Q.12. is equal to :

Q: What is the correct answer for JEE Main Mathematics Q12?

The correct answer is 2. Sol. Let $E = 16\sin^3 \frac{x}{2} \cos^2 \frac{x}{2}$. We know $\sin^2 \frac{x}{2} = \frac{1-\cos x}{2}$ and $\cos^2 \frac{x}{2} = \frac{1+\cos x}{2}$. $E = 16 \sin \frac{x}{2} \sin^2 \frac{x}{2} \cos^2 \frac{x}{2} = 16 \sin \frac{x}{2} (\frac{1-\cos x}{2}) (\frac{1+\cos x}{2})$ $E = 4 \sin \frac{x}{2} (1-\cos^2 x) = 4 \sin \frac{x}{2} \sin^2 x$. This transformation seems incorrect. Let's use $t

Q: Q.12. is equal to :

The correct answer is 2. Sol. Let $E = 16\sin^3 \frac{x}{2} \cos^2 \frac{x}{2}$. We know $\sin^2 \frac{x}{2} = \frac{1-\cos x}{2}$ and $\cos^2 \frac{x}{2} = \frac{1+\cos x}{2}$. $E = 16 \sin \frac{x}{2} \sin^2 \frac{x}{2} \cos^2 \frac{x}{2} = 16 \sin \frac{x}{2} (\frac{1-\cos x}{2}) (\frac{1+\cos x}{2})$ $E = 4 \sin \frac{x}{2} (1-\cos^2 x) = 4 \sin \frac{x}{2} \sin^2 x$. This transformation seems incorrect. Let's use $t

Q: What are the options for this JEE Main Mathematics question?

A. $3\sqrt{3}/2$ · B. $3\sqrt{3}$ · C. $4\sqrt{3}$ · D. $6\sqrt{3}$

Question

Q.12.  is equal to :

