Trigonometric Function MCQ Quiz in मराठी - Objective Question with Answer for Trigonometric Function - मोफत PDF डाउनलोड करा

Concept:

Formulae

• sinθ = 2 sin\(\frac{\theta}{2}\) cos\(\frac{\theta}{2}\)
• 1 + cosθ = 2 cos²\(\frac{\theta}{2}\)

Calculation:

The given equation is y = tan⁻¹​\(\big (\frac{sin\ x}{1+cos\ x}\big) \)

⇒ y = tan⁻¹\(\big (\frac{sin \ (\frac{2x}{2})}{2cos^2\ \frac{x}{2}}\big) \)

⇒ y = tan⁻¹\(\big (\frac{2\ sin \ (\frac{x}{2})\ cos \ (\frac{x}{2})}{2\ cos^2\ \frac{x}{2}}\big) \)

⇒ y = tan⁻¹\(\big (\frac{\ sin \ (\frac{x}{2})}{\ cos\ \frac{x}{2}}\big) \)

⇒ y = tan⁻¹tan(\(\frac{x}{2}\))

⇒ y = \(\frac{x}{2}\)

Differentiate with respect to x on both sides we get

⇒ \(\frac{dy}{dx}=\frac{1}{2}\)​

∴ \(\frac{dy}{dx}=\frac{1}{2}\)

Explanation:

f(x) = cos(tan 3x) + sin(tan 3x)

Differentiate f(x) is

\(\frac{d}{dx}f(x)=\frac{d}{dx}[(cos(tan3x)+sin(tan3x))]\)

\(=\frac{d}{dx}(cos(tan3x))+\frac{d}{dx}(sin(tan3x))\)

\(=-sin(tan3x)\frac{d}{dx}(tan3x)+cos(tan3x)\frac{d}{dx}(tan3x)\)

= -3sin(tan 3x).sec² 3x + 3cos(tan 3x).sec² 3x

= 3sec² 3x(cos(tan 3x) - sin(tan 3x))

Concept:

• Chain Rule: If y = f(g(x)), then \(\frac{{dy}}{{dx}}\)= f '(g(x))g'(x)
• \(1+ \tan^2 x= \sec^2 x\)
• tan(tan-1x) = x

Calculation:

If y = sec(tan-1x)

⇒ \(y = \sqrt{1 + \tan^2(\tan^{-1}x)}\) 

⇒ \(y = \sqrt{1 +x^2}\) 

⇒ \(\frac{{dy}}{{dx}} ={1 \over 2 \sqrt{1 + x^2}} (1+x^2)'\)

⇒ \(\frac{{dy}}{{dx}} ={2x \over 2 \sqrt{1 + x^2}} \)

⇒ \(\frac{{dy}}{{dx}} ={x \over \sqrt{1 + x^2}} \)

At x = 1, \(\frac{{dy}}{{dx}}\)= \(\frac{1}{{\sqrt 2 }}\)

∴ The correct answer is option (4).

CONCEPT:

• \(\frac{{d\left( {{{\sin }^{ - 1}}x} \right)}}{{dx}} = \frac{1}{{\sqrt {1 - {x^2}} }}\)
• \(\frac{{d\left( {{e^x}} \right)}}{{dx}} = {e^x}\)

CALCULATION:

Let \(y = e^{sin^{-1} x}\)

Here, we have to find \(\frac{dy}{dx}\)

So, differentiating both the sides of \(y = e^{sin^{-1} x}\) with respect to x we get,

⇒ \(\frac{dy}{dx} = \frac{d(e^{sin^{-1} x})}{dx}\)

⇒ \(\frac{dy}{dx} = e^{sin^{-1} x} \ \cdot \frac{d(sin^{-1} x)}{dx}\)

As we know that, \(\frac{{d\left( {{{\sin }^{ - 1}}x} \right)}}{{dx}} = \frac{1}{{\sqrt {1 - {x^2}} }}\)

⇒ \(\frac{dy}{dx} = \frac{e^{sin^{-1} x}}{\sqrt {1 - x^2}}\)

Hence, correct option is 4.

