JEE MATHEMATICS MCQ: LIMITS

1 Hard

1 of 1

Analyzing +4 | -1

g\left( x \right)

is a polynomial satisfying

g\left( x \right)g\left( y \right) = g\left( x \right) + g\left( y \right) + g\left( {xy} \right) - 2

for all real

x

and

y

and

g\left( 2 \right) = 5

, then

\lim\limits_{x \to 3} g\left( x \right)

View answer and explanation

Correct option: B

Option B (10) is correct because by substituting the given condition into the functional equation and solving for g(x), we find that g(x) = x² + 1, and thus

\lim\limits_{x \to 3} g(x) = 10

More Questions from LIMITS

Practice All Chapter MCQs →

\lim \limits_{{x \rightarrow 0}}\left(\frac{1}{x^2}-\operatorname{cosec}^2 x\right)

is equal to

\lim _{x \rightarrow \frac{\pi}{2}} \frac{\cos x}{(1-\sin x)^{2 / 3}}=

The value of

\lim\limits_{h \to 0} 2\left\{ {\frac{{\sqrt 3 \sin \left( {\frac{\pi }{6} + h} \right) - \cos \left( {\frac{\pi }{6} + h} \right)}}{{\sqrt 3 h\left( {\sqrt 3 \cos \,h - \sin \,h} \right)}}} \right\}

\lim \limits_{{x \rightarrow \infty}} x\left(e^{1 / x}-1\right)

is equal to If

f(x)=\frac{(x+1)^7 \sqrt{1+x^2}}{\left(x^2-x+1\right)^6}

, then the value of

f^{\prime}(0)

is equal to

\lim\limits_{n \to \infty } {\left( {1 + \sin \frac{a}{n}} \right)^n}

equals