FirstInTest · Accepted Answer

Sol. Let $E = 16\sin^3 \frac{x}{2} \cos^2 \frac{x}{2}$.
We know $\sin^2 \frac{x}{2} = \frac{1-\cos x}{2}$ and $\cos^2 \frac{x}{2} = \frac{1+\cos x}{2}$.
$E = 16 \sin \frac{x}{2} \sin^2 \frac{x}{2} \cos^2 \frac{x}{2} = 16 \sin \frac{x}{2} (\frac{1-\cos x}{2}) (\frac{1+\cos x}{2})$
$E = 4 \sin \frac{x}{2} (1-\cos^2 x) = 4 \sin \frac{x}{2} \sin^2 x$.
This transformation seems incorrect. Let's use $t = \sin(x/2)$ and $1-t^2 = \cos^2(x/2)$.
$E = 16 t^3 (1-t^2)$. Let $f(t) = 16(t^3-t^5)$.
Since $0 \le x \le \pi$, we have $0 \le x/2 \le \pi/2$. So $0 \le \sin(x/2) \le 1$.
Thus, $0 \le t \le 1$.
To find the maximum value, we differentiate $f(t)$ with respect to $t$:
$f'(t) = 16(3t^2 - 5t^4) = 16t^2(3 - 5t^2)$.
Set $f'(t) = 0$ to find critical points:
$16t^2(3 - 5t^2) = 0$.
$t=0$ or $3-5t^2=0 \Rightarrow t^2 = 3/5 \Rightarrow t = \sqrt{3/5}$ (since $t \ge 0$).
Evaluate $f(t)$ at the critical points and endpoints:
$f(0) = 16(0) = 0$.
$f(1) = 16(1-1) = 0$.
$f(\sqrt{3/5}) = 16((\sqrt{3/5})^3 - (\sqrt{3/5})^5) = 16(\frac{3\sqrt{3}}{5\sqrt{5}} - \frac{9\sqrt{3}}{25\sqrt{5}})$
$= 16 \frac{3\sqrt{3}}{5\sqrt{5}} (1 - \frac{3}{5}) = 16 \frac{3\sqrt{3}}{5\sqrt{5}} \frac{2}{5} = \frac{96\sqrt{3}}{25\sqrt{5}} = \frac{96\sqrt{15}}{125}$. This is not among the options.
Let's re-evaluate the expression $E = 16\sin^3 \frac{x}{2} \cos^2 \frac{x}{2}$.
Let $y = \frac{x}{2}$. Then $E = 16\sin^3 y \cos^2 y$. We need to maximize $f(y) = 16\sin^3 y \cos^2 y$ for $0 \le y \le \pi/2$.
$f'(y) = 16 [3\sin^2 y \cos y \cdot \cos^2 y + \sin^3 y \cdot 2\cos y (-\sin y)]$
$= 16 [3\sin^2 y \cos^3 y - 2\sin^4 y \cos y]$
$= 16 \sin^2 y \cos y [3\cos^2 y - 2\sin^2 y]$.
Set $f'(y) = 0$.
$\sin y = 0 \Rightarrow y=0$. $f(0)=0$.
$\cos y = 0 \Rightarrow y=\pi/2$. $f(\pi/2)=16(1)^3(0)^2 = 0$.
$3\cos^2 y - 2\sin^2 y = 0 \Rightarrow 3\cos^2 y = 2\sin^2 y \Rightarrow 	an^2 y = \frac{3}{2}$.
Since $0 \le y \le \pi/2$, $	an y = \sqrt{3/2}$.
If $	an y = \sqrt{3/2}$, then $\sin y = \frac{\sqrt{3}}{\sqrt{3+2}} = \frac{\sqrt{3}}{\sqrt{5}}$ and $\cos y = \frac{\sqrt{2}}{\sqrt{3+2}} = \frac{\sqrt{2}}{\sqrt{5}}$.
Substitute these values into $E = 16\sin^3 y \cos^2 y$:
$E = 16 (\frac{\sqrt{3}}{\sqrt{5}})^3 (\frac{\sqrt{2}}{\sqrt{5}})^2 = 16 \frac{3\sqrt{3}}{5\sqrt{5}} \frac{2}{5} = \frac{96\sqrt{3}}{25\sqrt{5}} = \frac{96\sqrt{15}}{125}$. This is still the same value.
Let's check the solution provided in the image. It uses $E = 4\sin x (1+\cos x)$.
This is $E = 16\sin^3(x/2)\cos^2(x/2)$.
We know $\sin(x/2)\cos(x/2) = \frac{1}{2}\sin x$.
So $E = 16 \sin(x/2) (\sin(x/2)\cos(x/2))^2 = 16 \sin(x/2) (\frac{1}{2}\sin x)^2 = 16 \sin(x/2) \frac{1}{4}\sin^2 x = 4\sin(x/2)\sin^2 x$.
This is not $4\sin x (1+\cos x)$.
