

$xyz=abc$ , then the least value of

$bcx+cay+abz$ is

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?

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

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:

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

$4(x - 1) \leq 8$

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

$2(x - 2) < 3$

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

$3x + 14 \geq 8$

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

$2(x + 7) \leq 9$

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.

$2x + 7 > 15$

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

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.

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

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

$\left ( \frac{1}{2} \right )^{log_3\left ( 2^{2a}-1 \right )} > \left ( \frac{1}{2} \right )^{log_3\left ( 2^{2a}+1 \right )}$

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

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.

$\left| x-1 \right| +\left| x-3 \right| \le 8$ , then the values of

$x$ lie in the interval

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

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)

$\Rightarrow 2x+2y=7x-7y$

$\Rightarrow 2x-7x=-7y-2y$

$\Rightarrow -5x==9y$

$\Rightarrow \dfrac{x}{y}=\dfrac{9}{5}$

$x:y=9:5$

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

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

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

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.

0%
1:2
0%
2:1
0%
2:3
0%
3:2

Explanation

Copper : Zinc

$= 5 : 2$

Copper 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

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

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?

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.
1553397_985337_ans_dbf23877875049ff98cfa348824e5dc0.jpg

The solution set of

$x$ for the inequations

$2x+3\ge 8$ and

$3x+1\le 12$ is

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)?

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

$\dfrac{2x}{2x^2+5x+2}>\dfrac{1}{x+1}$

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.

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

$B$

$\tan \theta =\dfrac{v}{u}=\dfrac{10}{5}=2$

$\Rightarrow \theta =tan^{-1}(2) N of E$

1039659_985311_ans_ef1c8de542464cfc9643d24a4d1c7dcf.PNG

CBSE Questions for Class 12 Commerce Applied Mathematics Quantification And Numerical Applications Quiz 10 - MCQExams.com

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Practice Class 12 Commerce Applied Mathematics Quiz Questions and Answers

CBSE Questions for Class 12 Commerce Applied Mathematics Quantification And Numerical Applications Quiz 10 - MCQExams.com

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Practice Class 12 Commerce Applied Mathematics Quiz Questions and Answers

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