The number of integral values of k for which the equation ex=(k/x-3) has exactly two solutions is

Question

The number of integral values of k for which the equation ex=(k/x-3) has exactly two solutions is (A) 5 (B) 6 (C) 7 (D) 9

Tardigrade Admin · Accepted Answer

ex(x-3)=k Let f(x)=ex(x-3) f prime(x)=ex-1+(x-3) ex=ex(x-2) From the graph it is clear that, for exactly two roots k ∈(-e2, 0) ∴ number of integral values of k is, 7 .

Q. The number of integral values of $k$ for which the equation $e^{x} = \frac{k}{x - 3}$ has exactly two solutions is

5

6

7

9

$e^{x} (x - 3) = k$
Let $f (x) = e^{x} (x - 3)$
$f^{'} (x) = e^{x} - 1 + (x - 3) e^{x} = e^{x} (x - 2)$
From the graph it is clear that, for exactly two roots $k \in (- e^{2}, 0)$
$∴$ number of integral values of $k$ is, 7 .

Q. The number of integral values of k for which the equation ex=x−3k​ has exactly two solutions is

5

6

7

9

ex(x−3)=k Let f(x)=ex(x−3) f′(x)=ex−1+(x−3)ex=ex(x−2) From the graph it is clear that, for exactly two roots k∈(−e2,0) ∴ number of integral values of k is, 7 .

Q. The number of integral values of $k$ for which the equation $e^{x} = \frac{k}{x - 3}$ has exactly two solutions is

$e^{x} (x - 3) = k$
Let $f (x) = e^{x} (x - 3)$
$f^{'} (x) = e^{x} - 1 + (x - 3) e^{x} = e^{x} (x - 2)$
From the graph it is clear that, for exactly two roots $k \in (- e^{2}, 0)$
$∴$ number of integral values of $k$ is, 7 .