Let's try to convert $E = 16\sin^3 \frac{x}{2} \cos^2 \frac{x}{2}$ to an expression in terms of $\cos x$.
$E = 16 \sin \frac{x}{2} (\sin^2 \frac{x}{2}) (\cos^2 \frac{x}{2}) = 16 \sin \frac{x}{2} (\frac{1-\cos x}{2}) (\frac{1+\cos x}{2})$
$E = 4 \sin \frac{x}{2} (1-\cos^2 x) = 4 \sin \frac{x}{2} \sin^2 x$.
This is correct. Now, let $u = \cos x$. Then $\sin x = \sqrt{1-u^2}$ (since $0 \le x \le \pi$, $\sin x \ge 0$).
Also $\sin(x/2) = \sqrt{\frac{1-\cos x}{2}} = \sqrt{\frac{1-u}{2}}$.
So $E = 4 \sqrt{\frac{1-u}{2}} (1-u^2) = 4 \sqrt{\frac{1-u}{2}} (1-u)(1+u)$.
$E = 4 \sqrt{\frac{1-u}{2}} (\sqrt{1-u})^2 (1+u) = 4 \frac{(1-u)^{3/2}}{\sqrt{2}} (1+u) = 2\sqrt{2} (1-u)^{3/2} (1+u)$.
Let $f(u) = (1-u)^{3/2} (1+u)$. We need to maximize this for $-1 \le u \le 1$ (since $0 \le x \le \pi$, $\cos x$ ranges from $-1$ to $1$).
$f'(u) = \frac{3}{2}(1-u)^{1/2}(-1)(1+u) + (1-u)^{3/2}(1)$
$= (1-u)^{1/2} [-\frac{3}{2}(1+u) + (1-u)]$
$= (1-u)^{1/2} [-\frac{3}{2} - \frac{3}{2}u + 1 - u]$
$= (1-u)^{1/2} [-\frac{1}{2} - \frac{5}{2}u] = -\frac{1}{2}(1-u)^{1/2} (1+5u)$.
Set $f'(u) = 0$.
$u=1$ (endpoint) or $1+5u=0 \Rightarrow u = -1/5$.
Evaluate $E$ at $u=1$, $u=-1$, $u=-1/5$.
If $u=1$ (i.e., $x=0$), $E = 0$.
If $u=-1$ (i.e., $x=\pi$), $E = 0$.
If $u=-1/5$ (i.e., $\cos x = -1/5$):
$E = 2\sqrt{2} (1 - (-1/5))^{3/2} (1 + (-1/5))$
$= 2\sqrt{2} (6/5)^{3/2} (4/5) = 2\sqrt{2} \frac{6\sqrt{6}}{5\sqrt{5}} \frac{4}{5} = \frac{48\sqrt{12}}{25\sqrt{5}} = \frac{48 \cdot 2\sqrt{3}}{25\sqrt{5}} = \frac{96\sqrt{3}}{25\sqrt{5}} = \frac{96\sqrt{15}}{125}$. Still the same value.
Let's look at the solution's approach: $E = 4\sin x (1+\cos x)$. This is not the original expression.
The solution seems to have made a substitution $E = 4\sin x (1+\cos x)$ from somewhere else.
Let's assume the solution's $E = 4\sin x (1+\cos x)$ is what they are maximizing.
Let $f(x) = 4\sin x (1+\cos x)$. For $0 \le x \le \pi$.
$f'(x) = 4 [\cos x (1+\cos x) + \sin x (-\sin x)]$
$= 4 [\cos x + \cos^2 x - \sin^2 x]$
$= 4 [\cos x + \cos^2 x - (1-\cos^2 x)]$
$= 4 [\cos x + 2\cos^2 x - 1]$.
Let $c = \cos x$. $f'(x) = 4(2c^2+c-1)$.
Set $f'(x) = 0 \Rightarrow 2c^2+c-1 = 0$.
$(2c-1)(c+1) = 0$.
So $c = 1/2$ or $c = -1$.
If $c = 1/2$, then $\cos x = 1/2 \Rightarrow x = \pi/3$ (since $0 \le x \le \pi$).
If $c = -1$, then $\cos x = -1 \Rightarrow x = \pi$.
Evaluate $f(x)$ at critical points and endpoints:
$f(0) = 4\sin(0)(1+\cos(0)) = 0$.
$f(\pi) = 4\sin(\pi)(1+\cos(\pi)) = 0$.
$f(\pi/3) = 4\sin(\pi/3)(1+\cos(\pi/3)) = 4(\frac{\sqrt{3}}{2})(1+\frac{1}{2}) = 2\sqrt{3}(\frac{3}{2}) = 3\sqrt{3}$.
This value $3\sqrt{3}$ is option (2).
So the question in the image is actually asking for the maximum of $4\sin x (1+\cos x)$, not $16\sin^3 \frac{x}{2} \cos^2 \frac{x}{2}$.
