Let \alpha = \mathop \max \limits_x \in R \ 8^2\sin 3x.4^4\cos 3x\ and \beta = \mathop \min \limits_x \in R \ 8^2\sin 3x.4^4\cos 3x\ . If 8x^2 + bx + c = 0 is a quadratic equation whose roots are \alpha 1/5 and \beta 1/5, then the value of c - b is equal to :

Quadratic Equation and Inequalities PYQ — JEE Main 2021 · 27 Jul · Shift 2 · Q33 with Solution

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Correct answer: A

Interpret the expression

We have

\alpha=\max_{x\in\mathbb R}\left\{8^{2\sin 3x}\cdot 4^{4\cos 3x}\right\}, \qquad \beta=\min_{x\in\mathbb R}\left\{8^{2\sin 3x}\cdot 4^{4\cos 3x}\right\}.

The notation in the question appears as $8^{2\sin 3x}.4^{4\cos 3x}$ , which means multiplication.

Rewrite in base 2

Since $8=2^3,\qquad 4=2^2,$ we get

8^{2\sin 3x}=(2^3)^{2\sin 3x}=2^{6\sin 3x},

4^{4\cos 3x}=(2^2)^{4\cos 3x}=2^{8\cos 3x}.

Therefore,

8^{2\sin 3x}\cdot 4^{4\cos 3x}=2^{6\sin 3x+8\cos 3x}.

So we need the maximum and minimum of $6\sin 3x+8\cos 3x.$

Find maximum and minimum of the exponent

For any expression of the form $a\sin\theta+b\cos\theta,$ its maximum value is $\sqrt{a^2+b^2}$ and minimum value is $-\sqrt{a^2+b^2}$ .

Here,

\sqrt{6^2+8^2}=\sqrt{36+64}=\sqrt{100}=10.

Hence,

\max(6\sin 3x+8\cos 3x)=10, \qquad \min(6\sin 3x+8\cos 3x)=-10.

Thus,

\alpha=2^{10}=1024, \qquad \beta=2^{-10}=\frac1{1024}.

Use the given roots

The roots of $8x^2+bx+c=0$ are given as $\alpha^{1/5},\quad \beta^{1/5}.$

So,

\alpha^{1/5}=1024^{1/5}=(2^{10})^{1/5}=2^2=4,

\beta^{1/5}=\left(2^{-10}\right)^{1/5}=2^{-2}=\frac14.

Therefore the roots are

Form relations using Vieta's formulas

For $8x^2+bx+c=0,$ if roots are $r_1,r_2$ , then

r_1+r_2=-\frac{b}{8}, \qquad r_1r_2=\frac{c}{8}.

Now,

r_1+r_2=4+\frac14=\frac{17}{4},

-\frac{b}{8}=\frac{17}{4} \implies b=-34.

Also,

r_1r_2=4\cdot \frac14=1,

\frac{c}{8}=1 \implies c=8.

Compute $c-b$

c-b=8-(-34)=42.

Check with options

$42$ corresponds to Option A.

Compare with stored answer

Stored correct answer: A

Our derived answer also is A.

More from Quadratic Equation and Inequalities

View written solutionFree

Correct answer: A

Interpret the expression

We have

\alpha=\max_{x\in\mathbb R}\left\{8^{2\sin 3x}\cdot 4^{4\cos 3x}\right\}, \qquad \beta=\min_{x\in\mathbb R}\left\{8^{2\sin 3x}\cdot 4^{4\cos 3x}\right\}.

The notation in the question appears as $8^{2\sin 3x}.4^{4\cos 3x}$ , which means multiplication.

Rewrite in base 2

Since $8=2^3,\qquad 4=2^2,$ we get

8^{2\sin 3x}=(2^3)^{2\sin 3x}=2^{6\sin 3x},

4^{4\cos 3x}=(2^2)^{4\cos 3x}=2^{8\cos 3x}.

Therefore,

8^{2\sin 3x}\cdot 4^{4\cos 3x}=2^{6\sin 3x+8\cos 3x}.

So we need the maximum and minimum of $6\sin 3x+8\cos 3x.$

Find maximum and minimum of the exponent

For any expression of the form $a\sin\theta+b\cos\theta,$ its maximum value is $\sqrt{a^2+b^2}$ and minimum value is $-\sqrt{a^2+b^2}$ .

Here,

\sqrt{6^2+8^2}=\sqrt{36+64}=\sqrt{100}=10.

Hence,

\max(6\sin 3x+8\cos 3x)=10, \qquad \min(6\sin 3x+8\cos 3x)=-10.

Thus,

\alpha=2^{10}=1024, \qquad \beta=2^{-10}=\frac1{1024}.

Use the given roots

The roots of $8x^2+bx+c=0$ are given as $\alpha^{1/5},\quad \beta^{1/5}.$

So,

\alpha^{1/5}=1024^{1/5}=(2^{10})^{1/5}=2^2=4,

\beta^{1/5}=\left(2^{-10}\right)^{1/5}=2^{-2}=\frac14.

Therefore the roots are

Form relations using Vieta's formulas

For $8x^2+bx+c=0,$ if roots are $r_1,r_2$ , then

r_1+r_2=-\frac{b}{8}, \qquad r_1r_2=\frac{c}{8}.

Now,

r_1+r_2=4+\frac14=\frac{17}{4},

-\frac{b}{8}=\frac{17}{4} \implies b=-34.

Also,

r_1r_2=4\cdot \frac14=1,

\frac{c}{8}=1 \implies c=8.

Compute $c-b$

c-b=8-(-34)=42.

Check with options

$42$ corresponds to Option A.

Compare with stored answer

Stored correct answer: A

Our derived answer also is A.

Quadratic Equation and Inequalities question

Quadratic Equation and Inequalities question

More from Quadratic Equation and Inequalities

Quadratic Equation and Inequalities question

Quadratic Equation and Inequalities question

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