COMPASSLearning

Talk to us Register Login

JEE Main · 2025 · 07 Apr 2025 · Shift 2 · Limits

For

t > - 1

, let

α_{t}

and

β_{t}

be the roots of the equation

((t + 2)^{\frac{1}{7}} - 1) x^{2} + ((t + 2)^{\frac{1}{6}} - 1) x + ((t + 2)^{\frac{1}{21}} - 1) = 0

.

If

lim_{t \to - 1^{+}} α_{t} = a

and

lim_{t \to - 1^{+}} β_{t} = b

, then

72 (a + b)^{2}

is equal to ______.

Answer: 98

Solution

By Vieta's formulas the sum of the roots is

α_{t} + β_{t} = - \frac{( t + 2 ) ^{1/6} - 1}{( t + 2 ) ^{1/7} - 1}

. Writing

y = t + 2

, as

t \to - 1^{+}

one has

y \to 1^{+}

, and

lim_{y \to 1} \frac{y ^{p} - 1}{y ^{q} - 1} = \frac{p}{q}

. Hence

a + b = - \frac{1/6}{1/7} = - \frac{7}{6}

. Therefore

72 (a + b)^{2} = 72 \cdot \frac{49}{36} = 98

.

JEE Main · 2025 · 07 Apr 2025 · Shift 2 · 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 →