Let's verify if $16\sin^3 \frac{x}{2} \cos^2 \frac{x}{2}$ can be simplified to $4\sin x (1+\cos x)$.
$16\sin^3 \frac{x}{2} \cos^2 \frac{x}{2} = 16 \sin \frac{x}{2} (\sin^2 \frac{x}{2}) (\cos^2 \frac{x}{2})$
$= 16 \sin \frac{x}{2} (\frac{1-\cos x}{2}) (\frac{1+\cos x}{2})$
$= 4 \sin \frac{x}{2} (1-\cos^2 x) = 4 \sin \frac{x}{2} \sin^2 x$.
And $4\sin x (1+\cos x) = 4 (2\sin \frac{x}{2} \cos \frac{x}{2}) (2\cos^2 \frac{x}{2}) = 16 \sin \frac{x}{2} \cos^3 \frac{x}{2}$.
These two expressions are different. The question text is $16\sin^3 \frac{x}{2} \cos^2 \frac{x}{2}$. The solution is for $16\sin \frac{x}{2} \cos^3 \frac{x}{2}$ (or $4\sin x (1+\cos x)$).
I will extract the question as written and provide the answer based on the solution's interpretation.
Let's assume the question meant $16\sin \frac{x}{2} \cos^3 \frac{x}{2}$.
Let $y = x/2$. Maximize $16\sin y \cos^3 y$ for $0 \le y \le \pi/2$.
$f(y) = 16\sin y \cos^3 y$.
$f'(y) = 16 [\cos y \cos^3 y + \sin y (3\cos^2 y)(-\sin y)]$
$= 16 [\cos^4 y - 3\sin^2 y \cos^2 y]$
$= 16 \cos^2 y [\cos^2 y - 3\sin^2 y]$.
Set $f'(y) = 0$.
$\cos y = 0 \Rightarrow y=\pi/2$. $f(\pi/2)=0$.
$\cos^2 y - 3\sin^2 y = 0 \Rightarrow \cos^2 y = 3\sin^2 y \Rightarrow 	an^2 y = 1/3$.
Since $0 \le y \le \pi/2$, $	an y = 1/\sqrt{3}$. So $y = \pi/6$.
At $y=\pi/6$ (i.e., $x=\pi/3$):
$f(\pi/6) = 16 \sin(\pi/6) \cos^3(\pi/6) = 16 (1/2) (\sqrt{3}/2)^3 = 8 \frac{3\sqrt{3}}{8} = 3\sqrt{3}$.
This matches the solution's answer. So the question text in the image is likely a typo and should have been $16\sin \frac{x}{2} \cos^3 \frac{x}{2}$ or $4\sin x (1+\cos x)$. I will use the question text as written, but the correct answer will be based on the solution's implicit question.
Given the discrepancy, I will lower the confidence slightly.
Let's stick to the question as written: $E = 16\sin^3 \frac{x}{2} \cos^2 \frac{x}{2}$.
We found the maximum value is $\frac{96\sqrt{15}}{125}$. This is approximately $\frac{96 	imes 3.87}{125} \approx \frac{371.52}{125} \approx 2.97$.
Option (2) is $3\sqrt{3} \approx 3 	imes 1.732 = 5.196$.
Option (1) is $3\sqrt{3}/2 \approx 2.598$.
Option (3) is $4\sqrt{3} \approx 6.928$.
Option (4) is $6\sqrt{3} \approx 10.392$.
Since $2.97$ is closest to $3\sqrt{3}/2$, if the options are correct for the written question, then (1) would be the answer. But the solution gives (2).
This is a clear discrepancy. I will follow the solution's answer and assume the question was intended to be $16\sin \frac{x}{2} \cos^3 \frac{x}{2}$ or $4\sin x (1+\cos x)$. I will note this in the explanation and lower confidence.

Answer

$3\sqrt{3}/2$

Answer

$3\sqrt{3}$

Answer

$4\sqrt{3}$

Answer

$6\sqrt{3}$

Related Questionsin Mathematics

More from JEE Main

Browse Engineering Questions by Subject

Practice More JEE Main Questions