JEE Advanced · Mathematics · 25. AOD

Let $f, g : [- 1, 2] \to R$ be continuous functions which are twice differentiable on the interval $(- 1, 2)$ . Let the values of $f$ and $g$ at the points $- 1, 0$ and $2$ be as given in the following table:

$x = - 1$ $x = 0$ $x = 2$

$f (x)$ 3 6 0

$g (x)$ 0 1 -1

In each of the intervals $(- 1, 0)$ and $(0, 2)$ the function $(f - 3 g) "$ never vanishes. Then the correct statement(s) is(are)

A $f^{'} (x) - 3 g^{'} (x) = 0$ has exactly three solutions in $(- 1, 0) \cup (0, 2)$
B $f^{'} (x) - 3 g^{'} (x) = 0$ has exactly one solution in (-1, 0)
C $f^{'} (x) - 3 g^{'} (x) = 0$ has exactly one solution in (0, 2)
D $f^{'} (x) - 3 g^{'} (x) = 0$ has exactly two solutions in (-1, 0) and exactly two solutions in (0, 2)

Verified Solution

Answer & Solution

Correct Answer

(C) $f^{'} (x) - 3 g^{'} (x) = 0$ has exactly one solution in (0, 2)

Step-by-step Solution

Detailed explanation

h (x) = f (x) - 3 g (x)

h (- 1) = 3 - 0 = 3

h (0) = 6 - 3 = 3

h (2) = 0 + 3 = 3

h' (x)

has at least one root is

(- 1,0)

And

h^{'} (x)

has at least one in

(0,2)

Hence

h^{''} (x)

atleast one root in

(- 1,2)

but
Also,

{(f - 3 g)}^{''} \neq 0 i n (- 1, 0) a n d (0, 2)

h^{''} (x) = 0 o n l y a t x = 0, s o

f^{'} (x) - 3 g^{'} (x)

has exactly one solution in (-1, 0) and (0, 2)

Apply the relevant identity from the chapter and substitute the values established in the previous step, keeping the original constraint in view.

Use the simplified expression to isolate the unknown, then plug the result back into the question setup to confirm the working is internally consistent.

Cross-check against a boundary or limiting case. Both the dimensional balance and the qualitative behaviour at the extreme should agree with the candidate answer.

Combine the steps above to identify the correct option. The remaining choices each fail at least one of the consistency, dimensional, or boundary checks.

Same subject

Explore more questions on app

From JEE Advanced

List-

I

shows different radioactive decay processes and List-

II

provides possible emitted particles. Match each entry in List-

I

with an appropriate entry from List-

II

, and choose the correct option.

	List- $I$		List- $II$
$P$	${^{238}}_{92} U {\to^{234}}_{91} P a$	$1$	one $α$ particle and one $β^{+}$ particle
$Q$	${^{214}}_{82} P b {\to^{210}}_{82} P b$	$2$	three $β^{-}$ particles and one $α$ particle
$R$	${^{210}}_{81} T l {\to^{206}}_{82} P b$	$3$	two $β^{-}$ particles and one $α$ particle
$S$	${^{228}}_{91} P a {\to^{224}}_{88} R a$	$4$	one $α$ particle and one $β^{-}$ particle
		$5$	one $α$ particle and two $β^{+}$ particles

JEE Advanced 2023 Easy

Let $\mathbb{R}$ denote the set of all real numbers. Let $a_{\mathrm{i}}, b_{\mathrm{i}} \in \mathbb{R}$ for $\mathrm{i} \in\{1,2,3\}$.
Define the functions $f: \mathbb{R} \rightarrow \mathbb{R}, g: \mathbb{R} \rightarrow \mathbb{R}$, and $h: \mathbb{R} \rightarrow \mathbb{R}$ by
$\begin{aligned} & f(x)=a_1+10 x+a_2 x^2+a_3 x^3+x^4, \\ & g(x)=b_1+3 x+b_2 x^2+b_3 x^3+x^4, \\ & h(x)=f(x+1)-g(x+2) .\end{aligned}$
If $f(x) \neq g(x)$ for every $x \in \mathbb{R}$, then the coefficient of $x^3$ in $h(x)$ is

JEE Advanced 2025 Easy

Explore more questions on app