Conclude from the following: $$t = (2y \times 10^3)+(3y \times 10^3)$$ $$y > 0$$ A: $$t$$ B: $$5y\times 10^3$$

The quantity A is greater than B.

The quantity B is greater than A.

The relationship cannot be determined from the information given.

MCQ Questions for Class 12 Commerce Applied Mathematics Quantification And Numerical Applications Quiz 14

$x : y = 5 : 2$ and

$y : z = 3 : 2$ , what is the ratio of

$x : z$ ?

0%
$3 : 1$
0%
$3 : 5$
0%
$5 : 3$
0%
$8 : 4$
0%
$15 : 4$

Explanation

We have,

$x : y = 5 : 2$ and

$y : z = 3 : 2$ ,

$\Rightarrow \dfrac{x}{y}=\dfrac{5}{2}$ and

$\dfrac{y}{z}=\dfrac{3}{2}$

Now multiply both

$\Rightarrow \dfrac{x}{y}\cdot \dfrac{y}{z}=\dfrac{5}{2}\cdot \dfrac{3}{2}$

$\Rightarrow \dfrac{x}{z}=\dfrac{15}{4}$ , simplify

$\therefore x:z=15:4$

Vinod recently opened a surfboard store at the beach. He buys each surfboard in wholesale for

$$80$ and has fixed monthly expenses of

$$3,500$ . If he sells each surfboard at

$\$120$ then how many surfboards does he need to sell in a month to get overall profit?

0%
$18$
0%
$30$
0%
$45$
0%
$90$

Explanation

Let the number of surfboards =

$x$

Given, he bought each surfboard at

$$80$

Hence, cost price of

$x$ surfboards =

$80x$

Given, he sold each surfboard at

$$120$

Hence, selling price of

$x$ surfboards =

$120x$

Given, his monthly expenses =

$$3,500$

To get profit,

$\text{selling price - cost price} \gt 3500$

$\implies 120x-80x \gt 3500$

$\implies 40x \gt 3500 \implies x \gt \dfrac{3500}{40}$

$\implies x\gt 87.5$

Therefore, the number of surfboards he need to sell to get profit should be

$> 87.5$

From the options,

$90>87.5$

Hence, the answer is

$90$ .

The minimum value of the sum of the lengths of diagonals of a cyclic quadrilateral of area

$a^2$ square units is

0%
$\sqrt 2$ a
0%
2 $\sqrt 2$ a
0%
2a
0%
none of these

Explanation

Let

$a,b,c,d$ be the length of the sides of a cyclic quadrilateral.

Given, Area of the quadrilateral is

$a^2$ square units.

Let

$e,f$ be the length of the diagonals.

For a cyclic quadrilateral, the relation between length of the sides and diagonals is given by

$e*f=ac+bd$

since,

$A.M.\geq G.M.$

therefore,

$\dfrac{e+f}{2}\geq \sqrt{e \times f}$

$\implies e+f\geq 2\sqrt{ac+bd}$

But,

$area=\dfrac{1}{2}(ac+bd)\sin\theta \implies a^2=\dfrac{1}{2}(ac+bd)$

(

$\sin\theta =1$ i.e., max value for sine is chooesen so that

$e+f$ should be greater than the maximum)

$\implies e+f\geq 2\sqrt{2a^2}$

$\implies e+f\geq 2\sqrt{2}a$

The minimum value of

$4^x+4^{1-x}$ ,

$x \epsilon R$ , is

0%
1
0%
2
0%
4
0%
none of these

Explanation

$AM\ge GM$

$\dfrac { { 4 }^{ x }+{ 4 }^{ 1-x } }{ 2 } \ge \sqrt { { 4 }^{ x }.{ 4 }^{ 1-x } }$

${ 4 }^{ x }+{ 4 }^{ 1-x }\ge 4$

The recommended daily calcium intake for a 20-year-old is 1,000 milligrams (mg). One cup of milk contains 299 mg of calcium and one cup of juice contains 261 mg of calcium. Which of the following inequalities represents the possible number of cups of milk m and cups of juice j a 20-year-old could drink in a day to meet or exceed the recommended daily calcium intake from these drinks alone?

0%
$\displaystyle 299m+261j\ge 1000$
0%
$\displaystyle 299m+261j>1000$
0%
$\displaystyle \frac { 299 }{ m } +\frac { 261 }{ j } \ge 1000$
0%
$\displaystyle \frac { 299 }{ m } +\frac { 261 }{ j } >1000$

Which of the following is a correct graph of

$x>1, x<4$ ?

0%
Line A
0%
Line B
0%
Line C
0%
Line D
0%
Line E

Conclude from the following:

$t = (2y \times 10^3)+(3y \times 10^3)$

$y > 0$

$t$

$5y\times 10^3$

0%
The quantity A is greater than B.
0%
The quantity B is greater than A.
0%
The two quantities are equal.
0%
The relationship cannot be determined from the information given.

