Let fx- -2xe1-t2 dt and gx-fhx where hx is defined for all x- R- If g-2-e4 and h-2-1 then absolute value of sum of all possible values of h2 is

The correct option is A 2 f(x)=∫_ 2^xe^(1+t)^2 dt #160; f'(x)=e^(1+x)^2 g(x)=f(h(x)) g'(x)=f'(h(x)).h'(x) g'(2)=f'(h(2)).h'(2) e ...

You visited us 1 times! Enjoying our articles? Unlock Full Access!

Byju's Answer

Standard XII

Mathematics

Chain Rule of Differentiation

Let fx=∫ -2...

Question

Let $f (x) = \int_{- 2}^{x} e^{{(1 + t)}^{2}} d t$ and $g (x) = f (h (x))$ where h(x) is defined for all $x \in R$ . If $g^{'} (2) = e^{4}$ and $h^{'} (2) = 1$ then absolute value of sum of all possible values of h(2) is

-2

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

No worries! We‘ve got your back. Try BYJU‘S free classes today!

-1

No worries! We‘ve got your back. Try BYJU‘S free classes today!

Open in App

Solution

The correct option is A -2
$f (x) = \int_{- 2}^{x} e^{(1 + t)^{2}}$ dt
$f^{'} (x) = e^{(1 + x)^{2}}$
$g (x) = f (h (x))$
$g^{'} (x) = f^{'} (h (x)) . h^{'} (x)$
$g^{'} (2) = f^{'} (h (2)) . h^{'} (2)$
$e^{4} = f^{'} (h (2)) .1$
$f^{'} (h (2)) = e^{4}$
$e^{(1 + h (2))^{2}} = e^{4}$
$\Rightarrow (1 + h (2))^{2} = 4$
$\Rightarrow 1 + h (2) = \pm 2$
$\Rightarrow h (2) = 1$ or $h (2) = - 3$
$\Rightarrow$ Sum $= - 2$

Suggest Corrections

Similar questions

Q. Let

f (x) = x^{2} + \frac{1}{x^{2}}

and

g (x) = x - \frac{1}{x}, x \in R - - 1, 0, 1

. If

h (x) = \frac{f (x)}{g (x)}

, then the local minimum value of the value of

h (x)

is:

Q. Let

g : R \to R

be a differentiable function with

g (0) = 0, g^{'} (0) = 0 and g^{'} (1) \neq 0.

Let

f (x) = ⎧ ⎨ ⎩ \begin{matrix} \frac{x}{| x |} g (x), x \neq 0 0, x = 0 \end{matrix}

and

h (x) = e^{| x |}

for all

x \in R .

Let

(f \circ h) (x)

denotes

f (h (x))

and

(h \circ f) (x)

denotes

h (f (x)) .

Then which of the following is (are) true?

Q. Let

f (x) = x^{2} - \frac{1}{x^{2}}

and

g (x) = x - \frac{1}{x}, x \in R - - 1, 0, 1

. If

h (x) = \frac{f (x)}{g (x)}

then the local minimum value of

h (x)

is:

Q. If f, g and h are three valued function defined by

f (x) = \sqrt{x - 1}

g (x) = \frac{1}{x}

h (x) = 2 x^{2} + 3

find the value of

2 f + g + h

x = 3 x - 2

Q. Let

g : R \to R

be a differentiable function with

g (0) = 0, g^{'} (0) = 0

and

g^{'} (1) \neq 0

. Let

f (x) = {\begin{matrix} \frac{x}{| x |} g (x), & x \neq 0 0, & x = 0 \end{matrix}

and

h (x) = e^{| x |}

for all

x \in R

. Let

(f \cdot h) (x)

denotes

f (h (x))

and

(h \cdot f) (x)

denote

h (f (x))

. Then which of the following is (are) true?

Join BYJU'S Learning Program

Let f(x)=∫x−2e(1+t)2dt and g(x)=f(h(x)) where h(x) is defined for all x∈R. If g′(2)=e4 and h′(2)=1 then absolute value of sum of all possible values of h(2) is

The correct option is A -2f(x)=∫x−2e(1+t)2 dt f′(x)=e(1+x)2g(x)=f(h(x))g′(x)=f′(h(x)).h′(x)g′(2)=f′(h(2)).h′(2)e4=f′(h(2)).1f′(h(2))=e4e(1+h(2))2=e4⇒(1+h(2))2=4⇒1+h(2)=±2⇒h(2)=1 or h(2)=−3⇒ Sum =−2

Let $f (x) = \int_{- 2}^{x} e^{{(1 + t)}^{2}} d t$ and $g (x) = f (h (x))$ where h(x) is defined for all $x \in R$ . If $g^{'} (2) = e^{4}$ and $h^{'} (2) = 1$ then absolute value of sum of all possible values of h(2) is