Q.24. For the functions , and , , let . If the first term of a G.P. is , its common ratio is and th

Q: What is the correct answer for JEE Main Mathematics Q24?

The correct answer is 1279. Sol. For $f(\theta) = a \tan^2 \theta + b \cot^2 \theta$. By AM-GM inequality, $a \tan^2 \theta + b \cot^2 \theta \ge 2\sqrt{(a \tan^2 \theta)(b \cot^2 \theta)} = 2\sqrt{ab}$. So, $\min f(\theta) = 2\sqrt{ab}$. For $g(\theta) = a \sin^2 \theta + b \cos^2 \theta = a \sin^2 \theta + b(1 - \sin^2 \theta) = (a-b)\sin^2 \theta + b$. Since $a > b > 0$, $a-b > 0$. For $\theta \in (0, \pi/2)$, $\sin^2 \th

Q: Q.24. For the functions , and , , let . If the first term of a G.P. is , its common ratio is and the sum of its first 10 terms is , , then is equal to ________.

The correct answer is 1279. Sol. For $f(\theta) = a \tan^2 \theta + b \cot^2 \theta$. By AM-GM inequality, $a \tan^2 \theta + b \cot^2 \theta \ge 2\sqrt{(a \tan^2 \theta)(b \cot^2 \theta)} = 2\sqrt{ab}$. So, $\min f(\theta) = 2\sqrt{ab}$. For $g(\theta) = a \sin^2 \theta + b \cos^2 \theta = a \sin^2 \theta + b(1 - \sin^2 \theta) = (a-b)\sin^2 \theta + b$. Since $a > b > 0$, $a-b > 0$. For $\theta \in (0, \pi/2)$, $\sin^2 \th

Question

Q.24. For the functions , and , , let . If the first term of a G.P. is , its common ratio is  and the sum of its first 10 terms is , , then  is equal to ________.

FirstInTest · Accepted Answer

Sol. For $f(	heta) = a 	an^2 	heta + b \cot^2 	heta$. By AM-GM inequality, $a 	an^2 	heta + b \cot^2 	heta \ge 2\sqrt{(a 	an^2 	heta)(b \cot^2 	heta)} = 2\sqrt{ab}$. So, $\min f(	heta) = 2\sqrt{ab}$. For $g(	heta) = a \sin^2 	heta + b \cos^2 	heta = a \sin^2 	heta + b(1 - \sin^2 	heta) = (a-b)\sin^2 	heta + b$. Since $a > b > 0$, $a-b > 0$. For $	heta \in (0, \pi/2)$, $\sin^2 	heta \in (0, 1)$. So, $\max g(	heta)$ occurs when $\sin^2 	heta$ is maximum, i.e., $\sin^2 	heta 	o 1$. So, $\max g(	heta) = (a-b)(1) + b = a$. Given $\min f(	heta) = \max g(	heta)$, so $2\sqrt{ab} = a$. Squaring both sides, $4ab = a^2$. Since $a > 0$, we can divide by $a$, so $4b = a$. The first term of the G.P. is $A = \frac{a}{2b}$. Since $a=4b$, $A = \frac{4b}{2b} = 2$. The common ratio is $r = \frac{2b}{a}$. Since $a=4b$, $r = \frac{2b}{4b} = \frac{1}{2}$. The sum of the first 10 terms of the G.P. is $S_{10} = A \frac{1-r^{10}}{1-r} = 2 \frac{1 - (1/2)^{10}}{1 - 1/2} = 2 \frac{1 - 1/1024}{1/2} = 2 \cdot 2 \cdot (\frac{1024-1}{1024}) = 4 \cdot \frac{1023}{1024} = \frac{1023}{256}$. This is given as $\frac{m}{n}$. So $m=1023$ and $n=256$. We need to check if $	ext{gcd}(m, n) = 1$. $1023 = 3 	imes 11 	imes 31$. $256 = 2^8$. They are coprime, so $	ext{gcd}(1023, 256) = 1$. We need to find $m+n = 1023 + 256 = 1279$.

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