Explanation

Given,

$t= (2y\times 10^3)+(3y\times 10^3)$

$\implies t=(2y+3y)\times 10^3$

$\implies t=5y\times 10^3$

Given

$A:t =5y \times 10^3$ and

$B =5y \times 10^3$

Therefore both the quatities are equal

The number of ordered tuples

$(x,y,z,w)$ where

$x,y,z,w \in [0,10]$ which satisfies the inequality

$2^{\sin ^2x}\times 3^{\cos ^2y}\times 4^{\sin ^2z}\times 5^{\cos ^2w}\geq 120$ , is

0%
$81$
0%
$144$
0%
$0$
0%
$\infty$

For

$\theta>\dfrac {\pi}3$ , the value of

$f(\theta)=\sec ^2 \theta+\cos ^2\theta$ always lies in the interval

0%
$(0,2)$
0%
$[0,1]$
0%
$(1,2)$
0%
$[2,\infty)$

Explanation

To find the optimum, we take the derivative and equate to zero.

given,

$f(\theta)=cos^2\theta + sec^2\theta$

taking derivative w.r.to

$\theta$ and equating to zero we get,

$\implies 2cos\theta(-sin\theta) + 2sec\theta(sec\theta\times tan\theta)=0$

$\implies cos\theta(-sin\theta) + sec\theta(sec\theta\times \dfrac{sin\theta}{cos\theta})=0$

$\implies cos\theta(-sin\theta) + sec\theta(sec\theta \times sin\theta \times sec\theta)=0$

$\implies cos\theta(-sin\theta) + sec^3\theta(sin\theta)=0$

$\implies sin\theta (-cos\theta+sec^3\theta)=0$

$\implies sin\theta=0$ and

$(-cos\theta+sec^3\theta)=0$

since,

$\theta >\dfrac{\pi}{3}$

Therefore,

$sin\theta=0 \implies \theta = n\pi$

$-cos\theta+sec^3\theta=0$

$\implies -cos \theta +\dfrac{1}{cos^3\theta}=0$

$\implies \dfrac{1-cos^4\theta }{cos^3\theta}=0$

$\implies 1-cos^4\theta =0$

$\implies cos^4\theta =1 \implies cos\theta =1$

since,

$\theta > \dfrac{\pi}{3}$

Therefore,

$cos\theta=1 \implies \theta = n\pi$

Substituting

$\theta =\pi$ we get

$(cos180)^2+(sec180)^2=1+1=2$

Therefore, for

$\theta=n\pi$ and

$\theta >\dfrac{\pi}{3}$ we get

$f(\theta)=[2,\infty)$

$0<x<\dfrac {\pi}2$ , then the minimum value of

$\dfrac {\cos ^4x}{\sin ^2x}+\dfrac {\sin ^4x}{\cos ^2x}$ is

0%
$\sqrt 3$
0%
$\dfrac 12$
0%
$\dfrac 13$
0%
$1$

Explanation

we know that,

$A.M. \geq G.M.$

$\implies \dfrac{a+b}{2}\geq \sqrt{ab}$

let

$a=\dfrac{cos^4x}{sin^2x}$ and

$b=\dfrac{sin^4x}{cos^2x}$

therefore,

$\dfrac{\dfrac{cos^4x}{sin^2x}+\dfrac{sin^4x}{cos^2x}}{2}\geq \sqrt{\dfrac{cos^4x}{sin^2x}*\dfrac{sin^4x}{cos^2x}}$

$\implies {\dfrac{cos^4x}{sin^2x}+\dfrac{sin^4x}{cos^2x}}\geq 2*\sqrt{{sin^2x*cos^2x}}$

$\implies {\dfrac{cos^4x}{sin^2x}+\dfrac{sin^4x}{cos^2x}}\geq 2*sinx*cosx$

$\implies {\dfrac{cos^4x}{sin^2x}+\dfrac{sin^4x}{cos^2x}}\geq sin2x$

for

$0<x<\dfrac{\pi}{2}$ we get

$0<sinx<1$

multiplying 2 we get

$0<2x<\pi \implies 0< sinx \leq 1$

therefore by substituting the maximum value of

$sin2x$ we get

${\dfrac{cos^4x}{sin^2x}+\dfrac{sin^4x}{cos^2x}}\geq 1$

therefore the minimum value of

${\dfrac{cos^4x}{sin^2x}+\dfrac{sin^4x}{cos^2x}}$ is

$1$

0%
Both Assertion and Reason are correct and Reason is the correct explanation for Assertion
0%
Both Assertion and Reason are correct but Reason is not the correct explanation for Assertion
0%
Assertion is correct but Reason is incorrect
0%
Both Assertion and Reason are incorrect

Explanation

$A.M. = \cfrac{a+b}{2}$ and

$G.M. = \sqrt{ab}$

Subtracting above equations we get

$A.M.- G.M. = \cfrac{a+b}{2}-\sqrt{ab}$

$= \cfrac{a+b-2\sqrt{ab}}{2}$ ------- (1)

$=\cfrac{(\sqrt{a}-\sqrt{b})^2}{2}\geq 0$ (since, square term is always

$\geq 0$ )

$\implies A.M.-G.M.\geq 0$

Therefore,

$A.M. \geq G.M.$

assume,

$a=1$ and

$b=x^4$ and substituting in eqn (1)

$\cfrac{1+x^4-2\sqrt{1 \times x^4}}{2} \geq 0$

$\implies 1+x^4-2x^2\geq 0$

$\implies 1+x^4\geq 2x^2$

$\implies 2x^2\leq 1+x^4$

$\implies \cfrac{2x^2}{1+x^4}\leq 1$

Given,

$\cfrac{2x^2}{1+x^4}\leq 1$ is true and the reason

$A.M.\geq G.M.$ is the correct explanation

If roots of the equation

$x^4-8x^3+bx^2+cx+16=0$ are positive, then

0%
$b=c=8$
0%
$b=-24,c=-32$
0%
$b=24,c=-32$
0%
$b=24,c=32$

Explanation

given that a polynomial

$x^4-8x^3+bx^2+cx+16=0$ has real roots.

let

$\alpha_1,\alpha_2,\alpha_3,\alpha_4$ be the roots of the polynomial.

