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.

**Q1.**If f''(x)=-f(x),where f(x)is a continuous double differentiable function and g(x)=f'(x).If F(x)=(f(x/2))^2+(g(x/2))^2 and F(5)=5 then F(10) is

Solution

(b) Given, f'' (x)=-f(x) ⇒g' (x)=-f(x) and f' (x)=g(x) …(i) Now, F(x)=(f(x/2))^2+(g(x/2))^2 ∴F'(x)=2(f(x/2)).f'(x/2).1/2 +2(g(x/2)).g'(x/2).1/2=0 [using Eq.(i)] ∴ F(x) is a constant ⇒ F(10)=F(5)=5

(b) Given, f'' (x)=-f(x) ⇒g' (x)=-f(x) and f' (x)=g(x) …(i) Now, F(x)=(f(x/2))^2+(g(x/2))^2 ∴F'(x)=2(f(x/2)).f'(x/2).1/2 +2(g(x/2)).g'(x/2).1/2=0 [using Eq.(i)] ∴ F(x) is a constant ⇒ F(10)=F(5)=5

**Q2.**x=t cost,y=t+sint, then (d^2 x)/(dy^2) at t=Ï€/2 is

Solution

(b) dx/dy=(dx/dt)/(dy/dt)=cost-t sint/(1+cost) ∴ (d^2 x)/(dy^2)=(d/dt (dx/dy))/(dy/dt) =(((-2 sint-t cost)(1+cost)-(cost-t sint)(-sint))/(1+cost)^2)/(1+cost) Now, put t=Ï€/2

(b) dx/dy=(dx/dt)/(dy/dt)=cost-t sint/(1+cost) ∴ (d^2 x)/(dy^2)=(d/dt (dx/dy))/(dy/dt) =(((-2 sint-t cost)(1+cost)-(cost-t sint)(-sint))/(1+cost)^2)/(1+cost) Now, put t=Ï€/2

**Q3.**Suppose the function f(x)-f(2x) has the derivative 5 at x=1 and derivative7 at x=2. The derivative of the function f(x)-f(4x) at x=1 has the value equal to

Solution

(a) y=f(x)-f(2x)⇒y'=f'(x)-2f'(2x) ⇒y'(1)=f'(1)-2f'(2)=5, and (1) y'(2)=f'(2)-2f'(4)=7 (2) Now let y=f(x)-f(4x) ⇒y'=f'(x)-4f'(4x) ⇒y'(1)=f'(1)-4f'(4) (3) Substituting the value of f'(2)=7+2f'(4) in(1), we get f'(1)-2(7+2f'(4))=5 f'(1)-4f'(4)=19

(a) y=f(x)-f(2x)⇒y'=f'(x)-2f'(2x) ⇒y'(1)=f'(1)-2f'(2)=5, and (1) y'(2)=f'(2)-2f'(4)=7 (2) Now let y=f(x)-f(4x) ⇒y'=f'(x)-4f'(4x) ⇒y'(1)=f'(1)-4f'(4) (3) Substituting the value of f'(2)=7+2f'(4) in(1), we get f'(1)-2(7+2f'(4))=5 f'(1)-4f'(4)=19

**Q5.**(d^20 y)/(dx^20 )(2 cosx cos3x) is equal to

Solution

(b) y= 2cosxcos3x = cos4x + cos2x ⇒(d^20 y)/(dx^20)=4^20cos4x + 2^20cos2x

(b) y= 2cosxcos3x = cos4x + cos2x ⇒(d^20 y)/(dx^20)=4^20cos4x + 2^20cos2x

**Q6.**Let y=t^10+1 and x=t^8+1, then (d^2 y)/(dx^2 ) is

Solution

(c) Here, y=t^10+1 and x=t^8+1 ∴t^8=x-1⇒t^2=(x-1)^(1/4) So, y=(x-1)^(5/4)+1 Differentiate both sides w.r.t. x, we get dy/dx=5/4(x-1)^(1/4) Again, differentiate both sides w.r.t., we get (d^2y)/(dx^2)=5/16(x-1)^(-3/4) ⇒(d^2y)/(dx^2)=5/(16(x-1)^(3/4))=5/(16(t^2 )^3)=5/(16t^6)

(c) Here, y=t^10+1 and x=t^8+1 ∴t^8=x-1⇒t^2=(x-1)^(1/4) So, y=(x-1)^(5/4)+1 Differentiate both sides w.r.t. x, we get dy/dx=5/4(x-1)^(1/4) Again, differentiate both sides w.r.t., we get (d^2y)/(dx^2)=5/16(x-1)^(-3/4) ⇒(d^2y)/(dx^2)=5/(16(x-1)^(3/4))=5/(16(t^2 )^3)=5/(16t^6)

**Q9.**Let g(x) be the inverse of an invertible function f(x) which is differentiable at x=c, then g'(f(c)) equals

Solution

(b) Since g(x) is the inverse of function f(x), therefore gof(x)=I(x) for all x Now gof(x)=I(x),∀ x ⇒(gof)'(x)=1,∀ x ⇒g'(f(x))f'(x)=1,∀ x ⇒g'(f(x))=1/(f'(x)),∀ x ⇒g'(f(c))=1/(f'(c)) (putting x=c)

(b) Since g(x) is the inverse of function f(x), therefore gof(x)=I(x) for all x Now gof(x)=I(x),∀ x ⇒(gof)'(x)=1,∀ x ⇒g'(f(x))f'(x)=1,∀ x ⇒g'(f(x))=1/(f'(x)),∀ x ⇒g'(f(c))=1/(f'(c)) (putting x=c)

**Q10.**The derivative of y=(1-x)(2-x)…(n-x) at x=1 is

Solution

(b) dy/dx=-[(2-x)(3-x)…(n-x)+(1-x)(3-x)…(n-x)+⋯(1-x)(2-x)…(n-1-x)] At x=1 dy/dx=-[(n-1)!+0+⋯+0]=(-1)(n-1)!

(b) dy/dx=-[(2-x)(3-x)…(n-x)+(1-x)(3-x)…(n-x)+⋯(1-x)(2-x)…(n-1-x)] At x=1 dy/dx=-[(n-1)!+0+⋯+0]=(-1)(n-1)!