COMPASSLearning

Talk to us Register Login

JEE Main · 2021 · 20 Jul 2021 · Shift 1 · Limits

If the value of

lim_{x \to 0} (2 - cos x cos 2 x)^{(\frac{x + 2}{x ^{2}})}

is equal to

e^{a}

, then a is equal to ______.

Answer: 3

Solution

Let

L = lim_{x \to 0} (2 - cos x cos 2 x)^{(x + 2) / x^{2}}

.

Taking logarithm:

ln L = lim_{x \to 0} \frac{x + 2}{x ^{2}} ln (2 - cos x cos 2 x)

.

Expand

cos x \approx 1 - \frac{x ^{2}}{2}

and

cos 2 x \approx 1 - 2 x^{2}

, so

cos 2 x \approx 1 - x^{2}

.

Thus

cos x cos 2 x \approx (1 - \frac{x ^{2}}{2}) (1 - x^{2}) \approx 1 - \frac{3 x ^{2}}{2}

.

So

2 - cos x cos 2 x \approx 1 + \frac{3 x ^{2}}{2}

.

Then

ln (2 - cos x cos 2 x) \approx \frac{3 x ^{2}}{2}

.

Thus

ln L = lim_{x \to 0} \frac{x + 2}{x ^{2}} \cdot \frac{3 x ^{2}}{2} = lim_{x \to 0} \frac{3 ( x + 2 )}{2} = 3

.

Therefore

L = e^{3}

, so

a = 3

.

JEE Main · 2021 · 20 Jul 2021 · Shift 1 · Integer answer

Practise more: all JEE Main PYQs on Limits →

More JEE Main questions on Limits

Let $f : R \to (0, \infty)$ be a twice differentiable function such that $f (3) = 18, f^{'} (3) = 0$ and $f^{''} (3) = 4$ .
Then …
JEE Main 2026View solution →
If $lim_{x \to 0} \frac{e ^{(a - 1) x} + 2 c o s b x + ( c - 2 ) e ^{- x}}{x c o s x - l o g _{e} ( 1 + x )} = 2$ , then $a^{2} + b^{2} + c^{2}$ is equal to :
JEE Main 2026View solution →
The value of
$lim_{x \to 0} \frac{l o g _{e} ( s e c ( e x ) \cdot s e c ( e ^{2} x ) \cdot \dots \cdot s e c ( e ^{10} x ) )}{e ^{2} - e ^{2 c o s x}}$
is equal to
JEE Main 2026View solution →
If $lim_{x \to 2} \frac{s i n ( x ^{3} - 5 x ^{2} + a x + b )}{( x - 1 - 1 ) l o g _{e} ( x - 1 )} = m$ , then $a + b + m$ is equal to:
JEE Main 2026View solution →
The product of all possible values of $α$ , for which $lim_{x \to 0} (\frac{1 - c o s ( α x ) c o s (( α + 1 ) x ) c o s (( α + 2 ) x )}{s i n ^{2} (( α + 1 ) x )}) = 2$ , is :
JEE Main 2026View solution →
Let $f (x) = lim_{y \to 0} \frac{( 1 - c o s ( x y )) t a n ( x y )}{y ^{3}}$ . Then the number of solutions of the equation $f (x) = sin x$ , $x \in R$ is:
JEE Main 2026View solution →