Therefore,

$\alpha_1\alpha_2\alpha_3\alpha_4=(-1)^*16=16$

$\alpha_1+\alpha_2+\alpha_3+\alpha_4=8$

$\alpha_1\alpha_2+\alpha_1\alpha_3+\alpha_1\alpha_4+\alpha_2\alpha_3+\alpha_2\alpha_4+\alpha_3\alpha_4=b$

$\alpha_1\alpha_2\alpha_3+\alpha_1\alpha_3\alpha_4+\alpha_1\alpha_2\alpha_4+\alpha_2\alpha_3\alpha_4=-c$

we know that

$A.M.\geq G.M.$

therefore,

$\dfrac{\alpha_1\alpha_2+\alpha_1\alpha_3+\alpha_1\alpha_4+\alpha_2\alpha_4+\alpha_2\alpha_3+\alpha_3\alpha_4}{6}\geq\sqrt[6]{\alpha_1^3*\alpha_2^3*\alpha_3^3*\alpha_4^3}$

$\implies \dfrac{b}{6}\geq\sqrt[6]{(\alpha_1\alpha_2\alpha_3\alpha_4)^3}$

$\implies \dfrac{b}{6}\geq\sqrt[6]{(16)^3}$

$\implies \dfrac{b}{6}\geq 16^{\dfrac{3}{6}}$

$\implies \dfrac{b}{6}\geq 16^{\dfrac{1}{2}}$

$\implies \dfrac{b}{6}\geq 4$

$\implies b\geq 24$

therefore the value of

$b$ is

$24$

similarly

$\dfrac{\alpha_1\alpha_2\alpha_3+\alpha_1\alpha_3\alpha_4+\alpha_1\alpha_2\alpha_4+\alpha_2\alpha_3\alpha_4}{4}\geq\sqrt[4]{\alpha_1^3*\alpha_2^3*\alpha_3^3*\alpha_4^3}$

$\implies \dfrac{-c}{4}\geq\sqrt[4]{(\alpha_1\alpha_2\alpha_3\alpha_4)^3}$

$\implies \dfrac{-c}{4}\geq\sqrt[4]{(16)^3}$

$\implies \dfrac{-c}{4}\geq 16^{\dfrac{3}{4}}$

$\implies \dfrac{-c}{4}\geq 2^{4*\dfrac{3}{4}}$

$\implies \dfrac{-c}{4}\geq 2^{3}$

$\implies \dfrac{-c}{4}\geq 8$

$\implies -c\geq 32$

$\implies c\leq -32$

therefore the value of

$c$ is

$-32$

$2^{\sin ^2x}+2^{\cos ^2x}$ is

0%
$>2$
0%
$\geq 2\sqrt 2$
0%
$\leq 2$
0%
$<1$

Explanation

we know that,

$A.M.\geq G.M.$

$\implies \dfrac{a+b}{2}=\sqrt{ab}$

let

$a=2^{sin^2x}$ and

$b=2^{cos^2x}$

therefore,

$\implies \dfrac{2^{sin^2x}+2^{cos^2x}}{2}\geq\sqrt{2^{sin^2x}*2^{cos^2x}}$

$\implies \dfrac{2^{sin^2x}+2^{cos^2x}}{2}\geq\sqrt{2^{sin^2x+cos^2x}}$

$\implies {2^{sin^2x}+2^{cos^2x}}\geq 2*\sqrt{2^1}$

$\implies {2^{sin^2x}+2^{cos^2x}}\geq 2\sqrt{2}$

The minimum value of the sum of the lengths of diagonals of a cyclic quadrilateral of area

$a^2$ sq. units is

0%
$\sqrt 2 a$
0%
$2a$
0%
$2\sqrt 2 a$
0%
None of the above.

Explanation

Let

$a,b,c,d$ be the length of the sides of a cyclic quadrilateral.

Given, Area of the quadrilateral is

$a^2$ square units.

Let

$e,f$ be the length of the diagonals.

For a cyclic quadrilateral, the relation between length of the sides and diagonals is given by

$e*f=ac+bd$

since,

$A.M.\geq G.M.$

therefore,

$\dfrac{e+f}{2}\geq \sqrt{e*f}$

$\implies e+f\geq 2\sqrt{ac+bd}$

But,

$area=\dfrac{1}{2}(ac+bd)sin\theta \implies a^2=\dfrac{1}{2}(ac+bd)$ (

$sin\theta =1$ i.e., max value for sine is chooesen so that

$e+f$ should be greater than the maximum)

$\implies e+f\geq 2\sqrt{2a^2}$

$\implies e+f\geq 2\sqrt{2}a$

$p,q,r$ be three distinct real numbers , then the value of

$(p+q)(q+r)(r+p)$ is

0%
$> 8pqr$
0%
$> 16pqr$
0%
$8pqr$
0%
$< 8pqr$

Explanation

$A.M. = \dfrac{a+b}{2}$ and

$G.M. = \sqrt{ab}$

Subtracting above equations we get

$A.M.- G.M. = \dfrac{a+b}{2}-\sqrt{ab}$

$= \dfrac{a+b-2\sqrt{ab}}{2}$ ------- (1)

$=\dfrac{(\sqrt{a}-\sqrt{b})^2}{2}\geq 0$ (since, square term is always

$\geq 0$ )

$\implies A.M.-G.M.\geq 0$

Therefore,

$A.M. \geq G.M.$

$\implies \dfrac{a+b}{2}\geq \sqrt{ab}$

squaring on both sides.

