Q.17. The product of all possible values of , for which , is :

Q: What is the correct answer for JEE Main Mathematics Q17?

The correct answer is C. Sol. The limit is of the form $\frac{0}{0}$ as $x \to 0$. We can use the standard limit $\lim_{x\to 0} \frac{1-\cos(kx)}{x^2} = \frac{k^2}{2}$. Also, $\sin^2(x) \approx x^2$ for small $x$. Let $f(x) = \cos(x)\cos((a+1)x)\cos((a+2)x)$. We know that $1 - \cos(Ax)\cos(Bx)\cos(Cx) \approx \frac{1}{2}(A^2+B^2+C^2)x^2$ for small $x$ if $A,B,C$ are coefficients of $x$. More generally, for $f(x) = 1 - \co

Q: Q.17. The product of all possible values of , for which , is :

The correct answer is C. Sol. The limit is of the form $\frac{0}{0}$ as $x \to 0$. We can use the standard limit $\lim_{x\to 0} \frac{1-\cos(kx)}{x^2} = \frac{k^2}{2}$. Also, $\sin^2(x) \approx x^2$ for small $x$. Let $f(x) = \cos(x)\cos((a+1)x)\cos((a+2)x)$. We know that $1 - \cos(Ax)\cos(Bx)\cos(Cx) \approx \frac{1}{2}(A^2+B^2+C^2)x^2$ for small $x$ if $A,B,C$ are coefficients of $x$. More generally, for $f(x) = 1 - \co

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

A. -2 · B. 1 · C. -1 · D. 5/4

Question

Q.17. The product of all possible values of , for which , is :

FirstInTest · Accepted Answer

Sol. The limit is of the form $\frac{0}{0}$ as $x 	o 0$. We can use the standard limit $\lim_{x	o 0} \frac{1-\cos(kx)}{x^2} = \frac{k^2}{2}$.
Also, $\sin^2(x) \approx x^2$ for small $x$.
Let $f(x) = \cos(x)\cos((a+1)x)\cos((a+2)x)$.
We know that $1 - \cos(Ax)\cos(Bx)\cos(Cx) \approx \frac{1}{2}(A^2+B^2+C^2)x^2$ for small $x$ if $A,B,C$ are coefficients of $x$.
More generally, for $f(x) = 1 - \cos(x)\cos((a+1)x)\cos((a+2)x)$, as $x 	o 0$, $f(x) \approx 1 - (1 - \frac{x^2}{2})(1 - \frac{(a+1)^2x^2}{2})(1 - \frac{(a+2)^2x^2}{2})$.
Ignoring higher order terms, this is $1 - (1 - \frac{x^2}{2} - \frac{(a+1)^2x^2}{2} - \frac{(a+2)^2x^2}{2}) = \frac{x^2}{2}(1 + (a+1)^2 + (a+2)^2)$.
So, $\lim_{x	o 0} \frac{1-\cos(x)\cos((a+1)x)\cos((a+2)x)}{\sin^2(x)} = \lim_{x	o 0} \frac{\frac{x^2}{2}(1 + (a+1)^2 + (a+2)^2)}{x^2}$
$= \frac{1}{2}(1 + (a+1)^2 + (a+2)^2)$.
Given that this limit is 2:
$\frac{1}{2}(1 + (a+1)^2 + (a+2)^2) = 2$
$1 + (a+1)^2 + (a+2)^2 = 4$
$1 + (a^2+2a+1) + (a^2+4a+4) = 4$
$2a^2 + 6a + 6 = 4$
$2a^2 + 6a + 2 = 0$
$a^2 + 3a + 1 = 0$.
This is a quadratic equation for $a$. The product of the roots (possible values of $a$) is given by $c/a = 1/1 = 1$.

Answer

-2

Answer

1

Answer

-1

Answer

5/4

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