If $\alpha$ is the positive root of the equation, $p(x)=x^{2}-x-2=0$, then $\lim _{x \rightarrow \alpha} \frac{\sqrt{1-\cos (p(x))}}{x+\alpha-4}$ is equal to:

JEE Main · 2020 · 05 Sep 2020 · Shift 1 · Limits

α

is the positive root of the equation,

p (x) = x^{2} - x - 2 = 0

, then

lim_{x \to α} \frac{1 - c o s ( p ( x ))}{x + α - 4}

is equal to:

Answer: (C)

The positive root of

x^{2} - x - 2 = 0

α = 2

, so the denominator

x + α - 4

becomes

x - 2

. Writing

1 - cos (p (x)) = 2 sin^{2} \frac{p ( x )}{2}

gives

1 - cos p (x) = 2 sin \frac{p ( x )}{2}

. As

x \to 2^{+}

p (x) = (x - 2) (x + 1) \to 0^{+}

, so

sin \frac{p ( x )}{2} \approx \frac{p ( x )}{2}

and the expression becomes

\frac{2 ( x - 2 ) ( x + 1 ) /2}{x - 2} = \frac{1}{2} (x + 1) \to \frac{3}{2}

JEE Main · 2020 · 05 Sep 2020 · Shift 1 · Single correct option

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 →