$\implies (\dfrac{a+b}{2})^2\geq {ab}$

$\implies(a+b)^2\geq {4ab}$

Let

$a=p$ and

$b=q$

$\implies (p+q)^2\geq 4pq$ -------(1)

Let

$a=q$ and

$b=r$

$\implies (q+r)^2\geq 4qr$ -------(2)

Let

$a=r$ and

$b=p$

$\implies (r+p)^2\geq 4rp$ -------(3)

multiplying (1) , (2) and (3)

$\implies (p+q)^2(q+r)^2(r+p)^2\geq 64(pqr)^2$

applying square root on both sides

$\implies (p+q)(q+r)(r+p)\geq (8pqr)$

The minimum value of

$|\sin x+\cos x+\tan x+\sec x+\mathrm{cosec} x+\cot x|$ is

0%
$2\sqrt 2-1$
0%
$2\sqrt 2+1$
0%
$\sqrt 2-1$
0%
$\sqrt 2+1$

Explanation

let

$p=|sinx+cosx+tanx+cotx+cosecx+secx|$

$p=|sinx+cosx+\dfrac{sinx}{cosx}+\dfrac{cosx}{sinx}+\dfrac{1}{sinx}+\dfrac{1}{cosx}|$

$=|sinx+cosx+\dfrac{sin^2x+cos^2x}{sinxcosx}+\dfrac{sinx+cosx}{sinxcosx}|$

$=|sinx+cosx+\dfrac{1}{sinxcosx}+\dfrac{sinx+cosx}{sinxcosx}|$

let

$t=sinx+cosx$ --------(1)

squaring on both sides we get

$t^2=(sinx+cosx)^2=sin^2x+cos^2x+2sinxcosx$

$\implies t^2=1+2sinxcosx$

$\implies sinxcosx=\dfrac{t^2-1}{2}$ -------(2)

substituting (1) and (2) in

$p$ we get

$p=|t+\dfrac{2}{t^2-1}+\dfrac{2t}{t^2-1}|$

$\implies p=|t+\dfrac{2+2t}{t^2-1}|$

$\implies p=|t+\dfrac{2(t+1)}{(t-1)(t+1)}|$

$\implies p=|t+\dfrac{2}{(t-1)}|$

let

$m=t-1$

$\implies p=|m-1+\dfrac{2}{m}|$

$\implies p=|-1+(m+\dfrac2m)|$ -----(3)

we know that

$A.M.\geq G.M.$

$\implies \dfrac{m+\dfrac 2m}2\geq \sqrt{m*\dfrac 2m}$

$\implies {m+\dfrac 2m}\geq2\sqrt{2}$

$\implies ({m+\dfrac 2m})-1\geq 2\sqrt{2}-1$

$\implies p\geq 2\sqrt 2-1$ (from (3))

therefore the minimum value of

$|sinx+cosx+tanx+cotx+cosecx+secx|$ is

$2\sqrt 2-1$

A stick of length

$20$ units is to be divided into

$n$ parts so that the product of the lengths of the parts is greater than unity.
The maximum possible value of

$n$ is

0%
$20$
0%
$19$
0%
$18$
0%
$21$

Explanation

Given, a stick is divided into

$n$ parts which is of length

$20$ units

Let the

$n$ parts be

$x_1, x_2, ...,x_n$

$x_1+x_2+x_3+...+x_n=20$

Given,

$x_1 *x_2* ...*x_n >1$ -------(1)

We know that

$A.M. \geq G.M.$

$A.M.$ of

$x_1, x_2, ...,x_n = \dfrac{x_1+x_2+...+x_n}{n}$

$G.M.$ of

$x_1, x_2, ...,x_n = \sqrt[n]{x_1*x_2*...*x_n}$

Therefore,

$\dfrac{x_1+x_2+...+x_n}{n}\geq \sqrt[n]{x_1*x_2*...*x_n}$

$\dfrac{20}{n}\geq \sqrt[n]{x_1*x_2*...*x_n}$ (since,

$x_1+x_2+x_3+...+x_n=20$ )

taking

$n^{th}$ power on both sides.

$(\dfrac{20}{n})^n\geq {x_1*x_2*...*x_n}$ -------(2)

From (1) and (2) we get

$(\dfrac{20}{n})^n> 1$

$\implies \dfrac{20}{n}>1$ (since,

$\sqrt[n]1=1$ )

$n<20$

Therefore, the maximum possible value of

$n = 19$

Equation

$x^4+ax^3+bx^2+cx+1=0$ has real roots (

$a,b,c$ are non-negative). Maximum value of

$a$ is

0%
$10$
0%
$9$
0%
$5$
0%
$-4$

Explanation

let

$\alpha_{1}, \alpha_{2}, \alpha_{3}, \alpha_{4}$ be the roots of the given equation

$x^4+ax^3+bx^2+cx+1=0$

Sum of the roots of the given polynomial is

$\alpha_{1} +\alpha_{2}+\alpha_{3}+\alpha_{4}=\dfrac{-a}{1}=-a$

product of the roots of the given polynomial is

$\alpha_{1} \times\alpha_{2}\times\alpha_{3}\times\alpha_{4}=\dfrac{(-1)^4\times1}{1}=1$

We know that,

$A.M.\geq G.M.$

therefore,

$\dfrac{\alpha_{1} +\alpha_{2}+\alpha_{3}+\alpha_{4}}{4}\geq \sqrt[4]{\alpha{1}\times\alpha{2}\times\alpha{3}\times\alpha{4}}$

