The slope of the tangent to the curve y=e^ x cos x is minimum at x=α, 0 ≤ a ≤ 2 π , then the value of α is

Correct answer: B. π Let m be the slope of the tangent to the curve y=e^ x x Then, m=(d y)/(d x)=e^ x ( x- x) Diff. w.r.t 'x' array r dm dx = ^ x ( x- x)+ ^ x (- x- x) =-2 ^ x x array and ^ 2 ~m dx^ 2 =-2 ^ x ( x+ x) Put dm dx =0 x=0 x=0, π, 2 π Clearly, ^ 2 ~m dx^ 2 0 for x=π Thus, y is minimum at x=π . Hence the value of α=π

BITSAT 2015

Exam: BITSAT
Year: 2015
Subject: Mathematics
Chapter: Application of Derivatives

BITSAT 2015 · Mathematics · Application of Derivatives

The slope of the tangent to the curve y=e^ x cos x is minimum at...

The slope of the tangent to the curve

y=e^{x}

cos

x

is minimum at

x=\alpha, 0 \leq a \leq 2 \pi

, then the value of

\alpha

Options

A $\underline{0}$
B $\pi$
C $2 \pi$
D $3 \pi 2$

Correct answer

B. $\pi$

Step-by-step solution

Let

m

be the slope of the tangent to the curve

y=e^{x} \cos x

Then,

m=\frac{d y}{d x}=e^{x}(\cos x-\sin x)

Diff. w.r.t 'x'

\begin{array}{r} \Rightarrow\frac{\mathrm{dm}}{\mathrm{dx}}=\mathrme^{\mathrm{x}}(\cos \mathrm{x}-\sin \mathrm{x})+\mathrme^{\mathrm{x}}(-\sin \mathrm{x}-\cos \mathrm{x})\\ =-2 \mathrme^{\mathrm{x}} \sin \mathrm{x} \end{array}

and

\frac{\mathrmd^{2} \mathrm{~m}}{\mathrm{dx}^{2}}=-2 \mathrme^{\mathrm{x}}(\sin \mathrm{x}+\cos \mathrm{x})

Put

\frac{\mathrm{dm}}{\mathrm{dx}}=0 \Rightarrow \sin \mathrm{x}=0 \Rightarrow \mathrm{x}=0, \pi, 2 \pi

Clearly,

\frac{\mathrmd^{2} \mathrm{~m}}{\mathrm{dx}^{2}} \gt 0

for

\mathrm{x}=\pi

Thus,

y

is minimum at

x=\pi

. Hence the value of

\alpha=\pi

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