Concept:

Formulae

• 1 + cos2θ = 2cos²θ
• 1 - cos2θ = 2sin²θ
• Sin(A+B) = sin A cos B + cos A sin B
• cos-1x = \(\frac{-1}{\sqrt {1-x^2}}\)

Calculation:

The given equation is \(\rm y = sin^{-1} \frac{\sqrt{ (1 + x)} + \sqrt {(1 - x)}}{2}\)

Putting x = cos2θ, θ = \(\frac{1}{2}\) cos^-1x

⇒ \(\rm y = sin^{-1} \frac{\sqrt{ (1 + cos2θ)} + \sqrt {(1 - cos2θ)}}{2}\)

⇒ \(\rm y = sin^{-1} \frac{\sqrt{ 2cos^2θ} + \sqrt {2sin^2θ}}{2}\)

⇒ \(\rm y = sin^{-1} \frac{\sqrt{2}cosθ + \sqrt {2}sinθ}{2}\)

⇒ \(\rm y = sin^{-1}\big(\frac{1}{\sqrt{2}}cosθ + \frac{1}{\sqrt {2}}sinθ\big)\)

⇒ \(\rm y = sin^{-1}\big(sin\frac{π}{4}cosθ + cos\frac{π}{4}sinθ\big)\)

⇒ \(\rm y = sin^{-1}sin\big(\frac{π}{4}+θ\big)\)

⇒ \(\rm y = \big(\frac{π}{4}+θ\big)\)

⇒ y = \(\frac{\pi}{4}\) + \(\frac{1}{2}\) cos-1x

Differentiating the above equation w.r.t x,

\(\rm \frac{dy}{dx}\) = \(\frac{1}{2} \frac{-1}{\sqrt {1-x^2}}\)

∴ \(\rm \frac{dy}{dx}\) = \(-\frac{1}{2\sqrt {1-x^2}}\)

Concept:

Derivatives of Trigonometric Functions:

\(\rm \frac{d}{dx}\sin x=\cos x\ \ \ \ \ \ \ \ \ \ \ \ \ \frac{d}{dx}\cos x=-\sin x\\ \frac{d}{dx}\tan x=\sec^2x\ \ \ \ \ \ \ \ \ \ \ \frac{d}{dx}\cot x=-\csc^2 x\\ \frac{d}{dx}\sec x=\tan x\sec x\ \ \ \ \frac{d}{dx}\csc x=-\cot x\csc x\)

Chain Rule of Derivatives:

• \(\rm \frac{d}{dx}f(g(x))=\frac{d}{d\ g(x)}f(g(x))× \frac{d}{dx}g(x)\).
• \(\rm \frac{dy}{dx}=\frac{dy}{du}× \frac{du}{dx}\).

Calculation:

We have y = cos (sin2 x).

Differentiating w.r.t. x, we get:

\(\rm\frac{dy}{dx}\) = - [sin (sin² x)] × (2 sin x) (cos x)

At x = \(\rm\frac{\pi}{2}\), we get:

\(\rm\frac{dy}{dx}\) = - [sin (sin2 \(\rm\frac{\pi}{2}\))] × (2 sin \(\rm\frac{\pi}{2}\)) (cos \(\rm\frac{\pi}{2}\)) = - [sin (1)] × (2 × 1) (0) = 0.

CONCEPT:

• \(\frac{{d\left( {\sin x} \right)}}{{dx}} = \cos x\)

CALCULATION:

Given: 2x + 3y = sin y

Here, we have to find \(\frac{dy}{dx}\)

So, let's differentiate the given equation with respect to x

⇒ \(\frac{d(2x + 3y)}{dx} = \frac{d(sin y)}{dx}\)

As we know that, \(\frac{{d\left( {\sin x} \right)}}{{dx}} = \cos x\)

⇒ \(2 + 3 \frac{dy}{dx} = cos y \frac{dy}{dx}\)

⇒ \(\frac{dy}{dx} = \frac{2}{cos \ y - 3}\)

Hence, correct option is 2.