$\implies \dfrac{-a}{4}\geq \sqrt[4]{1}$

$\implies-a\geq 4$

$\implies a\leq -4$

from this

$a$ value is always negative with a maximum of

$-4$

$f(x)=\dfrac {(x-2)(x-1)}{x-3}, \forall x>3$ . The minimum value of

$f(x)$ is equal to

0%
$3+2\sqrt 2$
0%
$3+2\sqrt 3$
0%
$3\sqrt 2+2$
0%
$3-2\sqrt 2$

Explanation

Given,

$f(x)=\dfrac{(x-1)(x-2)}{x-3}$ and

$x>3$

Taking the derivative and equating to zero. We get

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

$\implies f'(x)=\dfrac{1}{x-3}(x-1+x-2-\dfrac{(x^2-2x-x+2)}{(x-3)})$

$\implies f'(x)=\dfrac{1}{x-3}(2x-3-\dfrac{(x^2-3x+2)}{(x-3)})$

$\implies f'(x)=\dfrac{1}{(x-3)}(\dfrac{(2x-3)(x-3)-(x^2-3x+2)}{(x-3)})$

$\implies f'(x)=\dfrac{1}{(x-3)^2}(2x^2-3x-6x+9-x^2+3x-2)$

$\implies f'(x)=\dfrac{1}{(x-3)^2}(x^2-6x+7)$

$f'(x)=0 \implies \dfrac{1}{(x-3)^2}(x^2-6x+7)=0$

therefore

$(x^2-6x+7)=0$ and

${(x-3)^2} \neq 0$ since, denominator cannot be equal to zero.

The roots of

$(x^2-6x+7)=0$ are

$\dfrac{6+\sqrt{6^2-7*4}}{2*1}$ and

$\dfrac{6-\sqrt{6^2-7*4}}{2*1}$

$\implies \dfrac{6+\sqrt{8}}{2}$ and

$\dfrac{6-\sqrt{8}}{2}$

$\implies \dfrac{6+2\sqrt{2}}{2}$ and

$\dfrac{6-2\sqrt{2}}{2}$

$\implies 3+\sqrt{2}$ and

$3-\sqrt{2}$

$x\notin 3-\sqrt{2}$ since

$x>3$

Therefore

$x=3+\sqrt{2}$

substituting

$x=3+\sqrt{2}$ in

$f(x)$ we get

$f(x)=\dfrac{(3+\sqrt{2}-1)(3+\sqrt{2}-2)}{3+\sqrt{2}-3}$

$=\dfrac{(2+\sqrt{2})(1+\sqrt{2})}{\sqrt{2}}$

$=\dfrac{(2+\sqrt{2}+2\sqrt{2}+2)}{\sqrt{2}}$

$=\dfrac{(4+3\sqrt{2})}{\sqrt{2}} = 3+2\sqrt{2}$

If positive numbers

$a,b,c$ be in H.P, then the equation

$x^2-kx+2b^{101}-a^{101}-c^{101}=0$

$(k\in R)$ has

0%
Both roots positive
0%
Both roots negative
0%
One positive and one negative root
0%
Both roots imaginary

Explanation

We know that,

$A.M.\geq G.M.$

$\implies \dfrac{a+b}{2}\geq \sqrt{ab}$

let

$a=a^{101}, b=c^{101}$

therefore,

$\dfrac{a^{101}+c^{101}}{2}\geq \sqrt{a^{101}c^{101}}$ --------(1)

given,

$a,b,c$ are in H.P.

$\implies b=\dfrac{2ac}{a+c}$

let

$a=a^{101}, b=b^{101}, c=c^{101}$

therefore,

$b^{101}=\dfrac{2a^{101}c^{101}}{a^{101}+c^{101}}$

$\implies \dfrac{a^{101}+c^{101}}{2}= \dfrac{a^{101}c^{101}}{b^{101}}$ -----(2)

substituting (2) in (1) we get

$\dfrac{a^{101}c^{101}}{b^{101}}\geq \sqrt{a^{101}c^{101}}$

$\sqrt{a^{101}c^{101}}\geq b^{101}$ ----(3)

substituting (3) in (1)

$\dfrac{a^{101}+c^{101}}{2}\geq b^{101}$

$a^{101}+c^{101}\geq 2b^{101}$

$2b^{101}-a^{101}-c^{101}\leq 0$

For the given polynomial, product of roots is

$2b^{101}-a^{101}-c^{101}$ which is negative (

$\leq 0$ )

therefore, one root must be positive and the other must be negative. (since, product of positive and negative gives negative)

$xy+yz+zx=12$ , where

$x,y,z$ are positive values, then the greatest value of

$xyz$ is

0%
$8$
0%
$18$
0%
$6$
0%
None of the above.

Explanation

we know that,

$A.M. \geq G.M.$

$\implies \dfrac{a+b+c}{3}\geq \sqrt[3]{abc}$

let

$a=xy, b=yz,c=zx$

$\implies \dfrac{xy+yz+zx}{3}\geq \sqrt[3]{xy\times yz\times zx}$

$\implies \dfrac{12}{3}\geq \sqrt[3]{(xyz)^2}$ (since, given that

$xy+yz+zx=12$ )

