If $Sin (x + y)+Cos(x + y) = Log (x + y),$ then $\frac {d^2y}{dx^2}$=
- 1
- -1
- 0
- $\frac {-y}{x}$
The Correct Option is C
Solution and Explanation
$\sin (x+y)+\cos (x+y)=\log (x+y)$
On differentiating both sides, we get
$\cos (x+y)\left(1+\frac{d y}{d x}\right)-\sin (x+y)\left(1+\frac{d y}{d x}\right)$
$=\frac{1}{x+y}\left(1+\frac{d y}{d x}\right) $
$\Rightarrow \frac{d y}{d x}+1=0 \\$
$\Rightarrow \frac{d y}{d x}=-1$
$\Rightarrow \frac{d^{2} y}{d x^{2}}=0$
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Concepts Used:
Trigonometric Equations
Trigonometric equation is an equation involving one or more trigonometric ratios of unknown angles. It is expressed as ratios of sine(sin), cosine(cos), tangent(tan), cotangent(cot), secant(sec), cosecant(cosec) angles. For example, cos2 x + 5 sin x = 0 is a trigonometric equation. All possible values which satisfy the given trigonometric equation are called solutions of the given trigonometric equation.
A list of trigonometric equations and their solutions are given below:
| Trigonometrical equations | General Solutions |
| sin θ = 0 | θ = nπ |
| cos θ = 0 | θ = (nπ + π/2) |
| cos θ = 0 | θ = nπ |
| sin θ = 1 | θ = (2nπ + π/2) = (4n+1) π/2 |
| cos θ = 1 | θ = 2nπ |
| sin θ = sin α | θ = nπ + (-1)n α, where α ∈ [-π/2, π/2] |
| cos θ = cos α | θ = 2nπ ± α, where α ∈ (0, π] |
| tan θ = tan α | θ = nπ + α, where α ∈ (-π/2, π/2] |
| sin 2θ = sin 2α | θ = nπ ± α |
| cos 2θ = cos 2α | θ = nπ ± α |
| tan 2θ = tan 2α | θ = nπ ± α |