May 2026

Range of f(x)=cos²x+sin⁴x Find the Range of \( f(x)=\cos^2x+\sin^4x \) Question: The function \[ f(x)=\cos^2x+\sin^4x \] is defined from \(R\to R\). Then, \[ f(R)=? \] (a) \(\left[\frac34,1\right)\) (b) \(\left(\frac34,1\right]\) (c) \(\left[\frac34,1\right]\) (d) \(\left(\frac34,1\right)\) Solution: Using \[ \cos^2x=1-\sin^2x \] \[ f(x)=1-\sin^2x+\sin^4x \] Let \[ \sin^2x=t \] where \[ 0\le t\le1 \] Then, \[ f(x)=t^2-t+1 \] \[ […]

Find f(2002) Find \( f(2002) \) Question: If \[ f(x)=\frac{\sin^4x+\cos^2x}{\sin^2x+\cos^4x}, \qquad x\in R \] then \( f(2002) \) is equal to (a) \(1\) (b) \(2\) (c) \(3\) (d) \(4\) Solution: Let \[ \sin^2x=a,\qquad \cos^2x=b \] Then, \[ a+b=1 \] Now, \[ f(x)=\frac{a^2+b}{a+b^2} \] Since \[ b=1-a \] numerator: \[ a^2+b = a^2+1-a \] denominator: \[

Find h(x) Find \( h(x) \) Question: Let \[ f(x)=x,\qquad g(x)=\frac1x \] and \[ h(x)=f(x)g(x) \] Then, \[ h(x)=1 \] for (a) \(x\in R\) (b) \(x\in Q\) (c) \(x\in R-Q\) (d) \(x\in R,\; x\ne0\) Solution: \[ h(x)=f(x)g(x) \] \[ =x\cdot\frac1x \] \[ =1 \] This is defined only when \[ x\ne0 \] Therefore, \[ h(x)=1

Find the Value of the Given Expression Find the Value of the Given Expression Question: If \[ f(x)=\cos(\log_e x) \] then \[ f\left(\frac1x\right) f\left(\frac1y\right) – \frac12 \left\{ f(xy)+f\left(\frac{x}{y}\right) \right\} \] is equal to (a) \(\cos(x-y)\) (b) \(\log(\cos(x-y))\) (c) \(1\) (d) \(\cos(x+y)\) Solution: \[ f\left(\frac1x\right) = \cos\left(\log\frac1x\right) = \cos(-\log x) = \cos(\log x) \] Similarly, \[

Find f(f(f(2))) Find \( f(f(f(2))) \) Question: If \[ f(x)=\frac{x+1}{x-1}, \qquad x\ne1 \] then \[ f(f(f(2))) \] is equal to (a) \(1\) (b) \(2\) (c) \(3\) (d) \(4\) Solution: \[ f(2)=\frac{2+1}{2-1}=3 \] \[ f(f(2))=f(3)=\frac{3+1}{3-1}=2 \] \[ f(f(f(2)))=f(2)=3 \] \[ \boxed{\text{Correct Answer: (c)}} \] Next Question / Full Exercise

Range of Rational Function Find the Range of the Function Question: The range of the function \[ f(x)=\frac{x^2-x}{x^2+2x} \] is (a) \(R\) (b) \(R-\{1\}\) (c) \(R-\left\{-\frac12,1\right\}\) (d) none of these Solution: Let \[ y=\frac{x^2-x}{x^2+2x} \] Then, \[ yx^2+2yx=x^2-x \] \[ (y-1)x^2+(2y+1)x=0 \] \[ x\left[(y-1)x+(2y+1)\right]=0 \] Since \(x=0\) is not in domain, \[ (y-1)x+(2y+1)=0 \] \[

Find f(2x)+f(−x)−f(x) Find \( f(2x)+f(-x)-f(x) \) Question: Let \( f:R\to R \) be defined by \[ f(x)=2x+|x| \] Then \[ f(2x)+f(-x)-f(x) \] is equal to (a) \(2x\) (b) \(2|x|\) (c) \(-2x\) (d) \(-2|x|\) Solution: \[ f(2x)=4x+|2x| =4x+2|x| \] \[ f(-x)=-2x+|-x| =-2x+|x| \] \[ f(x)=2x+|x| \] Therefore, \[ f(2x)+f(-x)-f(x) \] \[ =(4x+2|x|)+(-2x+|x|)-(2x+|x|) \] \[ =2|x| \]

Find f(2) Find \( f(2) \) Question: If \[ 2f(x)-3f\left(\frac1x\right)=x^2 \qquad (x\ne0) \] then \( f(2) \) is equal to (a) \(-\frac74\) (b) \(\frac52\) (c) \(-1\) (d) none of these Solution: Replace \(x\) by \(\frac1x\), \[ 2f\left(\frac1x\right)-3f(x)=\frac1{x^2} \] Given, \[ 2f(x)-3f\left(\frac1x\right)=x^2 \] Solving, \[ -5f(x)=\frac3{x^2}+2x^2 \] \[ f(x)=-\frac15\left(\frac3{x^2}+2x^2\right) \] Put \(x=2\), \[ f(2) = -\frac15\left(\frac34+8\right)

Find f(x+y)f(x−y) Find \( f(x+y)f(x-y) \) Question: If \[ f(x)=\frac{2^x+2^{-x}}{2} \] then \[ f(x+y)f(x-y) \] is equal to (a) \(\frac12\{f(2x)+f(2y)\}\) (b) \(\frac12\{f(2x)-f(2y)\}\) (c) \(\frac14\{f(2x)+f(2y)\}\) (d) \(\frac14\{f(2x)-f(2y)\}\) Solution: \[ f(x+y)=\frac{2^{x+y}+2^{-(x+y)}}{2} \] \[ f(x-y)=\frac{2^{x-y}+2^{-(x-y)}}{2} \] Multiplying, \[ f(x+y)f(x-y) \] \[ = \frac14 \left(2^{2x}+2^{-2x}+2^{2y}+2^{-2y}\right) \] \[ = \frac12 \left[ \frac{2^{2x}+2^{-2x}}{2} + \frac{2^{2y}+2^{-2y}}{2} \right] \] \[ = \frac12\{f(2x)+f(2y)\} \]

Find the Value of the Given Expression Find the Value of the Given Expression Question: If \( f(x)=\cos(\log x) \), then find the value of \[ f(x)f(4)-\frac12\left\{ f\left(\frac{x}{4}\right)+f(4x) \right\} \] (a) \(1\) (b) \(-1\) (c) \(0\) (d) \(\pm1\) Solution: Using \[ 2\cos A\cos B=\cos(A-B)+\cos(A+B) \] \[ 2f(x)f(4) = 2\cos(\log x)\cos(\log 4) \] \[ = \cos(\log