$\implies 4\geq \sqrt[3]{(xyz)^2}$

$\implies \sqrt[3]{(xyz)^2} \leq 4$

cubing on both sides

$\implies {(xyz)^2} \leq 4^3$

$\implies {(xyz)^2} \leq 64$

applying square root on both sides

$xyz\leq 8$

therefore, the maximum value of

$xyz = 8$

$0<a,b,c<1$ and

$a+b+c=2$ , then the greatest value of

$\dfrac a{1-a}\times\dfrac b{1-b}\times\dfrac c{1-c}$ is

0%
$7$
0%
$4$
0%
$8$
0%
$12$

Explanation

Given,

$a+b+c=2$

we know that,

$A.M.\geq G.M.$

$\implies \dfrac{a+b+c}{3}\geq \sqrt[3]{abc}$

$\implies \dfrac{2}{3}\geq \sqrt[3]{abc}$

$\implies \sqrt[3]{abc}\leq \dfrac{2}{3}$

cubing on both sides

$\implies {abc}\leq \dfrac{8}{27}$ --------------(1)

Similarly

$\implies \dfrac{1-a+1-b+1-c}{3}\geq \sqrt[3]{(1-a)(1-b)(1-c)}$

$\implies \dfrac{3-(a+b+c)}{3}\geq \sqrt[3]{(1-a)(1-b)(1-c)}$

$\implies \dfrac{3-2}{3}\geq \sqrt[3]{(1-a)(1-b)(1-c)}$

$\implies \dfrac{1}{3}\geq \sqrt[3]{(1-a)(1-b)(1-c)}$

cubing on both sides

$\implies \dfrac{1}{(1-a)(1-b)(1-c)}\leq 27$ ---------------(2)

multiplying (1) and (2) we get

$\implies \dfrac{abc}{(1-a)(1-b)(1-c)}\leq \dfrac{8}{27}*27$

$\implies \dfrac{abc}{(1-a)(1-b)(1-c)}\leq 8$

The greatest value of

$\dfrac{abc}{(1-a)(1-b)(1-c)}$ is

$8$

Equation

$x^4+ax^3+bx^2+cx+1=0$ has real roots (

$a,b,c$ are non-negative). Minimum non-negative real value of

$b$ is

0%
$12$
0%
$15$
0%
$6$
0%
$10$

Explanation

Given that a polynomial

$x^4+ax^3+bx^2+cx+1=0$ has real roots.

Let

$\alpha_1,\alpha_2,\alpha_3,\alpha_4$ be the roots of the polynomial.

Therefore,

$\alpha_1\alpha_2\alpha_3\alpha_4=1$

$\alpha_1+\alpha_2+\alpha_3+\alpha_4=-a$

$\alpha_1\alpha_2+\alpha_1\alpha_3+\alpha_1\alpha_4+\alpha_2\alpha_3+\alpha_2\alpha_4+\alpha_3\alpha_4=b$

$\alpha_1\alpha_2\alpha_3+\alpha_1\alpha_3\alpha_4+\alpha_1\alpha_2\alpha_4+\alpha_2\alpha_3\alpha_4=-c$

We know that

$A.M.\geq G.M.$

therefore,

$\dfrac{\alpha_1\alpha_2+\alpha_1\alpha_3+\alpha_1\alpha_4+\alpha_2\alpha_4+\alpha_2\alpha_3+\alpha_3\alpha_4}{6}\geq\sqrt[6]{\alpha_1^3 \alpha_2^3 \alpha_3^3 \alpha_4^3}$

$\implies \dfrac{b}{6}\geq\sqrt[6]{(\alpha_1\alpha_2\alpha_3\alpha_4)^3}$

$\implies \dfrac{b}{6}\geq\sqrt[6]{(1)^3}$

$\implies \dfrac{b}{6}\geq 1$

$\implies b\geq 6$

Therefore the minimum value of

$b$ is

$6$

$\Delta ABC$ , the least value of

$\mathrm{cosec} \dfrac A2+\mathrm{cosec} \dfrac B2+\mathrm{cosec} \dfrac C2$ is

0%
$4\sqrt 2$
0%
$0$
0%
$\dfrac 3{\sqrt 2}$
0%
$6$

Explanation

For a triangle, the maximum or minimum occurs when the triangle is equilateral.

Therefore,

$\angle{A}=\angle{B}=\angle{C}=60^o$

$cosec(\dfrac A2)=cosec(\dfrac B2)=cosec(\dfrac C2)=\dfrac{1}{sin\dfrac{60}{2}}=\dfrac{1}{sin30}=2$

we know that

$A.M.\geq G.M.$

$\implies \dfrac{cosec\dfrac A2+cosec\dfrac B2+cosec\dfrac C2}{3}\geq \sqrt[3]{cosec\dfrac A2*cosec\dfrac B2*cosec\dfrac C2}$

$\implies \dfrac{cosec\dfrac A2+cosec\dfrac B2+cosec\dfrac C2}{3}\geq \sqrt[3]{2*2*2}$

$\implies {cosec\dfrac A2+cosec\dfrac B2+cosec\dfrac C2}\geq 3*\sqrt[3]{2^3}$

$\implies {cosec\dfrac A2+cosec\dfrac B2+cosec\dfrac C2}\geq 3*2$

$\implies {cosec\dfrac A2+cosec\dfrac B2+cosec\dfrac C2}\geq 6$

Therefore, the least value for

${cosec\dfrac A2+cosec\dfrac B2+cosec\dfrac C2}= 6$

Equation

$x^4+ax^3+bx^2+cx+1=0$ has real roots (

$a,b,ca,b,c$ are non-negative). Maximum real value of

$c$ is

0%
$-10$
0%
$-9$
0%
$-6$
0%
$-4$

Explanation

Given that a polynomial

$x^4+ax^3+bx^2+cx+1=0$ has real roots.

Let

$\alpha_1,\alpha_2,\alpha_3,\alpha_4$ be the roots of the polynomial.

