Important topics for Maths has been designed in such a way that it offers very practical and application-based learning to further make it easier for students to understand every concept or topic by correlating it with the day-to-day experiences.
Q4. If sin^(-1)((x^2-y^2)/(x^2+y^2 ))=loga, then dy/dx is equal to
Solution
(d) We have sin^(-1)((x^2-y^2)/(x^2+y^2 ))=loga ⇒(x^2-y^2)/(x^2+y^2 )=sin(loga) ⇒(1-tan^2θ)/(1+tan^2θ )=sin(loga), on putting y=x tanθ ⇒cos2θ=sin(loga) ⇒2θ=cos^(-1)(sin(loga)) θ=1/2 cos^(-1)(sin(loga)) ⇒tan^(-1)(y/x)=1/2 cos^(-1)(sin(loga)) ⇒y/x=tan(1/2 cos^(-1)(sin(loga))) Differentiating w.r.t. x ⇒ (x dy/dx-y)/x^2 =0 ⇒x dy/dx-y=0 ⇒dy/dx=y/x
(d) We have sin^(-1)((x^2-y^2)/(x^2+y^2 ))=loga ⇒(x^2-y^2)/(x^2+y^2 )=sin(loga) ⇒(1-tan^2θ)/(1+tan^2θ )=sin(loga), on putting y=x tanθ ⇒cos2θ=sin(loga) ⇒2θ=cos^(-1)(sin(loga)) θ=1/2 cos^(-1)(sin(loga)) ⇒tan^(-1)(y/x)=1/2 cos^(-1)(sin(loga)) ⇒y/x=tan(1/2 cos^(-1)(sin(loga))) Differentiating w.r.t. x ⇒ (x dy/dx-y)/x^2 =0 ⇒x dy/dx-y=0 ⇒dy/dx=y/x
Q6. If f(x)=|sinx- |cosx | |, then the value f'(x) at x=7Ï€/6 is
Solution
(a) In the neighbourhood of x=7Ï€/6, we have f(x)=|sinx+cosx |=-sinx-cosx ⇒f^' (x)=-cosx+sinx⇒f'(7Ï€/6)= -cos(7Ï€/6) +sin(7Ï€/6)=(√3-1)/2
(a) In the neighbourhood of x=7Ï€/6, we have f(x)=|sinx+cosx |=-sinx-cosx ⇒f^' (x)=-cosx+sinx⇒f'(7Ï€/6)= -cos(7Ï€/6) +sin(7Ï€/6)=(√3-1)/2
Q7.If f(0)=0,f^' (0)=2, then the derivative of
y=f(f(f(f(x))) at x=0 is
Solution
(c) y^' (x)=f^' (f(f(f(x)))) f^' (f(f(x))) f^' (f(x))f'(x) ⇒y^' (0)=f'(f(f(f(0)))) f'(f(f(0)))f'(f(0))f'(0) =f^' (f(f(0))) f^' (f(0)) f^' (0)f'(0) =f^' (f(0)) f^' (0) f^' (0)f'(0) =f^' (0) f^' (0) f^' (0)f'(0) =(f^' (0))^4=2^4=16
(c) y^' (x)=f^' (f(f(f(x)))) f^' (f(f(x))) f^' (f(x))f'(x) ⇒y^' (0)=f'(f(f(f(0)))) f'(f(f(0)))f'(f(0))f'(0) =f^' (f(f(0))) f^' (f(0)) f^' (0)f'(0) =f^' (f(0)) f^' (0) f^' (0)f'(0) =f^' (0) f^' (0) f^' (0)f'(0) =(f^' (0))^4=2^4=16
Q8.If y=(sinx )^tanx , then dy/dx=
Solution
(a) y=(sinx )^tanx ⇒logy=tanx logsinx Differentiating w.r.t. x, we get 1/y dy/dx=sec^2x logsinx+tanx 1/sinx .cosx ⇒dy/dx=(sinx )^tanx [1+sec^2x logsinx ]
(a) y=(sinx )^tanx ⇒logy=tanx logsinx Differentiating w.r.t. x, we get 1/y dy/dx=sec^2x logsinx+tanx 1/sinx .cosx ⇒dy/dx=(sinx )^tanx [1+sec^2x logsinx ]
Q10. The nth derivative of xe^x vanishes when
Solution
(c) f(x)=xe^x f'(x)=e^x+xe^x f''(x)=e^x+e^x+xe^x f'''(x)=2e^x+e^x+xe^x=3e^x+xe^x … … f^n (x)=ne^x+xe^x Now, f^n (x)=0 ⇒ne^x+xe^x=0⇒x=-n
(c) f(x)=xe^x f'(x)=e^x+xe^x f''(x)=e^x+e^x+xe^x f'''(x)=2e^x+e^x+xe^x=3e^x+xe^x … … f^n (x)=ne^x+xe^x Now, f^n (x)=0 ⇒ne^x+xe^x=0⇒x=-n