CONCEPT:

• \(\frac{{d\left( {\sin x} \right)}}{{dx}} = \cos x\)
• \(\frac{{d\left( {\cos x} \right)}}{{dx}} = \; - \sin x\)
• \(\frac{{d\left( {\ln x} \right)}}{{dx}} = \frac{1}{x}\)

CALCULATION:

Let f(x) = sin(ln x) + cos(ln x)

As we know that, \(\frac{{d\left( {\sin x} \right)}}{{dx}} = \cos x, \frac{{d\left( {\cos x} \right)}}{{dx}} = \; - \sin x \ and \ \frac{{d\left( {\ln x} \right)}}{{dx}} = \frac{1}{x}\)

⇒ \(f'\left( x \right) = \frac{{\cos \left( {\ln x} \right)}}{x} - \frac{{\sin \left( {\ln x} \right)}}{x}\)

⇒ \(f'\left( e \right) = \frac{{\cos \left( {\ln e} \right)}}{e} - \frac{{\sin \left( {\ln e} \right)}}{e}\)

As we know that, ln(e) = 1

⇒ \(f'(e) = \frac{cos1 - sin1}{e}\)

Hence, option 1 is correct.

Concept:

Suppose that we have two functions f(x) and g(x) and they are both differentiable.

• Chain Rule: \(\frac{{\rm{d}}}{{{\rm{dx}}}}\left[ {{\rm{f}}\left( {{\rm{g}}\left( {\rm{x}} \right)} \right)} \right] = {\rm{\;f'}}\left( {{\rm{g}}\left( {\rm{x}} \right)} \right){\rm{g'}}\left( {\rm{x}} \right)\)
• Product Rule: \(\frac{{\rm{d}}}{{{\rm{dx}}}}\left[ {{\rm{f}}\left( {\rm{x}} \right){\rm{\;g}}\left( {\rm{x}} \right)} \right] = {\rm{\;f'}}\left( {\rm{x}} \right){\rm{\;g}}\left( {\rm{x}} \right) + {\rm{f}}\left( {\rm{x}} \right){\rm{\;g'}}\left( {\rm{x}} \right)\)

Formula:

\(\rm \frac{d\tan x}{dx} = \sec^2 x\)

Calculation:

Let y = \(\rm \sqrt{\tan x}\)

Differentiating with respect to x, we get

\(\rm \Rightarrow \frac{dy}{dx} = \frac{d\sqrt{\tan x}}{dx}\\=\frac{d\sqrt{\tan x}}{d \tan x} \times \frac{d\tan x}{dx}\\=\frac{1}{2\sqrt{\tan x}} \times \sec^2 x \\\therefore \frac{dy}{dx} =\frac{\sec^2 x}{2\sqrt{\tan x}}\)

Explanation -

\(y(\alpha)= \sqrt{2\left(\frac{\tan \alpha+\cot \alpha}{1+\tan ^2 \alpha}\right)+\frac{1}{\sin ^2 \alpha}}\)

\(y(\alpha)=\sqrt{2 \frac{1}{\sin \alpha \cos \alpha \times \frac{1}{\cos ^2 \alpha}}+\frac{1}{\sin ^2 \alpha}}\)

\(y(\alpha)=\sqrt{2 \cot \alpha+\operatorname{cosec}^2 \alpha}\)

\(y(\alpha)=\sqrt{ (1+\cot \alpha)^2}\)

y(α) = -1 – cotα

dy/dα = 0 + cosec2α

dy/dα = cosec2 5π/6

dy/dα = 4

Hence Option (3) is correct.