Therefore,

$\alpha_1\alpha_2\alpha_3\alpha_4=1$

$\alpha_1+\alpha_2+\alpha_3+\alpha_4=\dfrac{-b}a=-a$

$\alpha_1\alpha_2+\alpha_1\alpha_3+\alpha_1\alpha_4+\alpha_2\alpha_3+\alpha_2\alpha_4+\alpha_3\alpha_4=\dfrac ca=b$

$\alpha_1\alpha_2\alpha_3+\alpha_1\alpha_3\alpha_4+\alpha_1\alpha_2\alpha_4+\alpha_2\alpha_3\alpha_4=\dfrac{-d}a=-c$

We know that

$A.M.\geq G.M.$

Therefore,

$\dfrac{\alpha_1\alpha_2\alpha_3+\alpha_1\alpha_3\alpha_4+\alpha_1\alpha_2\alpha_4+\alpha_2\alpha_3\alpha_4}{4}\geq\sqrt[4]{\alpha_1^3\ \alpha_2^3\ \alpha_3^3\ \alpha_4^3}$

$\implies \dfrac{-c}{4}\geq\sqrt[4]{(\alpha_1\alpha_2\alpha_3\alpha_4)^3}$

$\implies \dfrac{-c}{4}\geq\sqrt[4]{(1)^3}$

$\implies \dfrac{-c}{4}\geq 1$

$\implies -c\geq 4$

$\implies c\leq -4$

The least value of the expression

$2\log_{10}x-\log_x(0.01)$ , for

$x>1$ , is

0%
$1$
0%
$2$
0%
$4$
0%
$8$

Explanation

We know that

$A.M. \geq G.M.$

$\implies \dfrac{a+b}{2}\geq \sqrt{ab}$

Let

$a=2\log_{10}x$ and

$b=-\log_x0.01$

Therefore,

$\dfrac{2\log_{10}x-\log_x0.01}{2}\geq \sqrt{(2\log_{10}x)(-\log_x0.01)}$

$\implies \dfrac{2\log_{10}x-\log_x0.01}{2}\geq \sqrt{(2\log_{10}x)(-\log_x10^{-2})}$

$\implies \dfrac{2\log_{10}x-\log_x0.01}{2}\geq \sqrt{(2\log_{10}x)(\log_x10^{-2})^{-1}}$

$\implies \dfrac{2\log_{10}x-\log_x0.01}{2}\geq \sqrt{(2\log_{10}x)(\log_x10^{2})}$

$\implies \dfrac{2\log_{10}x-\log_x0.01}{2}\geq \sqrt{(2\dfrac{\log x}{\log 10})(\dfrac{\log 10^2}{\log x})}$

$\implies \dfrac{2\log_{10}x-\log_x0.01}{2}\geq \sqrt{(2\dfrac{\log x}{\log 10})(\dfrac{2\log 10}{\log x})}$

$\implies {2\log_{10}x-\log_x0.01}\geq 4$

Therefore the minimum value of

${3\log_{10}x-\log_x0.01}$ is

$4$

$\frac{4\alpha}{\alpha^2+1}\geq 1$ and

$\alpha+\frac{1}{\alpha}$ is an odd integer then number of possible values of

$\alpha$ is

0%
1
0%
2
0%
3
0%
4

Explanation

We know that,

$A.M. \geq G.M. \implies \dfrac{a+b}{2}\geq \sqrt{ab}$

Let

$a=1$ and

$b=\alpha^2$

$\implies \dfrac{1+\alpha^2}{2}\geq \sqrt{1*\alpha ^2}$

$\implies \dfrac{1+\alpha^2}{2}\geq \alpha$

$\implies \dfrac{1+\alpha^2}{2\alpha}\geq 1$

$\implies \dfrac{2\alpha}{1+\alpha^2}\leq 1$

multiplying 2 on both sides

$\implies \dfrac{4\alpha}{1+\alpha^2}\leq 2$

given that

$\dfrac{4\alpha}{1+\alpha^2}\geq 1$

and

$\alpha +\dfrac{1}{\alpha}=\dfrac{1+\alpha^2}{\alpha}$ is odd integer

$\implies \dfrac{4}{odd }\geq 1$ and

$\dfrac{4}{odd}\leq 2$

that odd integer can only be

$3$ satisfying above conditions.

Therefore,

$\alpha$ can take only one value .

The ratio of inradius and circumradius of a square is :

0%
$1\ :\ \sqrt { 2 }$
0%
$\sqrt { 2 } \ :\ \sqrt { 3 }$
0%
$1\ :\ 3$
0%
$1\ :\ 2$

Explanation

R.E.F image

let a be side of square

$r_{1}$ be radius of incircle

We have

$AB = CD$

$\Rightarrow a = 2r_{1}...(1)$

let

$r_{2}$ be radius of circumcircle

Here diameter

$=$ diagonal

$\Rightarrow 2r_{2} = \sqrt{2}a$

$\Rightarrow r_{2} = \dfrac{a}{\sqrt{2}}...(2)$

from (1) & (2)

$\dfrac{r_{1}}{r_{2}} = \dfrac{(\dfrac{a}{2})}{(\dfrac{a}{\sqrt{2}})} = \dfrac{(\dfrac{1}{2})}{(\dfrac{1}{\sqrt{2}})} = \dfrac{\sqrt{2}}{2} = \dfrac{1}{\sqrt{2}} = 1:\sqrt{2}$

1172649_804464_ans_f2b3ab59a952408ebc95f0a5d88e43e7.jpg

Number of integers satisfying the inequalities

$\sqrt {\log_{3}x - 1} + \dfrac {\dfrac {1}{2} \log_{3}x^{3}}{\log_{3}\dfrac {1}{3}} + 2 > 0$ , is

0%
$5$
0%
$6$
0%
$7$
0%
$8$

Explanation

Given

$\sqrt{\log_3x-1}+\dfrac{\dfrac 12\log_3x^3}{\log_3\dfrac 13}+2>0$

$\implies \sqrt{\log_3x-1}+\dfrac{\dfrac 32\log_3x}{\log_31-\log_33}+2>0$

$\implies \sqrt{\log_3x-1}+\dfrac{\dfrac 32\log_3x}{0-1}+2>0$

$\implies \sqrt{\log_3x-1}-{\dfrac 32\log_3x}+2>0$

$\implies \sqrt{\log_3x-1}>{\dfrac 32\log_3x}-2$

Squaring on both sides

$\implies {\log_3x-1}>\left({\dfrac 32\log_3x}-2\right)^2$

Let

$\log_3 x=t$

$\implies {t-1}>\left({\dfrac 32t}-2\right)^2$

$\implies {t-1}>{\dfrac 94t^2}+4-2\left(\dfrac{3}{2}\right)2t$

$\implies {t-1}>{\dfrac 94t^2}+4-6t$

$\implies {\dfrac 94t^2}+4-6t-t+1<0$

$\implies {\dfrac 94t^2}-7t+5<0$

Roots of the quadratic equation

$ax^2+bx+c=0$ is given by

$\dfrac{-b\pm\sqrt{b^2-4ac}}{2a}$

Therefore, roots are

$\dfrac{-(-7)\pm\sqrt{7^2-4\left(\dfrac{9}{4}\right)5}}{2\left(\dfrac {9}{4}\right)}$

$\implies \dfrac{7\pm\sqrt{49-45}}{\left(\dfrac 92\right)}$

$\implies \dfrac{7\pm2}{\left(\dfrac 92\right)}$

$\implies \dfrac{7+2}{\left(\dfrac 92\right)}$ and

$\dfrac{7-2}{(\dfrac 92)}$

$\implies 2$ and

$\dfrac {10}9$ are the roots

therefore

${\dfrac 94t^2}-7t+5<0 \implies (t-2)\left(t-\dfrac{10}{9}\right)<0$

$\implies (t-2)<0$ and

$\left(t-\dfrac{10}{9}\right)>0$ or

$(t-2)>0$ and

$\left(t-\dfrac{10}{9}\right)<0$ doesn't have a solution

therefore

$t<2$ and

$t>\dfrac{10}{9}$

$\implies \log_3x<2$ and

$\log_3x>\dfrac{10}{9}$

$\implies x<3^2$ and

$x>3^{\tiny \dfrac{10}{9}}$

$\implies x<9$ and

$x>3.39$

$\implies x\in (3.39,9)$

$\{4,5,6,7,8\}$ are integers which belong to the set

$(3.39,9)$

Therefore the total number of integers

$=5$

On a linear escalator running between two points

$A$ and

$B$ , a boy takes time

${t}_{1}$ , to move from

$A$ to

$B$ , if the boy runs with a constant speed on the escalator. If the boy runs from

$B$ to

$A$ , he takes time

${t}_{2}$ to reach

$B$ from

$A$ . The time taken by the boy to move from

$A$ to

$B$ if he stands still on the escalator will be (The escalator moves from

$A$ to

$B$ )

0%
$\cfrac { { t }_{ 1 }{ t }_{ 2 } }{ { t }_{ 2 }-{ t }_{ 1 } }$
0%
$\cfrac { { 2t }_{ 1 }{ t }_{ 2 } }{ { t }_{ 2 }-{ t }_{ 1 } }$
0%
$\cfrac { { t }_{ 1 }^{ 2 }+{ t }_{ 2 }^{ 2 } }{ { t }_{ 1 }{ t }_{ 2 } }$
0%
$\cfrac { { t }_{ 1 }^{ 2 }-{ t }_{ 2 }^{ 2 } }{ { t }_{ 1 }{ t }_{ 2 } }$

Explanation

Let the distance between A and B=

$d$ , Speed of boy =

$u$ and speed of escalator =

$v$ .

Let time

$t$ taken by boy if he stands still on escalator.

Relative speed of boy when he moves from A to B=

$u+v$ ,

Relative speed of boy when he moves from B to A=

$u-v$ ,

$\therefore t_1= \dfrac{d}{u+v}$

$\Rightarrow u+v= \dfrac{d}{t_1}$ ...1

$\therefore t_2= \dfrac{d}{u-v}$

$\Rightarrow u-v= \dfrac{d}{t_2}$ ...2

Subtracting equation 2 from 1,

$\Rightarrow 2v=d(\dfrac{1}{t_1}-\dfrac{1}{t_2})$

$\Rightarrow \dfrac{d}{v}= \dfrac{2t_1t_2}{t_2-t_1}$ ....3

$\therefore t= \dfrac{d}{v}= \dfrac{2t_1t_2}{t_2-t_1}$

CBSE Questions for Class 12 Commerce Applied Mathematics Quantification And Numerical Applications Quiz 14 - MCQExams.com

0:0:1

Practice Class 12 Commerce Applied Mathematics Quiz Questions and Answers

CBSE Questions for Class 12 Commerce Applied Mathematics Quantification And Numerical Applications Quiz 14 - MCQExams.com

0:0:1

Practice Class 12 Commerce Applied Mathematics Quiz Questions and Answers

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